content_image/299467.jpg content_image/299468.jpg content_image/299469.jpg content_image/299470.jpg content_image/299471.jpg
Likewise, there exists ˜ π$_{r}$ ∈ B$_{Π}$ ( π ⋆ $_{r}$, r ) such that 

<img src='content_image/299467.jpg'>

where the above equality holds by the definition of the limit superior (the o$_{r}$ (1) above rep- resents the term’s behavior as r → 0). In light of (A.5) and (A.7), we thus have 

<img src='content_image/299468.jpg'>

By the ‖·‖ -uniform continuity of z , | z ( f$_{π}$$_{r}$ ) − z ( f$_{˜}$ $_{π$_{r}$}$) | = o$_{r}$ (1) if ‖ f$_{π}$$_{r}$ − f$_{˜}$ $_{π$_{r}$}$‖ = o$_{r}$ (1). Let us show that the latter condition is met. Because ˜ π$_{r}$ ∈ B$_{Π}$ ( π ⋆ $_{r}$, r ), the triangle inequality, (A.6) and (A.11) imply that 

<img src='content_image/299469.jpg'>

Because Π$_{r}$ ⊂ { π ∈ Π : R ( π ) ≤ r } , Lemma A.4 (which applies because Π is totally bounded) implies that 

<img src='content_image/299470.jpg'>

from which we deduce that the RHS of (A.9) is o$_{r}$ (1). In summary, ‖ f$_{π}$$_{r}$ − f$_{˜}$ $_{π$_{r}$}$‖ = o$_{r}$ (1), hence | z ( f$_{π}$$_{r}$ ) − z ( f$_{˜}$ $_{π$_{r}$}$) | = o$_{r}$ (1) and consequently, by (A.8), 

<img src='content_image/299471.jpg'>

Taking r = r$_{m}$ and using the above result reveals that the RHS expression in (A.4) is o (1) indeed. This completes the proof of Lemma A.5. 

