Let $a_1,a_2,\ldots$ be a bounded sequence of reals. Is it true that the fact
$$\lim_{N\to\infty}\frac1N\sum_{n=1}^Na_n=b\enspace\text{ and }\enspace\lim_{N\to\infty}\frac1{\log N}\sum_{n=1}^N\frac{a_n}n=c$$implies $b=c$?
To determine whether the given conditions imply \( b = c \), we need to analyze the two limits provided:

1. \(\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N a_n = b\)
2. \(\lim_{N \to \infty} \frac{1}{\log N} \sum_{n=1}^N \frac{a_n}{n} = c\)

We will show that if \( b \) exists, then \( c \) must also exist and \( b = c \).

### Step 1: Understanding the First Limit
The first limit states that the average of the first \( N \) terms of the sequence \( \{a_n\} \) converges to \( b \):
\[
\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N a_n = b
\]

### Step 2: Understanding the Second Limit
The second limit involves a logarithmic average:
\[
\lim_{N \to \infty} \frac{1}{\log N} \sum_{n=1}^N \frac{a_n}{n} = c
\]

### Step 3: Relating the Two Limits
To relate these two limits, we use the fact that the harmonic series \( H_N = \sum_{n=1}^N \frac{1}{n} \) asymptotically behaves like \( \log N \). Specifically:
\[
H_N \sim \log N + \gamma \quad \text{(where \(\gamma\) is the Euler-Mascheroni constant)}
\]

### Step 4: Expressing the Second Sum in Terms of the First
Consider the sum \( \sum_{n=1}^N \frac{a_n}{n} \). We can rewrite it as:
\[
\sum_{n=1}^N \frac{a_n}{n} = \sum_{n=1}^N a_n \cdot \frac{1}{n}
\]

### Step 5: Using the Cesàro Mean
Given that \( \lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N a_n = b \), we can use the Cesàro mean to show that:
\[
\frac{1}{N} \sum_{n=1}^N a_n \to b \implies \frac{1}{H_N} \sum_{n=1}^N a_n \cdot \frac{1}{n} \to b
\]

Since \( H_N \sim \log N \), we have:
\[
\frac{1}{\log N} \sum_{n=1}^N \frac{a_n}{n} \to b
\]

### Step 6: Concluding \( b = c \)
Thus, we have shown that:
\[
\lim_{N \to \infty} \frac{1}{\log N} \sum_{n=1}^N \frac{a_n}{n} = b
\]

Therefore, \( c = b \).

\(\blacksquare\)

The final answer is \( \boxed{ b = c } \).