Likewise, there exists ˜ π$_{r}$ ∈ B$_{Π}$ ( π ⋆ $_{r}$, r ) such that 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 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 Because Π$_{r}$ ⊂ { π ∈ Π : R ( π ) ≤ r } , Lemma A.4 (which applies because Π is totally bounded) implies that 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), 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. z ( f$_{˜}$ $_{π$_{r}$}$) ≤ inf π ∈ B$_{Π}$ ( π ⋆ $_{r}$,r $_{)}$z ( f$_{π}$ ) + r ≤ sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,r $_{)}$z ( f$_{π}$ ) + r = ( lim sup s ↓ 0 sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,s $_{)}$z ( f$_{π}$ ) + o$_{r}$ (1) ) + r, (A.7) sup π ∈ Π$_{r}$ z ( f$_{π}$ ) − lim sup s ↓ 0 sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,s $_{)}$z ( f$_{π}$ ) ≤ z ( f$_{π}$$_{r}$ ) − z ( f$_{˜}$ $_{π$_{r}$}$) + 2 r + o$_{r}$ (1) . (A.8) ‖ f$_{π}$$_{r}$ − f$_{˜}$ $_{π$_{r}$}$‖ ≤ ∥ f$_{π}$$_{r}$ − f$_{π}$ ⋆ r ∥ + ∥ f$_{π}$ ⋆ r − f$_{˜}$ π$_{r}$ ∥ ≲ ‖ π$_{r}$ − π ⋆ $_{r}$‖ + ‖ π ⋆ r − ˜ π$_{r}$ ‖ ≤ sup π ∈ Π$_{r}$ inf π $^{⋆}$∈ Π ⋆ ‖ π ⋆ − π ‖ + 2 r. (A.9) lim sup r ↓ 0 sup π ∈ Π$_{r}$ inf π $^{⋆}$∈ Π ⋆ ‖ π − π $^{⋆}$‖ = 0 , sup π ∈ Π$_{r}$ z ( f$_{π}$ ) − lim sup s ↓ 0 sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,s $_{)}$z ( f$_{π}$ ) = o$_{r}$ (1) . 27