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Stanford FBC: Stories of the Season
All the news about Stanford Women's Basketball
How Stanford Can Play Better Against Oregon
Stanford suffered another lopsided loss to Oregon on Thursday, only slightly less humiliating than last year's loss at Maples. Give Oregon lots of credit. But Stanford can do better.
Oregon can be beaten. It's already happened twice this season, once to a then unranked ASU team. When Oregon is on its game, however, they are a formidable team. They have amazing chess pieces to work with. They bested team USA; and they played with the same relentless energy in dispatching Stanford on Thursday.
Sabrina Ionescu should win the national MVP award and will likely be the first player drafted by the WNBA. Ruthy Hebard is a world class post player. Hebard can be doubled, but only by leaving other players open. And Satau Sabally is power forward who, when her game is on, can play with the very best. She too will be a high draft choice.
Stanford has to do two things (at least) to improve the odds in the next matchup. First, Stanford has to cut way down on turnovers. Stanford had 18 turns, 10 of them during the second half massacre. A turnover is the absolute worst way to end a possession. I'd prefer a shot clock violation to a turnover.
To be sure, shot clock violations are frustrating and worthy of some head banging. But shot clock violations take a full 30 seconds off the game clock, time that the opponent cannot score. And when the shot buzzer sounds, play stops and the opponent must inbound the ball. This gives the defense time to organize and tends to negate any transition offense by the opponent.
Turnovers invite transition offense. There's no more fearsome sight than Ionescu heading down court in transition mode. Just for the record, Oregon scored 22 points off turnovers (Stanford had only 2). Of course, defensive rebounds can also lead to a transition offense, but turnovers are the absolute worst because they are unexpected and often occur in the back court.
When Oregon was in its half court offense, their productivity was impressive, but their offensive efficiency fell short of their transition offense. In the second half, for example, Ionescu, on her way to a career high 37 points, was merely 2-5 from the three point line, but 7-8 inside the arc. She was converting pull up jumpers and acrobatic layups, many of them in transition.
The other thing that Stanford must change is its own half court offense. Stanford in transition can be the equal of Oregon. Kiana Williams can be awesome, and other mobile players like Jones, the Hull sisters, and even the posts (Fingall and Prechtel) can run the court and score in transition. Stanford had no transition offense against Oregon, partly because Oregon was making so many shots. That left everything on the half court offense, and that was pretty miserable in the last 18 minutes of the game.
The last two times Stanford has bested Oregon, in the Pac 12 tournament last year and in Eugene two seasons back, Stanford had fewer turnovers and a more than respectable half court offense. In the first of those wins, Brittany McPhee went wild in the second half and humiliated the Ducks. Her offense was largely off the dribble, with the help of some well-placed screens. In last season's victory, Alanna Smith, DiJonai Carrington, and Kiana Williams took productive turns in the half court sets. Of course, when Stanford scored, the Ducks lost transition opportunities and had to generate their own half court game.
In this most recent loss, Stanford lost patience in the second half, attempting to force interior passes that simply weren't there. That led to turnovers and Oregon's formidable transition offense. Give the Oregon defense credit. They were ready for Stanford's interior passing and cuts to the basket. The Stanford team, however, has to be sufficiently flexible to adjust to that kind of defensive pressure. The team can rely more on screens, but it also needs a go to player like Brittany McPhee or Alanna Smith: a player who can consistently create in the paint. Stanford has players with these capabilities, including the absent Dijonai Carrington, but they will have to step up when Stanford next plays Oregon.
Actually, Stanford can't wait that long. The team will have to step up its game for all the remaining conference games, starting with Oregon State on Sunday. Learn from the loss, and move on!
Posted by Warren Grimes at 1/17/2020 12:05:00 PM 0 comments
Stanford WBB: Scoring by Committee?
Warren Grimes
Stanford finished the first days in January with an overall 13-1 record. With over one-third of the season already behind us, it's worth a look back and a peek forward.
Losing only one of the first 14 games is an achievement. This was not a soft preseason schedule. The wins were accomplished without any one player consistently leading the scoring. The wins came, in words of Tara VanDerveer, through "scoring by committee." In the fourteen games, seven different players were the leading scorers for the game. The team's top three scorers (Lexie Hull - 13.2 average; Kiana Williams - 12.4 average; and Haley Jones - 10.8 average) each had team highs in three games, but Nadia Fingall (in two games) and Lacie Hull, Ashten Prechtel, and Hannah Jump (in one game each) also achieved team highs. That's impressive. None of these players may receive an MVP award at the end of the season. But collectively, they are vital parts of a teamwork offense that creates nightmares for the defense.
I never have liked committees. Meetings are too long and too tedious, with some members talking just to hear themselves talk. But that's not what VanDerveer meant by committee scoring. She meant sharing the scoring burden, and doing it through teamwork. Teamwork can be measured in part through assists, and Stanford excelled in this category. In the fourteen games, Stanford had 220 assists to opponents' 119. Perhaps not surprisingly, the top three scoring players were also the top three in assists (Williams 53, Jones 32, and Lexie Hull 28). When your top scorers are also your top assist makers, that spells excellence in basketball at any level.
Stanford is team oriented not only on offense, but also on defense. Stanford's carefully scouted help defense has long been a trademark for VanDerveer. But this team, even with inexperienced freshmen, seems destined to play the help defense with aplomb. Low opponent scores, numerous steals, and a lot of blocks are all measures of an efficient help defense. On scoring, Stanford has an average 20.8 scoring margin (good but not great), has out-stolen oppenents 113 to 103 (better than Stanford teams of the past) and has out blocked opponents by a very impressive margin (81 to 22).
For the steals, Lexie Hull has been most impressive, with 23 steals, but five other players have double digit steals in the first 14 games. There's a lesson here. Hold onto your wallet when the Stanford players are around. And beware: this theft racket is definitely a team effort.
Its time to talk about the freshmen. When, if ever, has Stanford brought in four recruits, all of whom, in their first year, have either found their way into the starting lineup (Jones has started nine games and Belibi has started four) or are getting meaningful minutes (Jump has averaged 12.8 minutes and Prechtel 12.5 minutes)? I can't recall that ever happening. A "Fab 4" for the Stanford women! Both Jump and Prechtel have been team high scorers in one game. Who has the highest points per minute average for the team? None other than Hannah Jump with .563 per minute. I noticed that Jump is learning to fake a shot and then drive the basket. She's also learning to better utilize screens. Watch out!
On a blocks per minute tally, it is Ashten Prechtel that leads the team, slightly ahead of her freshman teammate Belibi. Prechtel is also the second most efficient rebounder (on a per minute basis). The wildcard is her three point shooting, which gives the defense fits. Prechtel has a lot to learn about playing the help defense and defending the post, but she, like the other freshmen, will improve with seasoning.
Haley Jones has been the high scorer in three separate games. Her performance is on the rise. If current trends continue, by the end of the season, Jones will be the team's top scorer in many, many games (and the second highest in assists). Jones has size, deceptive speed, basketball smarts and the competitive spirit to contribute inside and out, defensively and offensively. Jones is showing why she was the #1 ranked high school recruit in the country. No need to say more.
Then there's Belibi. She's had adjustments, including issues with turnovers, fouling, and learning the scout defense. But Belibi has started the last three games, for good reason. Belibi is the team's leading rebounder on a per minute basis. She's also the team leader in field goal percentage, primarily because of her under the basket performance. Belibi gets boards that other players on the team would not get (a healthy DiJonai Carrington might be an exception). Belibi is not yet a reliable and creative shot maker when the shot clock is running down, but the potential is there for her to be that and more.
Finally, not enough is written about the team's critical upperclassmen. Seniors Nadia Fingall and Anna Wilson are playing the best basketball of their careers. Fingall was known for her post skills on offense and defense, but surprised WSU with her three point shooting. When Fingall is in the game, the defense cannot ignore her outside shooting. All the more likely that a zone defense will be problematic for the opponent. Anna Wilson is a defensive stalwart with 11 steals. Her 24 assists, because of her more limited minutes, are on a par with Kiana Williams' assist per minute performance. She can spell Williams without any noticable offensive loss, or play with Williams to tighten the defense.
Kianna Williams is one of the conference's (and nation's) top point guards. She'd had some down games recently, but against Washington, Williams took over the game in the second half with her game high 20 points and multiple assists. Williams is a team leader who steps up when it matters most.
After Friday's home game against Cal, Stanford will be on the road for the next three games, including games in Oregon where the top five ranked home teams will be favored. Really tough assignments, but I look forward to watching this young and talented Stanford team take on the very best.
Away games update
For those already planning on traveling to away games, tickets are on sale NOW for all away games EXCEPT Utah (and their ticket office says single game tickets will be available shortly).
We will be using this space to provide information for fans wanting to connect with other Stanford fans at away sites.
CAL:
The Cal campus website indicates that on campus disabled parking is ONLY available for people with BOTH a disability placard AND a CAL permit -- which means probably not any of us. There are a number of paid city and university lots with parking at least theoretically available.
For public transportation to the CAL game, you can get to Haas Pavilion by Bus, BART or Train. These are the lines and routes that have stops nearby - Bus: 36, 51B, 52, 6 Train: CC BART: ORANGE, RED
At this time there is no planned pre or post game gathering location for Stanford fans and no plans for a bus from campus. Any suggestions?
Utah:
Fastbreak Club fan Stephen Perlman (sperlman.wbb@gmail.com) is planning on attending the Utah game and would like to coordinate with Stanford fans. He has provided his cell phone (# 310-600-0580). He recommends the Marriott across from the Utah campus as a place to stay.
Posted by Nancy Baker at 12/14/2019 10:50:00 AM 0 comments
What's Next for Stanford WBB: My Interview With Two Fantasy Fans
When I heard that Stanford had been given the #1 ranking in the AP poll, I was excited to tell Mort and Mary. They are Stanford fans, but a bit of an odd couple. Mort is, well – a cynic, and a glass-is-half-empty kind of guy. Mary, on the other hand, is upbeat – her glass is always half full.
When I spilled the news, Mort didn't hesitate.
"What's the big deal?" he said. "A month from now, no one will remember who had the #1 ranking in early December. What matters is who has it in April!"
"Well," I asked, "Isn't it still an achievement? Stanford has gone 7 years without a number 1 ranking."
Mary jumped in. "It is an accomplishment." Mary said. "Stanford won the Victoria tournament, defeating two top 20 teams on neutral turf."
Mort was unconvinced. "Stanford eeked out the narrowest of victories against unranked California Baptist, and gave up 78 points to that team. And," Mort continued, "Stanford barely beat Mississippi State."
Mary wouldn't be quieted, "I was impressed. Stanford won these games without three players who were pivotal to last year's team. Alana Smith graduated and neither Maya Dodson nor Dijonai Carrington, two key players from last year, played a single minute in the tournament."
I'll go with Mary on that last point. Stanford fielded a tournament team that lacked two of the three top scorers from last year. Only Kiana Williams played. This was a young team, with freshmen and sophomores playing absolutely critical roles. Senior Nadia Fingall (tournament MVP) and junior Kiana Williams played very well. But here are some of the young players' accomplishments in the Victoria tournament.
Sophomore Lexie Hull averaged over 10 points and 6 boards while converting 9 of 11 free throws. Her sister, Lacie Hull, played substantial minutes and contributed through defense. Lacie had six blocks and 3 steals during the tournament.
Freshman Haley Jones made her first two starts (against Syracuse and Mississippi State) and, in those two games, played more minutes (36 per game) than any other player. Jones averaged 13.3 points and 4.5 boards for the tournament. Against Mississippi State, Jones had a team leading 4 assists.
Freshman Hannah Jump had a break out game against Syracuse, playing 29 minutes and contributing a team leading 24 points, all of them on three point attempts (for the tournament, Jump converted 10 out of 17, or 59%, from three point range).
Freshman Fran Belibi, playing just over 12 minutes per game, averaged 8 points and 3.7 boards. That's actually quite impressive. For the season, Belibi has the highest field goal percentage (62.5%) and the highest rebounds per minute (.419) of any player. If Belibi can address turnovers and fouling, she is likely to see a lot more minutes.
Freshman Ashten Prechtel averaged only 4 minutes per game. Her potential, however, remains substantial. Prechtel is the team's second most efficient rebounder, trailing only Belibi on a per minute basis, and has averaged 6.5 points per game with a high field goal conversion percentage (52.3%).
Of course, the team's depth and experience will be greatly enhanced if Dijonai Carrington and Maya Dodson can return. In their absence, the Stanford team has demonstrated grit and determination. That's a critical intangible.
Stanford's next tree games, against Ohio State, Texas, and Tennessee, will be a challenge. Thinking about all of this, Mary said: "If we can win all three, Stanford will be well deserving of the number one rank."
Looking dour, Mort responded: "I'll believe it when I see it."
I say "Watch this team! They are great fun and can only get better."
Posted by Warren Grimes at 12/09/2019 02:30:00 PM 0 comments
Send your travel stories, tips, and updates
Hi Stanford Women's Basketball fans!
Feel free to send me your travel reports, tips, and news — I'll do a (probably) tiny bit of editing and post it here for others to read.
We have a fabulous team and following them will be fun!
Nancy Baker
Nlbaker1971@gmail.com
A Post Gonzaga Assessment
One shouldn't read too much into the first four games of the season – but commentators are for commenting. So I'll let fly.
Gonzaga is a very well coached team, with some talented, and hard working players. Their talent is not what Stanford brings, but it is still impressive. The Zags played really hard. They were physical and focused. I was impressed with their transition offense. I was impressed with their rebounding. I was impressed with their defense. When Stanford rebounded a ball, the Zags kept battling, and more than a couple of times knocked the ball loose from Belibi or Lexie Hull. Towards the end of the game, Lexie grabbed a rebound and cradled it hard – she had learned.
The Zags were playing to win, and had a two point lead with 16 seconds left in regulation. Stanford is a big game on their schedule. I haven't checked, but they don't often get to play the #3 team in the nation.
Meanwhile, Stanford
-got tested
-got molested
and, almost,
-got bested.
Stanford shot the ball miserably. Miserable from the three point line (22.7%); miserable from the foul line (58.6%). Stanford's two shooting stars – Kiana Williams and Lexie Hull, along with Anna Wilson – made some end of game free throws that brought the free throw shooting percentage above 50%. The free throw doldrums were sponsored by Belibi, Brown, Carrington, and Jones, who collectively shot 20% from the stripe. This team is capable of better. Had they shot a very reachable 70% of their free throws, or 30% of their three balls, the game would not have gone into overtime.
Stanford won ugly. But there were things to like, and motivation to be won from the experience gained by the four freshmen, all of whom saw minutes and scored points. In the first four games, the freshmen have made a real dent in team statistics. Right now, Prechtel is the third highest scorer on the team (after Lexie Hull and Williams) at 8.8 points per game. Jones is not far behind: she's the fifth highest scorer at 8 points per game. The two best rebounders on the team (so far) are Prechtel and Belibi. Based on rebounds per minute, Belibi is in first place at .33 boards/minute with Prechtel not far behind at .32 boards/minute. No one else is close. There's more: Belibi has the highest field goal percentage (.632) of any of the 12 players who have participated in all four games.
Three point shooting percentage? Of those who've launched 8 or more shots, the team leader is Prechtel at 44.4%, followed by Lexie Hull's 43.8% and Hannah Jump's 40%. Each of the four freshmen has made a mark in at least one statistical category.
There were other pluses from the Gonzaga game. I liked the way Williams' game rose to the occasion in the second half. I liked the way both Williams and Lexie Hull made critical shots when it mattered most. I liked the way Stanford defended – defense kept them in the game. And I liked the contributions of Prechtel (in the first half) and Jones (in the second half). Belibi was also a presence on the boards and defending with steals and blocks. All four of the freshman played.
Actually, 12 different players saw time in this game. In a close and hard fought game, a more typical TVD pattern would be to play a small rotation of maybe just 8 players. But this is early season, and this is a Stanford team with a very deep roster. Also, this was a game in which one of the big three scorers, Carrington, had a notably off night (Carrington is credited with three boards and not much else). She missed three shots from the floor and two from the free throw line. The reasons for Carrington's off night are unclear. Health and injuries permitting, she will be back and critical for the team's success.
Jones contributed substantially in the second half when it mattered most. Her 11 points made her the third highest scorer. Two plays stood out. In the first she was standing to the side of the key and prepared to hand the ball off as Williams scooted by. But the hand off was a fake: Jones spun and headed to the basket for an uncontested lay up. In the second play, in action under the hoop, she converted what appeared to be a two-handed reverse lay up without eyes on the basket. Jones showed she is capable of being a go to player, with nifty passes in transition and an ability to execute post moves under the basket.
Jones, and the other freshmen, will leave their mark on the team this season, and remain a great reason why we can expect exciting basketball from the 2019-20 edition of Stanford women's hoops. Meanwhile, it was two returning veterans who were pivotal in the win against the Zags. Kiana Williams is money. Lexie Hull is the team's leading scorer (16.5 points per game). That's a bit of a surprise. But when Lexie graduated from high school, I recall her high school coach saying that Lexie was more advanced at that point in her career than Brittany McPhee. I've seen nothing so far this year to question that assessment.
Posted by Warren Grimes at 11/19/2019 10:44:00 AM 2 comments
Posted by Warren Grimes at 8/21/2019 07:27:00 AM 0 comments
Good Vibrations in August
It's still a couple of months until women's college hoops takes center stage, but August is an appropriate time to think about what's coming. Stanford has an amazing group of returning veterans, eight of whom started three or more games last season. Oh, ----
and those four amazing incoming freshmen.
Let's start with one of those freshmen – Fran Belibi, who just returned from Bangkok, where she played on the FIBA U19 USA team that won the gold medal in the 16-team tournament. Belibi is not yet the complete player. She had no dunks, did not start, did not excel at the free throw line, did not garner lots of assists, did not take or make a single three point shot, and did have some turnovers. She played the post much of the time, but did not demonstrate a consistent ability to cash in on post moves. Belibi was not a go to player in crucial moments, and had her most impressive performances against lesser opponents. But just about everything else about Fran Belibi was impressive. No, actually, it was
Averaging just 16.1 minutes off the bench, Belibi was still the team's fifth highest scorer with 7.7 points per game. She was also, by a wide margin, the team's most efficient rebounder. FIBA keeps statistics on how many boards a player averages in a 40 minute span of time (the length of a game). Belibi averaged 18.2 boards in 40 minutes, well above team USA's second place Hillman Baker, who averaged 12 boards over a 40-minute span. There's more. Belibi's 18.2 rebound average was actually the best of any player in the 16-team tournament (there were well in excess of 170 players in the tournament). In addition to snatching more than her share of boards, Belibi leaps to keep the ball alive in other instances, allowing teammates an opportunity to capture the ball.
Steals are another indicator of Belibi's strengths – she averaged five steals during a 40 minute period, well above Hailey Van Lith's second place three steal average. This team leading stat is a good indicator of Belibi's impact. She has a long reach, leaping ability, and quickness that makes her a very effective defender. She is all over the court, she is disruptive, she blocks shots, she alters shots, she is a giant headache to the opposing team. Belibi's steals often led to fast break conversions. Although still lacking consistent post moves, Belibi is a very opportunistic scorer, with lots of putbacks from offensive boards.
FIBA also keeps statistics on a player's efficiency rating (a measure of how well the team performs when the player is on the court). Belibi's rating of 13.9 was second highest on the team, just below starting point guard Page Bueckers 14.3.
Belibi's shot blocking average (she was fourth on the team with an average of 2.1 blocks over a 40 minute span. But that that's not all. Belibi's quickness and long reach results in altered shots that are not included in the blocking total. On the subject of blocks, the team leader in blocks per minute was Cameron Brink (an average of 4.1 blocks in 40 minutes). Brink is a rising senior in high school who has committed to Stanford for the Fall of 2020. She averaged only 8 minutes per game in the FIBA tournament, but her performance was impressive.
Let's get back to Stanford's veterans. Even without the highly touted freshman class, Stanford would be an impressive team with the potential to win the conference and more. Nine players started three or more games for the team last year, and eight of those are returning. Also returning is Jenna Brown, a high school All American guard who averaged 8.1 minutes last year. That's nine players with substantial experience. The big losses for Stanford are in the post positions. Alanna Smith's leadership, three point shooting, post moves, boarding, shot blocking, and defending cannot be replaced by one player. Another substantial loss in the post is Shannon Coffee, who did not start a single game but was a key contributor in the rotation.
To replace these two post players, absent the freshmen class, Stanford would have to rely on three players who started in the post positions last year: senior Nadia Fingall, and juniors Maya Dodson and Alyssa Jerome. Last season, Fingall and Dodson both were hobbled by injuries. Neither played a complete season. Their healthy return will be crucial. Both Fingall and Jerome can convert from the three point line and help to offset the loss of Smith's outside shooting. Dodson is an athlete who could become a very effective offensive post player. Her defense is already impressive. I loved watching her defend the post in the conference championship game against Oregon.
The perimeter players that Stanford returns include already established all conference contributors Kiana Williams and DiJonai Carrington. They were the second and third highest scoring players last season. They know how to convert when it matters. Then there's Anna Wilson, who keeps battling injuries but whose game really stepped up at the end of last season. Last, but hardly least, three sophomores are likely to be in the mix. The Hull sisters started games and had an impact last season. They are focused players with talent and intensity. The Hulls are unlikely to take a back seat to anyone in terms of hard work and basketball smarts. They will contribute, as will their sophomore teammate Jenna Brown (who participated this summer in the trials for the U19 team).
Honestly, even without the freshman class, Stanford would have reason for optimism for the coming season. The biggest question mark would be post play. Health permitting, Stanford veterans could have a very productive post game.
Imagining Stanford without its highly touted freshmen is a way to focus on the talent of Stanford's returning veterans. They will show the way for the exciting new freshmen.
Stanford may be unique in its ability to recruit high school all Americans and keep almost all of them on its roster, even if some never achieve starting roles. That's a tribute to the academic reputation of the school and to Coach VanDerveer. This year, the challenge of giving everyone a role may be greater than ever. I expect VanDerveer will continue doing what she has done in the past: Give playing time to any player who plays hard and contributes in practice. And when a player enters a game and does contribute, give that player even more playing time.
Imagine the intensity of Stanford practices during the coming season.
The Jordan Brand Classic and the Future of Women's College Hoops
A week or so back, 24 of the best high school seniors played an all star game in Las Vegas: The Jordan Brand Classic. These high school all star games don't always show the players at their best. There's "hot-doggery" and "run-and-gunnery" – and usually not much focus on defense.
This all star game was well above average. There was some defense, lots of blocks and steals, and some notable intensity in the second half as the home and away teams competed to win a close game. Here are two examples, both involving the top-rated recruit in the country (and Stanford bound) Haley Jones. The first involved Jones defending a would be dunk by her future Stanford teammate, Fran Belibi. In that play, Belibi ran to the hoop to accept a high and very dunkable pass from a teammate. Jones kept pace with her on the cut, staying between Belibi and the hoop. Belibi lept above Jones to gain control of the pass, but could not get close enough to the hoop to convert.
In the second play, a loose ball was bounding in the direction of the basket. Jones dove to the floor to gain possession, scooting the ball ahead to a teammate heading for the hoop. Then Jones popped up in time and was there to tip in the miss.
These plays showed focus and intensity, and Jones was not the only player to demonstrate this. All three of the future Stanford players (Jones, Belibi, and Ashten Prechtel) acquitted themselves well, as did many others on the court.
On a broader level, the 24 players selected for the game suggest the continuing dominance of the big-name schools. South Carolina and Stanford had three players each while Notre Dame, Ohio State, and UCLA had two each. Other power teams had representatives, including Baylor, Louisville, Maryland, Mississippi State, Oregon State, Tennessee, and Texas.
Perhaps a bit of a surprise: two final four teams, Connecticut and Oregon, had no representatives in the Jordan Brand. But don't expect either of these teams to falter. U Conn has had numerous players in recent Jordan Brand games, and will surely have one or more in next year's game. Oregon, which lost only one starter, has brought in two outstanding players from Australia and Germany (neither could be considered for the Jordan Brand Classic).
Overall, the lineups for the Classic showed the preeminence of the Pac-12 conference. Seven of the 24 players (that's 29% of players) have chosen to play for Pac-12 schools (three for Stanford, two for UCLA, and one each for Oregon State and USC). This suggests that the conference's strong performance in recent NCAA tournaments is likely to continue.
During the run up to the game, Tennessee was busy finding a replacement coach. They chose Kellie Harper, a former star Tennessee point guard and experienced coach who had just taken Missouri State to the Sweet Sixteen, only to be bumped by Stanford. Harper's Missouri State team was well organized and played very disciplined defense against Stanford. I expect Harper to have Tennessee, with its roster of highly rated recruits, back in the top ten.
Harper will be Tennessee's second attempt to replace Pat Summit. The problem of replacing a legacy coach is one that will confront a number of other top programs in the next years. Among the top programs with relatively senior coaches are Baylor, U Conn., Notre Dame, and Stanford. When these replacements occur, there is a possibility of shaking up the established order.
But now let's return to the Stanford story. With three players among the 24 (12 on each team), Stanford, along with South Carolina, had preeminence in this game. South Carolina's recruits were rated the best in the country, above Stanford. One can argue about this. Much of the dispute centers on how highly Fran Belibi is rated. She was not in the top fifteen in recruiting rankings, but was chosen as a member of USA Today's first team high school All American list, suggesting a top five ranking.
In the game, Belibi played roughly half the minutes and still had a double-double: 16 points and 15 rebounds. The rebounding performance was, by my count, the best of any player. Belibi's strength and slashing style drew a number of fouls, and she converted 8 of 10 at the charity stripe. Belibi also had 2 assists, 2 blocks, and 2 steals. I don't care where the rating services place Belibi: I'm glad she'll be playing for Stanford next year. She'll get lots of minutes, much to the chagrin of opposing coaches.
As for Jones, she won the MVP for her team, with a game high 17 points, 5 boards, 4 assists, and 2 steals. Her leadership, smarts, and intensity were impressive. She's a player. And Prechtel was no slouch. In her time on the court, she had 5 points, 5 boards, and 3 blocks. Prechtel is a mobile post player who can run the court, shoot the three, block shots, board, and pass the ball. She's for real. Maybe a Shannon Coffee with a bit of Alanna Smith mixed in.
Overall, Stanford's three players put up 38 points. South Carolina's group of three contributed 34 points and had their own MVP (Zia Cooke). I'll let the experts debate the ranking of these two recruiting classes. As for me, I wouldn't do the trade. Stanford's class includes Hannah Jump, who, according to her high school coach, was likely the most reliable three point shooter in the high school game. If she can play defense, she'll get minutes too, maybe lots of them.
To quote the Stanford band, "Alright now!"
Prognostications Into the Future Can Provide Some Fun Fodder
Suspend your disbelief for a few minutes and pretend we're in this same moment, but in 2020 and then in 2021.
It turns out that Sabrina made the right decision and in April of 2020, Oregon beat Baylor and won the national championship. The first PAC12 national championship since Stanford won it in 1992. In the Final Four, Oregon beat UConn and Baylor beat Stanford in two classic battles. Of course, Stanford was disappointed to lose in the semis, but everyone knew how far they had come after incorporating their phenomenal freshman class.
In 2020, the lay of the land in the WBB world was massively changed from 2019. Baylor lost Kalani Brown and Chloe Jackson to graduation, but with Lauren Cox, Didi Richards, Juicy Landrum, Queen Egbo, etc. they still had enough firepower to get past Stanford and reach the final game of the tournament. Notre Dame had lost all five of their 2019 starters and were rebuilding. Louisville was without Asia Durr, Sam Fuehring, and Arica Carter and were also rebuilding. Mississippi State lost Teaira McCowan, and Jordan Danberry and most of their power. UCLA and Oregon State both fought their way deep into the tournament fray.
In 2020, Stanford says goodbye to Dijonai, Anna, and Nadia who were instrumental in reaching the Final Four in April. At the same time, Stanford was welcoming Cameron Brink (the number #2 recruit in the nation) and Jana Van Gytenbeek (also a top-50 recruit). Stanford moved Haley Jones to the point guard position at the start of the 2019/2020 season and Kiana to the two guard. Haley became the most successful freshman PG in school history and Kiana became the leading three point shooter in the nation. Stanford's women broke a dozen records in the 2019/2020 season - averaged ten three's per game and 40% as a team.
In the 2020/2021 season Stanford goes undefeated and wins the NCAA tournament championship. They also announce the number one recruiting class in the nation for 2021. The magnetic field of the WBB world has been reversed and the best players in the nation are now coming to the PAC12 and to Stanford, which will be where the action is in NCAA women's basketball for the next decade.
Well, what do you think? Too far out? Anyway, that's; my story and I'm sticking with it!!
Posted by craftsman at 4/21/2019 08:25:00 PM 0 comments
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Beenden wir ein Jahr mit einigen internationalen Grinds. Frohe Weihnachten und ein tolles 2013 von sk8boarding4life.
Let´s end the year with a few international grinds.
Merry Christmas and a Happy New Year 2013.
Danke, all denen die Photos geschickt haben.
Thanks to all those who sent photos.
A very good 2013 to you, your skating buddies and family !


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text:
\section{Introduction}

\medskip

The problem of approximating a smooth function and its derivatives by a
function of higher order smoothness on a Banach space $X$ has been
investigated by several authors, although the number of such results is
limited. When $X$ is finite dimensional excellent results are known, and in
fact Whitney in his classical paper \cite{W} provides essentially a complete
answer by showing: for every $C^{k}$ function $f:\mathbb{R}^{n}\rightarrow
\mathbb{R}^{m}$ and every continuous $\varepsilon:\mathbb{R}^{n
\rightarrow(0,+\infty)$ there exists a real analytic function $g$ such that
$\Vert D^{j}g(x)-D^{j}f(x)\Vert\leq\varepsilon(x)$ for all $x\in\mathbb{R
^{n}$ and $j=1,...,k$. This is the so-called $C^{k}$ fine approximation of $f$.

\medskip

The first results for $X$ infinite dimensional concern the smooth,
non-analytic case, and are due to Moulis \cite{M}. She proves, in particular,
a $C^{1}$ fine approximation theorem; namely, that for $X=c_{0}\ $or $l_{p}$
with $1<p<\infty,$ and $Y$ an arbitrary Banach space, given a $C^{1}$ map
$f:X\rightarrow Y,$ and a continuous function $\varepsilon:X\rightarrow\left(
0,\infty\right)  ,$ there exists a $C^{k}$ smooth map $g:X\rightarrow Y$
(where the optimal value of $k\geq1$ depends on the choice of $X$) such that
$\left\vert g\left(  x\right)  -f\left(  x\right)  \right\vert <\varepsilon
\left(  x\right)  $ and $\left\Vert g^{\prime}\left(  x\right)  -f^{\prime
}\left(  x\right)  \right\Vert <\varepsilon\left(  x\right)  .$ This result
was later extended in \cite{AFGJL} to the case where $X$ has an unconditional
basis and admits a Lipschitz, $C^{k}$ smooth bump function. Further work along
this line can be found in \cite{HJ}, where using ideas from \cite{F1} it is
shown that for certain range spaces $Y,$ one can relax the conditions on $X$
in \cite{AFGJL} and, for example, take $X$ to be merely separable.

\medskip

It is important to note that all the results mentioned above require, in a
very essential way, a theorem concerning the approximation of Lipschitz
functions $f$ by more regular, Lipschitz functions $g$, where the Lipschitz
constant of $g$ is fixedly proportional to the Lipschitz constant of $f,$
regardless of the precision in the approximation. In \cite{M} and \cite{AFGJL}
this is achieved by reducing the problem to the finite dimensional case using
the unconditional basis, but otherwise without this reduction traditional
methods of smooth approximation, such as smooth partitions of unity, do not
work in addressing this problem. A new approach was found in \cite{F1}, and
further developed in \cite{AFM}, \cite{F2}, \cite{AFK2}, and \cite{HJ}. This
technique has been called the method of \textit{sup-partitions of unity} in
\cite{HJ}. It seems that $C^{k}$ fine approximation must rely on such results.

\medskip

Concerning $C^{k}$ fine approximation by analytic functions for $X$ infinite
dimensional, nothing is known. In view of the remarks in the preceding
paragraph, it would appear that first one needs the ability to approximate
Lipschitz functions by Lipschitz, analytic functions with good control over
the Lipschitz constant. That is, one requires a kind of analytic sup-partition
of unity. Only very recently has this been possible with the work of
\cite{AFK1}, where it is proven that if $X$ is separable and admits a
separating polynomial, then for every Lipschitz function $f:X\rightarrow
\mathbb{R}$ and $\varepsilon>0$ there exists a Lipschitz, analytic function
$g:X\rightarrow\mathbb{R}$ with $\left\vert f-g\right\vert <\varepsilon$ and
Lip$\left(  g\right)  \leq C$Lip$\left(  f\right)  ,$ where the constant $C>1$
depends only on $X$ (for a precursor to this work see \cite{FK}). Using this,
we are able in this note to give the first results on the $C^{1}$ fine
analytic approximation problem in infinite dimensions. We remark that this
work is new even for $X$ a separable Hilbert space. We establish,

\begin{theorem}
\label{Main Theorem}Let $X$ be a separable Banach space which admits a
separating polynomial. Let $f:X\rightarrow\mathbb{R}$ be bounded and
Lipschitz, with uniformly continuous derivative, and $\varepsilon>0.$ Then
there exists an analytic function $g:X\rightarrow\mathbb{R}$ such that
$\left\vert f-g\right\vert <\varepsilon$ and $\left\Vert f^{\prime}-g^{\prime
}\right\Vert <\varepsilon.$
\end{theorem}

\medskip

\noindent Our notation is standard, with $X$ denoting a Banach space, and an
open ball with centre $x$ and radius $r$ denoted $B_{r}(x).$ If $\left\{
f_{j}\right\}  _{j}$ is a sequence of Lipschitz functions defined on $X,$ then
we will at times say this family is \textit{equi-Lipschitz} if there is a
common Lipschitz constant for all $j.$ A \textit{homogeneous polynomial of
degree }$n$ is a map,\textit{\ }$P:X\rightarrow\mathbb{R},$ of the form
$P\left(  x\right)  =A\left(  x,x,...,x\right)  ,$ where $A:X^{n
\rightarrow\mathbb{R}$ is $n-$multilinear and continuous. For $n=0$ we take
$P$ to be constant. A \textit{polynomial of degree }$n$ is a sum $\sum
_{i=0}^{n}P_{i}\left(  x\right)  ,$ where $i\geq1$ the $P_{i}$ are
$i$-homogeneous polynomials.

\medskip

\noindent Let $X$ be a Banach space, and $G\subset X$ an open subset. A
function $f:G\rightarrow\mathbb{R}$ is called \textit{analytic }if for every
$x\in G,$ there are a neighbourhood $N_{x},$ and homogeneous polynomials
$P_{n}^{x}:X\rightarrow\mathbb{R}$ of degree $n$, such that
\[
f\left(  x+h\right)  =\sum_{n\geq0}P_{n}^{x}\left(  h\right)
\;\text{provided\ }x+h\in N_{x}.
\]
\noindent Further information on polynomials may be found, for example, in
\cite{SS}$.$

For a Banach space $X,$ we define its (Taylor) complexification $\widetilde
{X}=X\bigoplus iX$ with norm
\[
\left\Vert x+iy\right\Vert _{\widetilde{X}}=\sup_{0\leq\theta\leq2\pi
}\left\Vert \cos\theta\ x-\sin\theta\ y\right\Vert _{X}=\sup_{T\in X^{\ast
},\Vert T\Vert\leq1}\sqrt{T(x)^{2}+T(y)^{2}}.
\]
If $L:E\rightarrow F$ is a continuous linear mapping between two real Banach
spaces then there is a unique continuous linear extension $\widetilde
{L}:\widetilde{E}\rightarrow\widetilde{F}$ of $L$ (defined by $\widetilde
{L}(x+iy)=L(x)+iL(y)$) such that $\Vert\widetilde{L}\Vert=\Vert L\Vert$. For a
continuous $k$-homogeneous polynomial $P:E\rightarrow\mathbb{R}$ there is also
a unique continuous $k$-homogeneous polynomial $\widetilde{P}:\widetilde
{E}\rightarrow\mathbb{C}$ such that $\widetilde{P}=P$ on $E\subset
\widetilde{E}$ (but the norm of $P$ is not generally preserved: one has that
$\Vert\widetilde{P}\Vert\leq2^{k-1}\Vert P\Vert$). It follows that if
$q\left(  x\right)  $ is a continuous polynomial on $X,$ there is a unique
continuous polynomial $\widetilde{q}\left(  z\right)  =\widetilde{q}\left(
x+iy\right)  $ on $\widetilde{X}$ where for $y=0$ we have $\widetilde{q}=q.$
For more information on complexifications (and polynomials) we recommend
\cite{MST}. In the sequel, all extensions of functions from $X$ to
$\widetilde{X},$ as well as subsets of $\widetilde{X},$ will be embellished
with a tilde.

\section{Main Results}

\subsection{The functions $\varphi_{n}$}

To prove Theorem 1, we start with a lemma which is a variation of \cite[Lemma
3]{AFK1}, where here we have made three changes: added part $\left(
4^{\prime}\right)  ;$ included constants $M_{n}$ for the estimate in $\left(
5\right)  ;$ and relaxed the condition that $r\geq1$ to $r>0.$ To obtain
$\left(  4^{\prime}\right)  ,$ we replace the function $b_{n}$ in the proof of
\cite[Lemma 3]{AFK1} with a $C^{1}$ version; the change in $\left(  5\right)
$ is easily handled; and requiring merely $r>0$ means that certain constants
will depend on $r,$ but as we shall apply the lemma with $r$ fixed throughout,
this causes no problem.

\medskip

\noindent First we need some definitions and notation. If $X$ posseses an
$n^{\text{th}}$ order separating polynomial, then it admits a $2n$-homogeneous
polynomial $q$ such that
\begin{equation}
\left\Vert x\right\Vert ^{2n}\leq q\left(  x\right)  \leq A\left\Vert
x\right\Vert ^{2n},
\end{equation}
for some $A>1$ (see e.g., \cite{AFK1}). In \cite[Lemma 2]{AFK1} it is proved
that the function $Q\left(  x\right)  =\left(  q\left(  x\right)  +1\right)
^{1/2n}-1$ satisfies:

\begin{enumerate}
\item $Q$ is (real) analytic on $X$,

\item $Q$ is Lipschitz on $X$, where we can take Lip$\left(  Q\right)  >1,$

\item $\inf\left\{  Q\left(  x\right)  :\left\Vert x\right\Vert \geq1\right\}
>0=Q\left(  0\right)  ,$

\item $Q\left(  x\right)  <4\rho\Rightarrow\left\Vert x\right\Vert <8\rho
\ $when $\rho\geq1$; otherwise $Q\left(  x\right)  <4\rho\Rightarrow\left\Vert
x\right\Vert <\delta\left(  \rho\right)  \equiv\left(  \left(  1+4\rho\right)
^{2n}-1\right)  ^{1/2n},$ this latter implication simply using $\left(
2.1\right)  $ and the definition of $Q$.

\item there exists $\delta>0$ such that $Q$ extends to a Lipschitz,
holomorphic map $\widetilde{Q}$ on the uniform strip $X\subset W_{\delta
}\subset\widetilde{X}$ given by

\[
W_{\delta}=\left\{  x+z:x\in X,\ z\in\widetilde{X},\ \left\Vert z\right\Vert
_{\widetilde{X}}<\delta\right\}  .
\]

\end{enumerate}

\medskip

\noindent We use the notion of a $Q$\textit{-body}, which for $\rho>0$ is
defined b

\[
D_{Q}\left(  x,\rho\right)  =\left\{  y\in X:Q\left(  y-x\right)
<\rho\right\}  .
\]

\noindent Let $\left\Vert \cdot\right\Vert _{c_{0}}$ be an equivalent analytic
norm on $c_{0},$ with $\left\Vert x\right\Vert _{\infty}\leq\left\Vert
x\right\Vert _{c_{0}}\leq A_{1}\left\Vert x\right\Vert _{\infty}$ for all
$x\in c_{0},$ and some $A_{1}>1$ (see e.g., \cite{FPWZ}, and also \cite{AFK1},
\cite{FK} where it is referred to as the Preiss norm). It may be worth
pointing out that the Preiss norm $\left\Vert \cdot\right\Vert _{c_{0}}$ is
obtained as the restriction of a holomorphic map $\widetilde{\lambda}$ defined
on a neighbourhood of $c_{0}\backslash\left\{  0\right\}  $ in $\widetilde
{c}_{0}.$

\medskip

\noindent For the remainder of the proof, we fix a dense sequence $\left\{
x_{n}\right\}  _{n=1}^{\infty}$ in $X.$

\medskip

\begin{lemma}
\label{lemma phi}Let $\widetilde{V}=W_{\delta}$ be an open strip around $X$ in
$\widetilde{X}$ in which the function $\widetilde{Q}$ given above is defined.
Given $\varepsilon\in(0,1),$ $r>0,$ and a covering $\left\{  D_{Q
(x_{n},r)\right\}  _{n=1}^{\infty}$ of $X$, there exists a sequence of
holomorphic functions $\widetilde{\varphi}_{n}=\widetilde{\varphi
}_{n,r,\varepsilon}:\widetilde{V}\rightarrow\mathbb{C}$, whose restrictions to
$X$ we denote by $\varphi_{n}=\varphi_{n,r,\varepsilon}$, with the following properties:

\begin{description}
\item[1] The collection $\{\varphi_{n,r,\varepsilon}:X\rightarrow
\lbrack0,2]\,|\,n\in\mathbb{N}\}$ is equi-Lipschitz on $X$, with Lipschitz
constant of the form $L_{\varphi}=L_{1}\text{Lip}(Q)/r>1$ (where $L_{1}\geq1$
is independent of $\varepsilon$ and $n$),

\item[2] $0\leq\varphi_{n,r,\varepsilon}(x)\leq1+\varepsilon$ for all $x\in X$.

\item[3] For each $x\in X$ there exists $m=m_{x,r}\in\mathbb{N}$ (independent
of $\varepsilon)$ with $\varphi_{m,r,\varepsilon}(x)>1/2$.

\item[4] $0\leq\varphi_{n,r,\varepsilon}(x)\leq\varepsilon$ for all $x\in
X\setminus D_{Q}(x_{n},5r)$.

\item[4$^{\prime}$] $\left\Vert \varphi_{n,r,\varepsilon}^{\prime
}(x)\right\Vert \leq\varepsilon$ for all $x\in X\setminus D_{Q}(x_{n},5r)$.

\item[5] For each $x\in X$ there exist $\delta_{x,r}>0$, $a_{x,r}>0,\ $and
$n_{x,r}\in\mathbb{N}$ (all independent of $\varepsilon$) such that
\[
|\widetilde{\varphi}_{n,r,\varepsilon}(x+z)|<\frac{1}{M_{n}n!a_{x,r}^{n
}\,\,\text{ for }\,n>n_{x,r},\ \,z\in\widetilde{X}\text{ with }\Vert
z\Vert_{\widetilde{X}}<\delta_{x,r},
\]

where $M_{n}=e^{2C^{2}\kappa}\left(  1+\left\Vert x_{n}\right\Vert \right)  ,$
and the $\kappa=\kappa\left(  r\right)  >1$ and $C>1$ are constants that will
be specified in the proof of Theorem 1.

\item[6] For each $x\in X,$ there exists $\delta_{x,r}>0$ (independent of
$\varepsilon$) and $n_{x,\varepsilon,r}\in\mathbb{N}$ such that for $\Vert
z\Vert_{\widetilde{X}}<\delta_{x,r}$ and $n>n_{x,\varepsilon,r}$ we have
$|\widetilde{\varphi}_{n,r,\varepsilon}(x+z)|<\varepsilon$.

\item[7] For each $x\in X,$ there exists $\delta_{x,\varepsilon,r}$ such that
\[
|\widetilde{\varphi}_{n,r,\varepsilon}(x+z)|\leq1+2\varepsilon\,\,\text{ for
}n\in\mathbb{N},\ \text{and }z\in\widetilde{X}\text{ with }\Vert
z\Vert_{\widetilde{X}}\leq\delta_{x,\varepsilon,r}.
\]

\end{description}
\end{lemma}

\medskip

\noindent\textbf{Proof.\ }We largely follow the proof of \cite[Lemma 3]{AFK1},
with the few noted changes. As the proof in \cite{AFK1} is rather long and
technical, we here indicate only the key constructions, referring the reader
to the above cited paper for full details. Note that because $r$ is fixed
throughout, for ease of notation, we shall often suppress dependences on $r.$
Define subsets $A_{1,r}=\{y_{1}\in\mathbb{R}:-1\leq y_{1}\leq4r\}$, and, for
$n\geq2$,
\begin{align*}
A_{n,r}  &  =\{y=\{y_{j}\}_{j=1}^{n}\in\ell_{\infty}^{n}:-1-r\leq y_{n
\leq4r,\,2r\leq y_{j}\\
&  \leq M_{n,r}+2r\text{ for }1\leq j\leq n-1\},
\end{align*
\begin{align*}
A_{n,r}^{\prime}  &  =\{y=\{y_{j}\}_{j=1}^{n}\in\ell_{\infty}^{n}:-1\leq
y_{n}\leq3r,\,3r\leq y_{j}\\
&  \leq M_{n,r}+r\text{ for }1\leq j\leq n-1\},
\end{align*
\[
\text{where }M_{n,r}=\sup\left\{  Q\left(  x-x_{j}\right)  :x\in D_{Q
(x_{n},4r),\ 1\leq j\leq n\right\}  .
\]

\medskip

\noindent Let $\mu\in C^{\infty}\left(  \mathbb{R},\left[  0,1+\varepsilon
\right]  \right)  $ be Lipschitz such that $\mu\left(  t\right)  =0$ iff
$t\geq1,$ and $\mu\left(  t\right)  =1+\varepsilon$ iff $t\leq1/2.$ Let
$b^{n}\in C^{\infty}\left(  \mathbb{R},\left[  0,1\right]  \right)  $ be
Lipschitz such that $b^{n}\left(  t\right)  =1$ iff $t\notin\left(
2r,M_{n,r}+2r\right)  ,$ and $b^{n}\left(  t\right)  =0$ iff $t\in\left[
3r,M_{n,r}+r\right]  .$ Let $\widehat{b}\in C^{\infty}\left(  \mathbb{R
,\left[  0,1\right]  \right)  $ be Lipschitz such that $\widehat{b}\left(
t\right)  =1$ iff $t\notin\left(  -1-r,4r\right)  ,$ and $\widehat{b}\left(
t\right)  =0$ iff $t\in\left[  -1,3r\right]  .$ Now define a Lipschitz,
$C^{\infty}\ $smooth map $b_{n}:c_{00}\subset c_{0}\rightarrow\left[
0,1\right]  $ by $b_{n}\left(  y_{1},...,y_{n}\right)  =\mu\left(  \left\Vert
\left(  b^{n}\left(  y_{1}\right)  ,...,b^{n}\left(  y_{n-1}\right)
,\widehat{b}\left(  y_{n}\right)  \right)  \right\Vert _{c_{0}}\right)  .$
Then support$\left(  b_{n}\right)  =\overline{A}_{n},$ and $b_{n
=1+\varepsilon$ on $A_{n}^{\prime}.$ Moreover, $b_{n}$ is Lipschitz with
constant of the form $L_{1}/r,$ where $L_{1}\geq1$ is independent of $n.$

\medskip

\noindent Now one defines, on $\mathbb{R}^{n},$ the ma
\begin{align*}
h_{n}\left(  x\right)   &  =\frac{1}{T_{n}}\int_{\mathbb{R}^{n}
b_{n}(y)e^{-k\sum_{j=1}^{n}2^{-j}(x_{j}-y_{j})^{2}}dy,\\
& \\
T_{n}  &  =\int_{\mathbb{R}^{n}}e^{-k\sum_{j=1}^{n}2^{-j}y_{j}{}^{2}}dy.
\end{align*}

\medskip

\noindent Because $b_{n}=b_{n,\varepsilon}$ has compact support, is bounded,
Lipschitz, and $C^{1},$ one can choose $k=k_{n,\varepsilon}>0$ sufficiently
large tha

\begin{equation}
|b_{n}(x)-h_{n}(x)|\leq\varepsilon/2\,\,\text{ for all }\,\,x\in\mathbb{R
^{n},
\end{equation}

\noindent an
\begin{equation}
|b_{n}^{\prime}(x)-h_{n}^{\prime}(x)|\leq\varepsilon/2\,\,\text{ for all
}\,\,x\in\mathbb{R}^{n}.
\end{equation}

\medskip

\noindent Next one defines (real) analytic maps $\varphi_{n}:X\rightarrow
\mathbb{R}$ by,
\[
\varphi_{n}\left(  x\right)  =h_{n}\left(  Q\left(  x-x_{1}\right)
,...,Q\left(  x-x_{n}\right)  \right)  =\frac{1}{T_{n}}\int_{\mathbb{R}^{n
}b_{n}\left(  y\right)  e^{-k_{n}\sum_{j=1}^{n}2^{-j}\left(  Q\left(
x-x_{j}\right)  -y_{j}\right)  ^{2}}.
\]

\noindent It is more or less standard to show that Lip$\left(  \varphi
_{n}\right)  \leq\frac{L_{1}}{r}$ Lip$\left(  Q\right)  .$ We can extend the
maps $\varphi_{n,r,\varepsilon}$ to complex valued maps defined on $W_{_{Q}}$
(see above) calling them $\widetilde{\varphi}_{n}$. Namely (where $x\in X$,
$z\in\widetilde{X}$),
\[
\widetilde{\varphi}_{n}\left(  x+z\right)  =\frac{1}{T_{n}}\int_{\mathbb{R
^{n}}b_{n}\left(  y\right)  e^{-k_{n}\sum_{j=1}^{n}2^{-j}\left(  \widetilde
{Q}\left(  x-x_{j}+z\right)  -y_{j}\right)  ^{2}}dy
\]
\noindent Note that the $\widetilde{\varphi}_{n}$ are well defined (as the
$b_{n}$ have compact supports) and are holomorphic where $\widetilde{Q}$ is
(namely on $\widetilde{W}_{\delta}$).

\medskip

\noindent To see $\left(  4\right)  $ and $\left(  4^{\prime}\right)  $, note
that if $Q\left(  x-x_{n}\right)  \geq5r,$ then there is a neighbourhood $N$
of $x$ for which $y\in N$ implies that the point,
\[
\widehat{x}=\left(  Q\left(  y-x_{1}\right)  ,...,Q\left(  y-x_{n}\right)
\right)  \in\mathbb{R}^{n}\backslash A_{n},
\]
from which we have $b_{n}\left(  \widehat{x}\right)  =0$ and $b_{n}^{\prime
}\left(  \widehat{x}\right)  =0.$ Hence, by $\left(  2.2\right)  $ and
$\left(  2.3\right)  ,$ we have, $\left\vert \varphi_{n}\left(  x\right)
\right\vert <\varepsilon/2$ and $\left\Vert \varphi_{n}^{\prime}\left(
x\right)  \right\Vert <\varepsilon/2.$

\medskip

\noindent The remaining parts are handled as in \cite{AFK1}, noting that for
$\left(  5\right)  $ we choose $\kappa_{n}$ larger if need be to ensure the
stated estimate involving the $M_{n}.$ $\square$

\bigskip

\noindent We return now to the proof of the theorem. Let $\varepsilon>0$ be
given and choose $\varepsilon^{\prime}$ satisfying
\[
0<\varepsilon^{\prime}<\min\{\frac{1}{8},1/(132C_{0}A_{1}^{2}L_{1
Lip(Q)),1/(10A_{1}r)\},
\]
where $L_{1}$ is as in part (1) of the preceding lemma, where \ is defined
immediately below, and where $C_{0}$ is a constant, only depending on $X$,
which will be fixed later on (see page 9 below). Because $f$ is bounded, we
may suppose that $1\leq f\leq2.$ As $f^{\prime}$ is uniformly continuous on
$X$, we can find a fixed $\rho>0$ so that for all $n,$ $x\in B_{\rho}\left(
x_{n}\right)  $ implies $\left\Vert f^{\prime}\left(  x_{n}\right)
-f^{\prime}\left(  x\right)  \right\Vert <\varepsilon^{\prime}.$ Now,
considering property $\left(  4\right)  $ of $Q,$ and noting that
$\delta\left(  r\right)  \rightarrow0^{+}$ as $r\rightarrow0^{+},$ we can
choose $r\in\left(  0,1\right)  $ (independent of $n$) so that $D_{Q}\left(
x_{n},5r\right)  \subset B_{\rho}\left(  x_{n}\right)  $ for all $n.$ It will
be convenient to write $D_{n}\equiv D_{Q}\left(  x_{n},5r\right)  .$ This $r$
shall be fixed for the remainder of the proof.

\medskip

\subsection{The functions $\nu_{n}$}

Next let $\overline{\nu}\in C^{\infty}\left(  \mathbb{R},\left[  0,1\right]
\right)  $ be Lipschitz such that $\overline{\nu}\left(  t\right)  =1$ iff
$\left\vert t\right\vert \leq5r,$ and $\overline{\nu}\left(  t\right)  =0$ iff
$\left\vert t\right\vert \geq\frac{11}{2}r.$ Put $L=$ Lip$\left(  f\right)  $.
Fix a sequence of functions $\left\{  \varphi_{n,r,\varepsilon_{1}}\right\}
_{n=1}^{\infty}$ with respect to the covering $\left\{  D_{Q}\left(
x_{n},r\right)  \right\}  _{n=1}^{\infty}$ of $X$ as given by Lemma
\ref{lemma phi}, where $r$ is fixed as above and the $\varepsilon$ of the
Lemma is chosen to be
\[
\varepsilon_{1}:=\min\left\{  \varepsilon^{\prime}r/3C_{0}L\text{Lip}\left(
\overline{\nu}\right)  ,\varepsilon^{\prime}r/25L\text{Lip}\left(
\overline{\nu}\right)  \right\}  ,
\]

\noindent We write $\varphi_{n,r,\varepsilon_{1}}$ as $\varphi_{n}$ for ease
of notation, and, as in Lemma \ref{lemma phi} $\left(  1\right)  ,$
$L_{\varphi}=$ Lip$\left(  \varphi_{n}\right)  =L_{1}$Lip$\left(  Q\right)
/r\geq1$, which we recall is independent of $n.$ Often we will subsume
dependence on $\varepsilon_{1}$ as dependence on $\varepsilon^{\prime}$ and
$L.$

\medskip

\noindent Put $\Delta\left(  t\right)  =\left(  \left(  \left\vert
t\right\vert +1\right)  ^{2n}-1\right)  ^{1/2n}\geq0.$ Now via convolution
integrals between $\overline{\nu}$ and Gaussian kernels, we can find
Lipschitz, analytic functions $\nu,$ with Lip$\left(  \nu\right)  =$
Lip$\left(  \overline{\nu}\right)  ,$ and which $C^{1}$-fine approximate
$\overline{\nu}$ in the following sense,

\medski

\begin{align}
\left\vert \nu\left(  t\right)  -\overline{\nu}\left(  t\right)  \right\vert
&  <\frac{\varepsilon^{\prime}r/2LL_{\varphi}}{1+\Delta\left(  t\right)
},\nonumber\\
& \\
\left\vert \nu^{\prime}\left(  t\right)  -\overline{\nu}^{\prime}\left(
t\right)  \right\vert  &  <\frac{\varepsilon^{\prime}r/2LL_{\varphi}
{1+\Delta\left(  t\right)  }.\nonumber
\end{align}

\medskip

\noindent Indeed, we can take $\nu$ to be of the form,

\medski

\begin{align*}
\nu\left(  t\right)   &  =\frac{1}{a}\int_{\mathbb{R}}\overline{\nu}\left(
s\right)  e^{-\kappa\left(  t-s\right)  ^{2}}ds,\\
& \\
a  &  =\int_{\mathbb{R}}e^{-\kappa s^{2}}ds,
\end{align*}

\noindent where $\kappa>1$ is chosen sufficiently large and is independent of
$t$ (although it does depend on $\max\left\{  \Delta\left(  t\right)
:t\in\ \text{supp}\left(  \overline{\nu}\right)  \right\}  <\infty$). This is
possible because $\overline{\nu}$ is $C^{\infty}$ with compact support, and
the function $t\rightarrow\frac{\varepsilon^{\prime}/2L}{1+\Delta\left(
t\right)  }$ is strictly positive, continuous and decreases slowly enough with
respect to $e^{-\kappa t^{2}}$ (namely, $\lim_{t\rightarrow\infty
\Delta(t)/e^{\kappa t^{2}}=0$). Moreover, since $\overline{\nu}$ has compact
support, $\nu$ has a holomorphic extension,
\[
\widetilde{\nu}\left(  z\right)  =\frac{1}{a}\int_{\mathbb{R}}\overline{\nu
}\left(  s\right)  e^{-\kappa\left(  z-s\right)  ^{2}}ds,
\]
to $\mathbb{C}.$ Next observe that for $t,s\in\mathbb{R}$ and $z\in\mathbb{C}$
with $\left\vert z\right\vert \leq\eta,$ we have

\begin{align*}
\operatorname{Re}\left(  t+z-s\right)  ^{2}  &  =\left(  t-s\right)
^{2}+2\left(  t-s\right)  \operatorname{Re}z+\operatorname{Re}\left(
z^{2}\right) \\
& \\
&  =\left(  t-s+\operatorname{Re}z\right)  ^{2}-\left(  \operatorname{Re
z\right)  ^{2}+\operatorname{Re}\left(  z^{2}\right) \\
& \\
&  \geq\left(  t-s+\operatorname{Re}z\right)  ^{2}-2\eta^{2}.
\end{align*}

\noindent Therefore when $\left\vert z\right\vert <\eta$ we get,
\begin{align}
\left\vert \widetilde{\nu}\left(  t+z\right)  \right\vert  &  =\frac{1
{a}\left\vert \int_{\mathbb{R}}\overline{\nu}\left(  s\right)  e^{-\kappa
\left(  t+z-s\right)  ^{2}}ds\right\vert \nonumber\\
& \nonumber\\
&  \leq\frac{1}{a}\int_{\mathbb{R}} e^{-\kappa\operatorname{Re}\left(
t+z-s\right)  ^{2}}ds\nonumber\\
& \nonumber\\
&  \leq\frac{1}{a}\int_{\mathbb{R}} e^{-\kappa\left(  t-s+\operatorname{Re
z\right)  ^{2}-2\eta^{2}}ds\\
& \nonumber\\
&  =\frac{e^{2\kappa\eta^{2}}}{a}\int_{\mathbb{R}} e^{-\kappa\left(
t+\operatorname{Re}z-s\right)  ^{2}}ds\nonumber\\
& \nonumber\\
&  =e^{2\kappa\eta^{2}},\nonumber
\end{align}

\medskip

\noindent where we have used a variable change to obtain the last line. Now
define Lipschitz, analytic functions $\nu_{n}:X\rightarrow\mathbb{R}$ by,
\[
\nu_{n}\left(  x\right)  =\nu\left(  Q\left(  x-x_{n}\right)  \right)  .
\]
Clearly $\nu_{n}$ has the holomorphic extension $\widetilde{\nu}_{n}\left(
z\right)  =\widetilde{\nu}\left(  \widetilde{Q}\left(  z-x_{n}\right)
\right)  . $ It will be convenient to put $\overline{\nu}_{n}\left(  x\right)
=\overline{\nu}\left(  Q\left(  x-x_{n}\right)  \right)  .$ Observe that,
writing $\widehat{D}_{n}=D_{Q}\left(  x_{n},6r\right)  ,

\begin{equation}
\left\vert \nu_{n}\left(  x\right)  \right\vert <\frac{\varepsilon^{\prime
}r/2LL_{\varphi}}{1+\Delta\left(  Q\left(  x-x_{n}\right)  \right)
},\ \ \text{for\ }x\notin\widehat{D}_{n},
\end{equation}

an

\begin{equation}
\left\vert \nu_{n}^{\prime}\left(  x\right)  \right\vert <\frac{\text{Lip
\left(  Q\right)  \varepsilon^{\prime}r/2LL_{\varphi}}{1+\Delta\left(
Q\left(  x-x_{n}\right)  \right)  },\ \ \text{for\ }x\notin\widehat{D}_{n}.
\end{equation}

\noindent Note that
\begin{align}
\frac{\varepsilon^{\prime}r/2LL_{\varphi}}{1+\Delta\left(  Q\left(
x-x_{n}\right)  \right)  }  &  =\frac{\varepsilon^{\prime}r/2LL_{\varphi
}{1+q\left(  x-x_{n}\right)  ^{1/2n}}\nonumber\\
& \\
&  \leq\frac{\varepsilon^{\prime}r/2LL_{\varphi}}{1+\left\Vert x-x_{n
\right\Vert }.\nonumber
\end{align}

\medskip

\noindent Now we estimate $\left\vert \widetilde{\nu}_{n}\left(  x+z\right)
\right\vert =\left\vert \widetilde{\nu}\left(  \widetilde{Q}\left(
x-x_{n}+z\right)  \right)  \right\vert ,$ for $\left\Vert z\right\Vert
_{\widetilde{X}}<\eta.$ From \cite[Lemma 2]{AFK1}, we can writ

\[
\widetilde{Q}\left(  x-x_{n}+z\right)  =Q\left(  x-x_{n}\right)  +Z_{n},
\]

\noindent where $Z_{n}\in\mathbb{C}$ with $\left\vert Z_{n}\right\vert \leq
C\left\Vert z\right\Vert _{\widetilde{X}},$ for some constant $C>1.$ Then from
the calculation $\left(  2.5\right)  $ we get, for $\left\Vert z\right\Vert
_{\widetilde{X}}<\eta,

\begin{align}
\left\vert \widetilde{\nu}_{n}\left(  x+z\right)  \right\vert  &  =\left\vert
\widetilde{\nu}\left(  \widetilde{Q}\left(  x-x_{n}+z\right)  \right)
\right\vert \nonumber\\
& \nonumber\\
&  =\left\vert \widetilde{\nu}\left(  Q\left(  x-x_{n}\right)  +Z_{n}\right)
\right\vert \\
& \nonumber\\
&  \leq e^{2C^{2}\kappa\eta^{2}}.\nonumber
\end{align}

\medskip

\noindent It is also worthwhile to note that $\nu\left(  t\right)
<1+\varepsilon^{\prime}$ for all $t.$

\medskip

\noindent Let $T_{n}\left(  x\right)  =f^{\prime}\left(  x_{n}\right)  \left(
x-x_{n}\right)  +f\left(  x_{n}\right)  $ be the first order Taylor polynomial
of $f$ at $x_{n}.$ Note that $\left\Vert T_{n}^{\prime}\left(  x\right)
\right\Vert =\left\Vert f^{\prime}\left(  x_{n}\right)  \right\Vert \leq L.$
Observe that $T_{n}-f$ is Lipschitz on $B_{\rho}\left(  x_{n}\right)  ,$ with
Lip$\left(  T_{n}-f\right)  \leq\left\Vert \left(  T_{n}-f\right)  ^{\prime
}\right\Vert =\left\Vert f^{\prime}\left(  x_{n}\right)  -f^{\prime}\left(
x\right)  \right\Vert \leq\varepsilon^{\prime}$ on $B_{\rho}\left(
x_{n}\right)  .$ It follows that $T_{n}-f$ is Lipschitz on $D_{n}\subset
B_{\rho}\left(  x_{n}\right)  $ with constant not greater than $\varepsilon
^{\prime}.$ Denote by $\overline{T_{n}-f}$ a bounded and Lipschitz extension
of $\left(  T_{n}-f\right)  \mid_{D_{n}}$ to all of $X,$ having the same bound
and Lipschitz constant. For example, one can take, temporarily writing
$h=\left(  T_{n}-f\right)  \mid_{D_{n}},
\[
\left(  \overline{T_{n}-f}\right)  \left(  x\right)  =\max\{-\Vert
h\Vert_{\infty},\min\{\Vert h\Vert_{\infty},\,\inf_{y\in D_{n}
\{h(y)+\text{Lip}(h)\Vert x-y\Vert\}\,\}\,\}.
\]

\medskip

\noindent Write $\epsilon_{n}\left(  x\right)  =\left(  \overline{T_{n
-f}\right)  \left(  x\right)  .$ We now apply \cite[Proposition 3]{AFK1} to
$\epsilon_{n}\left(  x\right)  ,$ along with the standard `scaling argument'
that appears at the very end of the proof of \cite[Theorem 1]{AFK1}, to obtain
the following: there exists a constant $C_{0}>1,$ depending only on $X,$ a
neighbourhood $X\subset\widetilde{W}\subset\widetilde{X},$ where
$\widetilde{W}=\widetilde{W}_{\varepsilon^{\prime},r}$ depends only on
$\varepsilon^{\prime}$ and $r$ (the dependence on $L_{\varphi}$ written as a
dependence on $r$), and an analytic map $\delta_{n}:X\rightarrow\mathbb{R}$
such that

\begin{enumerate}
\item $\left\vert \epsilon_{n}\left(  x\right)  -\delta_{n}\left(  x\right)
\right\vert <\varepsilon^{\prime}r/L_{\varphi}$ for all $x\in X,$

\item Lip$\left(  \delta_{n}\right)  \leq C_{0}$ Lip$\left(  \epsilon
_{n}\right)  \leq C_{0}\varepsilon^{\prime},$

\item the map $\delta_{n}$ extends to a holomorphic map $\widetilde{\delta
}_{n}$ on $\widetilde{W}$ (where in particular, $\widetilde{W}$ is independent
of $n$)$,$

\item $|\widetilde{\delta}_{n}(x+iy)-\delta_{n}(x)|\leq M_{\Delta}$ for all
$x+iy\in\widetilde{W},$ where $M_{\Delta}$ depends on $\varepsilon^{\prime}$
and is independent of $n.$
\end{enumerate}

\medskip

\noindent Now we define analytic functions on $X$ by

\[
\psi_{n}\left(  x\right)  =\left(  T_{n}\left(  x\right)  \nu_{n}\left(
x\right)  -\delta_{n}\left(  x\right)  \right)  \varphi_{n}\left(  x\right)
.
\]

\medskip

\noindent Observe that from property $\left(  3\right)  $ of $\delta_{n}$ and
Lemma \ref{lemma phi}, $\psi_{n}$ extends to a holomorphic map $\widetilde
{\psi}_{n}\left(  z\right)  =\left(  \widetilde{T}_{n}\left(  z\right)
\widetilde{\nu}_{n}\left(  z\right)  -\widetilde{\delta}_{n}\left(  z\right)
\right)  \widetilde{\varphi}_{n}\left(  z\right)  ,$ where
\[
\widetilde{T}_{n}\left(  z\right)  =\widetilde{T}_{n}\left(  x+iy\right)
=\widetilde{f^{\prime}\left(  x_{n}\right)  }\left(  x+iy-x_{n}\right)
+f\left(  x_{n}\right)
\]
($\widetilde{f^{\prime}\left(  x_{n}\right)  }$ being the canonical extension
of $f^{\prime}\left(  x_{n}\right)  $ to all of $\widetilde{X}$), on a
neighbourhood $X\subset\widetilde{W}\subset\widetilde{X},$ where
$\widetilde{W}$ is independent of $n.$

\medskip

\subsection{The approximating function $g$}

Let us define the function $g:X\rightarrow\mathbb{R}$ by
\[
g\left(  x\right)  =\frac{\left\Vert \left\{  \psi_{n}\left(  x\right)
\right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}}{\left\Vert \left\{  \varphi
_{n}\left(  x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}
\]

\medskip

\noindent We next show that $g$ is analytic. Since the norm $\|\cdot\|_{c_{0
}$ is real analytic on $c_{0}\setminus\{0\}$, it is sufficient to check that
the mappings $x\mapsto\{\varphi_{n}(x)\}_{n=1}^{\infty}$ and $x\mapsto
\{\psi_{n}(x)\}_{n=1}^{\infty}$ are real analytic from $X$ into $c_{0}$ and do
not take the value $0\in c_{0}$. Using Lemma \ref{lemma phi} it is easy to
show that the function $x\mapsto\{\varphi_{n}(x)\}_{n=1}^{\infty}$ has such
properties (see \cite[Lemma 4]{AFK1}).

As for the function $x\mapsto\{\psi_{n}(x)\}_{n=1}^{\infty}$, let us first
show that it does not take the value $0$. In fact we show that for each $x\in
X$ there exists an $n$ so that the number $\left(  T_{n}\left(  x\right)
\nu_{n}\left(  x\right)  -\delta_{n}\left(  x\right)  \right)  \varphi
_{n}\left(  x\right)  $ is bounded above $1/4$. Indeed, for each $x\in X,$
there is a minimal $n=n_{x}$ with $x\in D_{Q}\left(  x_{n_{x}},3r\right)  ,$
and via the proof of \cite[Lemma 3 $\left(  3\right)  $]{AFK1}, for such
$n_{x}$ we have $\varphi_{n_{x}}\left(  x\right)  \geq1/2.$ Note also that
$D_{Q}\left(  x_{n_{x}},3r\right)  \subset D_{Q}\left(  x_{n_{x}},5r\right)
=D_{n_{x}},$ and $\epsilon_{n_{x}}\left(  x\right)  =T_{n_{x}}\left(
x\right)  -f\left(  x\right)  $ on $D_{n_{x}}.$ So, from this and property
$\left(  1\right)  $ of $\delta_{n},$ we have $\left\vert T_{n_{x}}\left(
x\right)  \overline{\nu}_{n_{x}}\left(  x\right)  -f\left(  x\right)
-\delta_{n_{x}}\left(  x\right)  \right\vert =\left\vert \epsilon_{n_{x
}\left(  x\right)  -\delta_{n_{x}}\left(  x\right)  \right\vert \leq
\varepsilon^{\prime}.$ Now to replace $\overline{\nu}_{n_{x}}$ with
$\nu_{n_{x}},$ we observe by $\left(  2.4\right)  \ $and $\left(  2.8\right)
,

\begin{align}
\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -T_{n}\left(
x\right)  \overline{\nu}_{n}\left(  x\right)  \right\vert  &  =\left\vert
T_{n}\left(  x\right)  \right\vert \left\vert \nu_{n}\left(  x\right)
-\overline{\nu}_{n}\left(  x\right)  \right\vert \nonumber\\
& \nonumber\\
&  \leq\left(  L\left\Vert x-x_{n}\right\Vert +\left\vert f\left(
x_{n}\right)  \right\vert \right)  \ \frac{\varepsilon^{\prime}r/2LL_{\varphi
}}{1+\left\Vert x-x_{n}\right\Vert }\nonumber\\
& \\
&  \leq\left(  L\left\Vert x-x_{n}\right\Vert +2\right)  \ \frac
{\varepsilon^{\prime}r/2LL_{\varphi}}{1+\left\Vert x-x_{n}\right\Vert
}\nonumber\\
& \nonumber\\
&  \leq\varepsilon^{\prime}r/2L_{\varphi}+\varepsilon^{\prime}r/LL_{\varphi
}\nonumber\\
& \nonumber\\
&  \leq3\varepsilon^{\prime}r/L_{\varphi}\leq3\varepsilon^{\prime}.\nonumber
\end{align}

\noindent Therefore, these estimates give, $\left\vert T_{n_{x}}\left(
x\right)  \nu_{n_{x}}\left(  x\right)  -f\left(  x\right)  -\delta_{n_{x
}\left(  x\right)  \right\vert \leq4\varepsilon^{\prime},$ and because
$f\geq1,$ we have our desired bound
\begin{align*}
\left\vert T_{n_{x}}\left(  x\right)  \nu_{n_{x}}\left(  x\right)
-\delta_{n_{x}}\left(  x\right)  \right\vert \varphi_{n_{x}}\left(  x\right)
&  \geq\left\vert T_{n_{x}}\left(  x\right)  \nu_{n_{x}}\left(  x\right)
-\delta_{n_{x}}\left(  x\right)  \right\vert \left(  1/2\right) \\
& \\
&  \geq\left(  f\left(  x\right)  -4\varepsilon^{\prime}\right)  \left(
1/2\right)  >1/4.
\end{align*}
We remark that it follows from this that for any $x,$
\begin{equation}
\left\Vert \left\{  \psi_{n}\left(  x\right)  \right\}  _{n}\right\Vert
_{c_{0}}\geq\left\Vert \left\{  \psi_{n}\left(  x\right)  \right\}
_{n}\right\Vert _{\infty}=\left\Vert \left\{  \left(  T_{n}\left(  x\right)
-\delta_{n}\left(  x\right)  \right)  \varphi_{n}\left(  x\right)  \right\}
_{n}\right\Vert _{\infty}\geq1/4.
\end{equation}

\medskip

\noindent Next, to show that the function $x\mapsto\{\psi_{n}(x)\}_{n=1
^{\infty}$ is real analytic from $X$ into $c_{0}$, we shall require that for
each $x,$ there exists $n_{x}$ and $\delta_{x}\in\left(  0,1\right)  $ so that
for $n\geq n_{x}$ and $\left\Vert z\right\Vert _{\widetilde{X}}<\delta_{x},$
we have
\begin{equation}
\frac{\left\vert \widetilde{T}_{n}\left(  x+z\right)  \widetilde{\nu
_{n}\left(  x+z\right)  -\widetilde{\delta}_{n}\left(  x+z\right)  \right\vert
}{M_{n}}\leq M_{x},
\end{equation}
where $M_{x}$ depends on $x,$ but is independent of $n.$

\medskip

\noindent Recalling that $\widetilde{T}_{n}\left(  w\right)  =$ $\widetilde
{f^{\prime}\left(  x_{n}\right)  }\left(  x-x_{n}+w\right)  +f\left(
x_{n}\right)  ,$ and using $\left(  2.9\right)  $ and property $\left(
1\right)  \ $and $\left(  4\right)  $ of $\delta_{n},$ where we may suppose
that $x+z\in\widetilde{W}$ when $\left\Vert z\right\Vert _{\widetilde{X
}<\delta_{x}<1,$ we obtain,
\begin{align*}
&  \left\vert \widetilde{T}_{n}\left(  x+z\right)  \widetilde{\nu}_{n}\left(
x+z\right)  -\widetilde{\delta}_{n}\left(  x+z\right)  \right\vert \\
& \\
&  \leq\left\vert \widetilde{T}_{n}\left(  x+z\right)  \right\vert \left\vert
\widetilde{\nu}_{n}\left(  x+z\right)  \right\vert +\left\vert \widetilde
{\delta}_{n}\left(  x+z\right)  -\delta_{n}\left(  x\right)  \right\vert
+\delta_{n}\left(  x\right) \\
& \\
&  \leq\left\vert \widetilde{f^{\prime}\left(  x_{n}\right)  }\left(
x-x_{n}+z\right)  +f\left(  x_{n}\right)  \right\vert e^{2C^{2}\kappa
\delta_{x}^{2}}+M_{\Delta}+\varepsilon^{\prime}\\
& \\
&  <\left(  L\left(  \left\Vert x-x_{n}\right\Vert +\left\Vert z\right\Vert
_{\widetilde{X}}\right)  +\left\vert f\left(  x_{n}\right)  \right\vert
\right)  e^{2C^{2}\kappa\delta_{x}^{2}}+2M_{\Delta}\\
& \\
&  <\left(  L\left(  \left\Vert x-x_{n}\right\Vert +1\right)  +2\right)
e^{2C^{2}\kappa}+2M_{\Delta}\\
& \\
&  \leq\left(  3L\left(  \left\Vert x-x_{n}\right\Vert +1\right)  \right)
e^{2C^{2}\kappa}+2M_{\Delta
\end{align*}

\medskip

\noindent Now recalling that $M_{n}=e^{2C^{2}\kappa}\left(  1+\left\Vert
x_{n}\right\Vert \right)  ,$ we see that
\begin{align*}
&  \frac{3L\left(  \left\Vert x-x_{n}\right\Vert +1\right)  e^{2C^{2}\kappa
}{M_{n}}\\
& \\
&  \leq\frac{3L\left(  \left\Vert x\right\Vert +\left\Vert x_{n}\right\Vert
+1\right)  }{1+\left\Vert x_{n}\right\Vert }\\
& \\
&  \leq3L\left(  1+\left\Vert x\right\Vert \right)  .
\end{align*}
Putting $M_{x}=2M_{\Delta}+3L\left(  1+\left\Vert x\right\Vert \right)  ,$ we
have established $\left(  2.12\right)  .$

\medskip

\noindent Now, to show the analyticity of $\left\{  \psi_{n}\left(  x\right)
\right\}  _{n=1}^{\infty}$, we first note that property $(5)$ of Lemma
\ref{lemma phi} together with $(2.12)$ yield
\[
|\widetilde{\psi}_{n}(x+z)|=|\widetilde{T}_{n}(x+z)\widetilde{\nu
_{n}(x+z)-\widetilde{\delta}_{n}(x+z)|\, |\widetilde{\varphi}_{n
(x+z)|\leq\frac{M_{x}}{n! a_{x,r}^{n}}
\]
whenever $n\geq n_{x}$ and $\|z\|_{\widetilde{X}}<\delta_{x}$.

Because the numerical series $\sum_{n=1}^{\infty} M_{x}/n! a_{x,r}^{n}$ is
convergent, we then have that the series of functions $\sum_{n=1}^{\infty}|
\widetilde{\psi}_{n}\left(  x+z\right)  | $ is uniformly convergent on the
ball $B_{\widetilde{X}}(0, \delta_{x})$, which clearly implies that the
series
\[
\sum_{n=1}^{\infty}\widetilde{\psi}_{n}(z)e_{n} =\{\widetilde{\psi
_{n}(z)\}_{n=1}^{\infty}
\]
is uniformly convergent for $z\in B_{\widetilde{X}}(x, \delta_{x})$. Then it
is clear that $\{\widetilde{\psi}_{n}(z)\}_{n=1}^{\infty}$, being a series of
holomorphic mappings which converges uniformly on the ball $B_{\widetilde{X
}(x, \delta_{x})$, is a holomorphic mapping on this ball. Since $x\in X$ is
arbitrary, this shows that $x\mapsto\{\psi_{n}(x)\}_{n=1}^{\infty}$ is real analytic.

\medskip

\subsection{Showing that $g$ does the job}

Now we move on to our final estimates; $\left\vert g-f\right\vert $ and
$\left\Vert g^{\prime}-f^{\prime}\right\Vert .$ Fix $x\in X,$ and put
$\mathcal{N}=\mathcal{N}_{x}=\left\{  n:x\in D_{n}\right\}  .$ Now we have
(using $f\geq1>0),$ that
\begin{align*}
\left\vert g\left(  x\right)  -f\left(  x\right)  \right\vert  &  =\left\vert
\frac{\left\Vert \left\{  \psi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty
}\right\Vert _{c_{0}}}{\left\Vert \left\{  \varphi_{n}\left(  x\right)
\right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}}-f\left(  x\right)  \right\vert
\\
& \\
&  =\left\vert \frac{\left\Vert \left\{  \psi_{n}\left(  x\right)  \right\}
_{n=1}^{\infty}\right\Vert _{c_{0}}}{\left\Vert \left\{  \varphi_{n}\left(
x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}}-\frac{\left\Vert
\left\{  f\left(  x\right)  \varphi_{n}\left(  x\right)  \right\}
_{n=1}^{\infty}\right\Vert _{c_{0}}}{\left\Vert \left\{  \varphi_{n}\left(
x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}}\right\vert \\
& \\
&  =\frac{1}{\left\Vert \left\{  \varphi_{n}\left(  x\right)  \right\}
_{n=1}^{\infty}\right\Vert _{c_{0}}}\left\Vert \left\{  \left(  T_{n}\left(
x\right)  \nu_{n}\left(  x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)
\varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}\\
& \\
&  \leq2\left\Vert \left\{  \left(  T_{n}\left(  x\right)  \nu_{n}\left(
x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)  \varphi_{n}\left(
x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0}},
\end{align*}

\medskip

\noindent the last line by Lemma \ref{lemma phi}\ $\left(  3\right)  .$ We
proceed in cases.

\medskip

\textbf{Case 1:\ \ }For $n\in\mathcal{N},$ we have $\epsilon_{n}\left(
x\right)  =T_{n}\left(  x\right)  -f\left(  x\right)  =T_{n}\left(  x\right)
\overline{\nu}_{n}\left(  x\right)  -f\left(  x\right)  ,$ and so by property
$\left(  1\right)  $ of $\delta_{n}$ and Lemma \ref{lemma phi} $\left(
7\right)  ,$ we obtain the estimate, $\left\vert T_{n}\left(  x\right)
\overline{\nu}_{n}\left(  x\right)  -f\left(  x\right)  -\delta_{n
(x)\right\vert \varphi_{n}\left(  x\right)  \leq\left(  \varepsilon^{\prime
}r/L_{\varphi}\right)  3.$ Then using $\left(  2.10\right)  ,$ we have
\begin{equation}
\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(  x\right)
-\delta_{n}(x)\right\vert \varphi_{n}\left(  x\right)  \leq6r\varepsilon
^{\prime}/L_{\varphi}.
\end{equation}

\medskip

\textbf{Case 2:\ \ }For $n\notin\mathcal{N},$ recall $\varphi_{n}\left(
x\right)  \leq\varepsilon_{1}\leq\varepsilon^{\prime}r/25L.$

\medskip

\noindent Now, for $n$ such that $x\in\widehat{D}_{n},$ we have $Q\left(
x-x_{n}\right)  <6r,$ and so $\left\Vert x-x_{n}\right\Vert \leq\left(
q\left(  x-x_{n}\right)  \right)  ^{1/2n}<\left(  \left(  6r+1\right)
^{2n}-1\right)  ^{1/2n}<7,$ as $r<1.$ Hence, for such $n$ we have

\begin{equation}
\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  \right\vert
\leq\left(  L\left\Vert x-x_{n}\right\Vert +\left\vert f\left(  x_{n}\right)
\right\vert \right)  \nu_{n}\left(  x\right)  \leq\left(  7L+2\right)  \left(
1+\varepsilon^{\prime}\right)  \leq18L.
\end{equation}

\medskip

\noindent On the other hand, for $n$ such that $x\notin\widehat{D}_{n},$ by
$\left(  2.6\right)  $ we obtain

\begin{align*}
\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  \right\vert  &
\leq\left(  L\left\Vert x-x_{n}\right\Vert +\left\vert f\left(  x_{n}\right)
\right\vert \right)  \frac{\varepsilon^{\prime}r/2LL_{\varphi}}{1+\left\Vert
x-x_{n}\right\Vert }\\
& \\
&  \leq\left(  L\left\Vert x-x_{n}\right\Vert +2\right)  \frac{\varepsilon
^{\prime}/2L}{1+\left\Vert x-x_{n}\right\Vert }\\
& \\
&  \leq\varepsilon^{\prime}/2+\varepsilon^{\prime}/L\leq2\varepsilon^{\prime}.
\end{align*}

\noindent In any event, for all $n$ we have,
\begin{equation}
\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  \right\vert
\leq18L.
\end{equation}

\medskip

\noindent Therefore, for $n\notin\mathcal{N},$ using again property $\left(
1\right)  $ of $\delta_{n},$ we have,
\begin{align}
\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(  x\right)
-\delta_{n}(x)\right\vert \varphi_{n}\left(  x\right)   &  \leq\left(
\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  \right\vert
+\left\vert f\left(  x\right)  \right\vert +\delta_{n}(x)\right)  \varphi
_{n}\left(  x\right) \nonumber\\
& \\
&  \leq\left(  18L+2+2\varepsilon^{\prime}\right)  \left(  \varepsilon
^{\prime}r/25L\right)  \leq\varepsilon^{\prime}r.\nonumber
\end{align}

\medskip

\noindent It follows that, $\left\vert g\left(  x\right)  -f\left(  x\right)
\right\vert \leq10A_{1}\varepsilon^{\prime}r<\varepsilon.$

\medskip

\noindent We now establish some derivative estimates. Fix $x$ and consider the
expression
\[
\left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  \right)  ^{\prime
}=T_{n}^{\prime}\left(  x\right)  \nu_{n}\left(  x\right)  +T_{n}\left(
x\right)  \nu_{n}^{\prime}\left(  x\right)  .
\]
From an estimate analogous to $\left(  2.15\right)  ,$ using $\left(
2.4\right)  $ and $\left(  2.7\right)  ,$ we have that for all $n$,
$\left\Vert T_{n}\left(  x\right)  \nu_{n}^{\prime}\left(  x\right)
\right\Vert \leq9L$Lip$\left(  Q\right)  $Lip$\left(  \nu\right)  .$ Also,
$\left\Vert T_{n}^{\prime}\left(  x\right)  \nu_{n}\left(  x\right)
\right\Vert \leq L\left(  1+\varepsilon^{\prime}\right)  \leq2L.$ Hence,
\[
\left\Vert \left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  \right)
^{\prime}\right\Vert \leq2L+9L\text{Lip}\left(  Q\right)  \text{Lip}\left(
\nu\right)  \leq11L\text{Lip}\left(  Q\right)  \text{Lip}\left(  \nu\right)
.
\]

\medskip

\noindent Using this, and property $\left(  2\right)  $ of $\delta_{n},$ we have,

\medski

\begin{equation}
\text{Lip}\left(  T_{n}\nu_{n}-f-\delta_{n}\right)  \leq11L\text{Lip}\left(
Q\right)  \text{Lip}\left(  \nu\right)  +L+C_{0}\varepsilon^{\prime
\leq13C_{0}L\text{Lip}\left(  Q\right)  \text{Lip}\left(  \nu\right)  .
\end{equation}

\medskip

\noindent Next, for $x\in D_{n},$ $\overline{\nu}_{n}\left(  x\right)  =1,$
and again by property $\left(  2\right)  $ of $\delta_{n},$
\begin{equation}
\text{Lip}\left(  \left(  T_{n}\overline{\nu}_{n}-f-\delta_{n}\right)
\mid_{D_{n}}\right)  =\ \text{Lip}\left(  \left(  T_{n}-f-\delta_{n}\right)
\mid_{D_{n}}\right)  \leq\varepsilon^{\prime}+C_{0}\varepsilon^{\prime
\leq2C_{0}\varepsilon^{\prime}.
\end{equation}
Next we compute, using $\left(  2.4\right)  ,$
\begin{align*}
\left\Vert \left(  T_{n}\left(  x\right)  \left(  \nu_{n}-\overline{\nu
_{n}\right)  \right)  ^{\prime}\right\Vert  &  =\left\Vert T_{n}^{\prime
}\left(  x\right)  \right\Vert \left\vert \nu_{n}\left(  x\right)
-\overline{\nu}_{n}\left(  x\right)  \right\vert +\left\vert T_{n}\left(
x\right)  \right\vert \left\Vert \nu_{n}^{\prime}\left(  x\right)
-\overline{\nu}_{n}^{\prime}\left(  x\right)  \right\Vert \\
& \\
&  \leq L\frac{\varepsilon^{\prime}r/2LL_{\varphi}}{1+\left\Vert
x-x_{n}\right\Vert }+\left(  L\left\Vert x-x_{n}\right\Vert +2\right)
\frac{\text{Lip}\left(  Q\right)  \varepsilon^{\prime}r/2LL_{\varphi
}{1+\left\Vert x-x_{n}\right\Vert }\\
& \\
&  \leq\varepsilon^{\prime}r/2+\text{Lip}\left(  Q\right)  \ \varepsilon
^{\prime}r/2+\text{Lip}\left(  Q\right)  \ \varepsilon^{\prime}r/L\\
& \\
&  \leq2\text{Lip}\left(  Q\right)  \varepsilon^{\prime}.
\end{align*}

\medskip

\noindent It follows from this and $\left(  2.18\right)  $ that

\begin{equation}
\text{Lip}\left(  \left(  T_{n}\nu_{n}-f-\delta_{n}\right)  \mid_{D_{n
}\right)  \leq2C_{0}\varepsilon^{\prime}+2\text{Lip}\left(  Q\right)
\varepsilon^{\prime}r\leq4C_{0}\text{Lip}\left(  Q\right)  \varepsilon
^{\prime}.
\end{equation}

\medskip

\noindent Finally we turn to $\left\Vert g^{\prime}\left(  x\right)
-f^{\prime}\left(  x\right)  \right\Vert $ with the help of the above
estimates. Again fix $x\in X,$ and we obtain

\begin{align*}
&  \left\Vert g^{\prime}\left(  x\right)  -f^{\prime}\left(  x\right)
\right\Vert \\
& \\
&  =\left(  \frac{\left\Vert \left\{  \left(  T_{n}\left(  x\right)  \nu
_{n}\left(  x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)  \varphi
_{n}\left(  x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0}
}{\left\Vert \left\{  \varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty
}\right\Vert _{c_{0}}}\right)  ^{\prime}\\
& \\
&  =\frac{1}{\left\Vert \left\{  \varphi_{n}\left(  x\right)  \right\}
_{n=1}^{\infty}\right\Vert _{c_{0}}^{2}}\ \times\\
& \\
&  \left(  \left\Vert \left\{  \varphi_{n}\left(  x\right)  \right\}
_{n=1}^{\infty}\right\Vert _{c_{0}}\left\Vert \left\{  \left(  T_{n}\left(
x\right)  \nu_{n}\left(  x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)
\varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0
}^{\prime}\right. \\
& \\
&  \left.  -\left\Vert \left\{  \varphi_{n}\left(  x\right)  \right\}
_{n=1}^{\infty}\right\Vert _{c_{0}}^{\prime}\left\Vert \left\{  \left(
T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(  x\right)
-\delta_{n}(x)\right)  \varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty
}\right\Vert _{c_{0}}\right)
\end{align*}

\medskip

\noindent Let us first consider
\begin{align*}
&  \left(  \left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(
x\right)  -\delta_{n}(x)\right)  \varphi_{n}\left(  x\right)  \right)
^{\prime}\\
& \\
&  =\left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(
x\right)  -\delta_{n}(x)\right)  ^{\prime}\varphi_{n}\left(  x\right)
+\left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(  x\right)
-\delta_{n}(x)\right)  \varphi_{n}^{\prime}\left(  x\right)  .
\end{align*}
For the first term, and $n\in\mathcal{N},$ we have, using property $\left(
2\right)  $ of Lemma \ref{lemma phi} and $\left(  2.19\right)  ,

\begin{align*}
\left\Vert \left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(
x\right)  -\delta_{n}(x)\right)  ^{\prime}\varphi_{n}\left(  x\right)
\right\Vert  &  \leq\left\Vert \left(  T_{n}\left(  x\right)  \nu_{n}\left(
x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)  ^{\prime}\right\Vert
\varphi_{n}\left(  x\right) \\
& \\
&  \leq4C_{0}\text{Lip}\left(  Q\right)  \varepsilon^{\prime}\left(
1+\varepsilon_{1}\right) \\
& \\
&  \leq8C_{0}\text{Lip}\left(  Q\right)  \varepsilon^{\prime}.
\end{align*}

\medskip

\noindent For $n\notin\mathcal{N},$ using Lemma \ref{lemma phi} $\left(
4^{\prime}\right)  $ and $\left(  2.17\right)  ,$ we obtain

\begin{align*}
\left\Vert \left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(
x\right)  -\delta_{n}(x)\right)  ^{\prime}\varphi_{n}\left(  x\right)
\right\Vert  &  \leq\left\Vert \left(  T_{n}\left(  x\right)  \nu_{n}\left(
x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)  ^{\prime}\right\Vert
\varphi_{n}\left(  x\right) \\
& \\
&  \leq13C_{0}L\text{Lip}\left(  Q\right)  \text{Lip}\left(  \nu\right)
\left(  \varepsilon^{\prime}/6C_{0}L\text{Lip}\left(  \nu\right)  \right) \\
& \\
&  \leq3\text{Lip}\left(  Q\right)  \varepsilon^{\prime}.
\end{align*}

\medskip

\noindent In any event, $\left\Vert \left(  T_{n}\left(  x\right)  \nu
_{n}\left(  x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)  ^{\prime
}\varphi_{n}\left(  x\right)  \right\Vert _{c_{0}}\leq8C_{0}A_{1}$Lip$\left(
Q\right)  \varepsilon^{\prime}.$

\medskip

\noindent Next we consider the second term, $\left(  T_{n}\left(  x\right)
\nu_{n}\left(  x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)
\varphi_{n}^{\prime}\left(  x\right)  .$ For $n\in\mathcal{N},$ from the
estimate giving $\left(  2.13\right)  ,$ we have
\begin{align*}
\left\Vert \left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(
x\right)  -\delta_{n}(x)\right)  \varphi_{n}^{\prime}\left(  x\right)
\right\Vert  &  \leq\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)
-f\left(  x\right)  -\delta_{n}(x)\right\vert \left\Vert \varphi_{n}^{\prime
}\left(  x\right)  \right\Vert \\
& \\
&  \leq6\varepsilon^{\prime}r/L_{\varphi}\left(  L_{\varphi}\right)
\leq6\varepsilon^{\prime}.
\end{align*}

\medskip

\noindent For $n\notin\mathcal{N},$ just as in $\left(  2.16\right)  $ we obtain

\begin{align*}
\left\Vert \left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(
x\right)  -\delta_{n}(x)\right)  \varphi_{n}^{\prime}\left(  x\right)
\right\Vert  &  \leq\left\vert T_{n}\left(  x\right)  \nu_{n}\left(  x\right)
-f\left(  x\right)  -\delta_{n}(x)\right\vert \left\Vert \varphi_{n}^{\prime
}\left(  x\right)  \right\Vert \\
& \\
&  \leq\varepsilon^{\prime}.
\end{align*}

\noindent Hence, altogether we see that,
\begin{align*}
\left\Vert \left(  \left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)
-f\left(  x\right)  -\delta_{n}(x)\right)  \varphi_{n}\left(  x\right)
\right)  ^{\prime}\right\Vert _{c_{0}}  &  \leq8C_{0}A_{1}\text{Lip}\left(
Q\right)  \varepsilon^{\prime}+6A_{1}\varepsilon^{\prime}\\
& \\
&  \leq14C_{0}A_{1}\text{Lip}\left(  Q\right)  \varepsilon^{\prime}.
\end{align*}

\medskip

\noindent Lastly, we examine $\left\Vert \left\{  \varphi_{n}\left(  x\right)
\right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}^{\prime}\left\Vert \left\{
\left(  T_{n}\left(  x\right)  \nu_{n}\left(  x\right)  -f\left(  x\right)
-\delta_{n}(x)\right)  \varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty
}\right\Vert _{c_{0}}.$ Recall our estimate of $\left\vert f-g\right\vert $
found $\left\Vert \left\{  \left(  T_{n}\left(  x\right)  \nu_{n}\left(
x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)  \varphi_{n}\left(
x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}\leq3A_{1
\varepsilon^{\prime}r.$ Therefore we have

\begin{align*}
&  \left\Vert \left\{  \varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty
}\right\Vert _{c_{0}}^{\prime}\left\Vert \left\{  \left(  T_{n}\left(
x\right)  \nu_{n}\left(  x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)
\varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}\\
& \\
&  \leq A_{1}L_{\varphi}\left\Vert \left\{  \left(  T_{n}\left(  x\right)
\nu_{n}\left(  x\right)  -f\left(  x\right)  -\delta_{n}(x)\right)
\varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{\infty}\\
& \\
&  \leq A_{1}\frac{L_{1}\text{Lip}\left(  Q\right)  }{r}\left(  5A_{1
\varepsilon^{\prime}r\right)  =5A_{1}^{2}L_{1}\text{Lip}\left(  Q\right)
\varepsilon^{\prime
\end{align*}

\medskip

\noindent Finally, because $\left\Vert \left\{  \varphi_{n}\left(  x\right)
\right\}  _{n=1}^{\infty}\right\Vert _{c_{0}}^{2}\geq\left\Vert \left\{
\varphi_{n}\left(  x\right)  \right\}  _{n=1}^{\infty}\right\Vert _{\infty
}^{2}\geq1/4$ as noted above, putting all the above estimates together
yields,
\begin{align*}
\left\Vert g^{\prime}\left(  x\right)  -f^{\prime}\left(  x\right)
\right\Vert  &  \leq\frac{\left(  2A_{1}\right)  14C_{0}A_{1}\text{Lip}\left(
Q\right)  \varepsilon^{\prime}+5A_{1}^{2}L_{1}\text{Lip}\left(  Q\right)
\varepsilon^{\prime}}{1/4}\\
& \\
&  \leq\left(  132C_{0}A_{1}^{2}L_{1}\text{Lip}\left(  Q\right)  \right)
\varepsilon^{\prime}<\varepsilon.\ \ \square
\end{align*}

\medskip

\medskip

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Tuesday July 11, Maria, Maria Cantina, 710 Camino Ramon Road, Danville CA, approx. 6-9pm. After about nine months(!), Rolando finally returns to Carlos Santana's Danville restaurant for Taco Tuesday. $1 tacos, baby! And Rolando on solo guitar and voice with his diverse repertoire performing out on the patio by the fire pits. Reservations: (925) 820-2366. See www.mariamariarestaurants.com.
Next PostNext Friday July 14, 2017 Rolando Morales performs at Pairings Cellars, Pleasanton.


text:
Q: replace text from docx using Python I have a little side project I'm doing at work to help me with the "manual" labor of creating welcome documents for employees.
What I'm trying to do is:
1) Extract cell contents from a CSV file (username, first name, password)
2) Take cell contents and replace the strings [FirstName], [Username] and [Password] in the DOCX file.
This is the code:
""" Dependencies """
import csv
import numpy as np
import re
import shutil
import docx
#import fileinput

""" Open CSV and add usernames to list """
with open('UploadUsers.csv', 'rU') as csvfile:
    readCSV = np.loadtxt(csvfile, delimiter=',', skiprows=1, dtype = 'str')
    first_names = []
    usernames = []
    passwords = []

for row in readCSV:
    username = row[1]
    first_name = row[2]
    password = row[5]

    usernames.append(username)
    first_names.append(first_name)
    passwords.append(password)

for username in usernames:

    doc = "Free_Tracking_Manager_US.docx"
    doc_copy = "Free_Tracking_Manager_US.docx."+username+".docx"

    """ Make file copies """
    shutil.copy(doc, doc_copy)

    """ Make changes to first name, username and password """

    document = docx.Document(doc_copy)

    for paragraph in document.paragraphs:
        if '[FirstName]' in paragraph.text:
            print(first_name)
        if '[Username]'  in paragraph.text:
            print(username)
        if '[Password]'  in paragraph.text:
            print(password)

    """ Save document """
    document.save(doc_copy)

The script runs without errors, it makes copies of the documents but doesn't replace the strings. Not sure where the issue is...
UPDATE:
made the changes Abass suggested. I replaced:
for paragraph in document.paragraphs:
        if '[FirstName]' in paragraph.text:
            print(first_name)
        if '[Username]'  in paragraph.text:
            print(username)
        if '[Password]'  in paragraph.text:
            print(password)

with:
for paragraph in document.paragraphs:
        if '[FirstName]' in paragraph.text:
            print(first_name)
            paragraph.text = paragraph.text.replace('[FirstName]',first_name)
        if '[Username]'  in paragraph.text:
            print(username)
            paragraph.text = paragraph.text.replace('[Username]',username)
        if '[Password]'  in paragraph.text:
            print(password)
            paragraph.text = paragraph.text.replace('[Password]',username)

Still doesn't work...

A: IIUC You are not replacing the keyword strings after checking for their presence:
Do this change and it shall have the strings replaced.
for paragraph in document.paragraphs:
        if '[FirstName]' in paragraph.text:
            print(first_name)
            paragraph.text = paragraph.text.replace('[FirstName]',first_name)
        if '[Username]'  in paragraph.text:
            print(username)
            paragraph.text = paragraph.text.replace('[Username]',username)
        if '[Password]'  in paragraph.text:
            print(password)
            paragraph.text = paragraph.text.replace('[Password]',username)

redpajama_set_name: RedPajamaStackExchange


text:
Тамира Пашек (), је аустријска тенисерка рођена 6. децембра 1990. године. Професионално игра од 26. октобра 2005. године, а прву и, за сада, једину титулу освојила је септембра 2006. године у Порторожу.

Деби на ВТА турнирима имала је у Линцу (добила 'вајлд-кард') и одмах је забележила прву победу. Жртва је била Украјинка Јелена Веснина (7:6, 1:6, 7:5), а у другом колу је пружила солидан отпор Ани Ивановић – 7:5, 6:1.

На Вимблдону 2005. и Ју-Ес опену 2006. у конкуренцији јуниорки доспела је до финала. На Вимблдону је изгубила од Агњешке Радвањске, а на Ју-Ес опену од Анастасије Пављученкове.

Пласман око 300. места на свету приморавао ју је да игра квалификације за већину турнира, а лед је пробила у Истанбулу. Прошла је три кола квалификација (победила првог носиоца Јулијану Федак), затим у првом колу главног жреба савладала Алишу Молик са 5:7, 6:4, 6:4, сјајно започела меч против Каталине Кастано из Колумбије, али ниије имала снаге за више – 3:6, 6:4, 6:4. Играла је 'челинџере', понекад и 'фјучерсе', а врхунац те године био је у Порторожу – Тамира је добила осам узастопних мечева, прошла квалификације и освојила титулу, а на путу до првог трофеја савладала је Ивету Бенешову, Емили Лоа и Марију Елену Камерин. Завршила је сезону на 181. месту и са великим амбицијама је ушла у наредну.

На Аустралијан опену је надиграла три боље рангиране играчице – Лисјак и Хсие у квалификацијама, а затим и Северин Бремон у првом колу главног жреба. Зауставила је Вера Звонарјова у другом колу – 6:1, 6:3. Прошла је квалификације и у Дубаију, затим победила Катарину Среботник и одиграла одличан меч са Жистин Енен. Меч је трајао два часа и 28 минута, Белгијанка је губила са 1:0 у сетовима, суочила се са брејк лоптом при резултату 5:5 у другом, али на крају ипак тријумфовала – 4:6, 7:5, 6:1. Још неколико пута је прошла квалификације, а на Ролан Гаросу је поново намучила Ененову – губила је са 5:0 у првом сету, стигла до 5:5, није успела да искористи две везане гем лопте за тај брејк, па је Белгијанка привела меч крају – 7:5, 6:1.

На свом првом турниру на трави у каријери у Бирмингему стигла је до четврфинала, где је одузела сет Шараповој, али ипак изгубила – 6:3, 4:6, 6:2. Бриљирала је на Вимблдону – елиминисала је Татјану Головин и Јелену Дементјеву, а поражена је тек у осмини финала од Светлане Кузњецове са 6:3, 6:2.

Пласман на ВТА листи на крају сезоне

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Спољашње везе

 Вебсајт Тамире Пашек
 Тамира Пасек ВТА профилј
 ИТФ профил
 Spielerprofil Juniortennis.com
 Spielerprofil Fed Cup
 Tamira Paszek Fansite

Рођени 1990.
Аустријски тенисери
Тенисери на Летњим олимпијским играма 2012.
Аустријски олимпијци на Летњим олимпијским играма 2012.

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