Let z ∈ ℓ $^{∞}$( F ) be ‖·‖ -uniformly continuous. If Π is totally bounded, then h is continuous at ( z, 0) . Proof of Lemma A.5 and Corollary A.6. Fix a sequence { ( z$_{m}$, r$_{m}$ ) }$_{m}$$_{≥}$$_{1}$ satisfying the condi- tions of the Lemma A.5. Observe that, for every m ≥ 1, where the above RHS expression is o (1) because z$_{m}$ → z in ℓ $^{∞}$( F ). Therefore, Let us show now that the RHS expression in (A.4) is o (1). For any r > 0, there exists a π$_{r}$ ∈ Π$_{r}$ such that Furthermore, there exists a π ⋆ r ∈ Π ⋆ such that | g ( z$_{m}$, r$_{m}$ ) − g ( z, r$_{m}$ ) | ≤ ∣ sup π ∈ Π$_{r}$$_{m}$ z$_{m}$ ( f$_{π}$ ) − sup π ∈ Π$_{r}$$_{m}$ z ( f$_{π}$ ) ∣ + ∣ lim sup s ↓ 0 sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,s $_{)}$z$_{m}$ ( f$_{π}$ ) − lim sup s ↓ 0 sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,s $_{)}$z ( f$_{π}$ ) ∣ ≤ 2 sup π ∈ Π | z$_{m}$ ( f$_{π}$ ) − z ( f$_{π}$ ) | g ( z$_{m}$, r$_{m}$ ) = g ( z$_{m}$, r$_{m}$ ) − g ( z, r$_{m}$ ) + g ( z, r$_{m}$ ) = g ( z$_{m}$, r$_{m}$ ) − g ( z, r$_{m}$ ) + [ sup π ∈ Π$_{r}$$_{m}$ z ( f$_{π}$ ) − lim sup s ↓ 0 sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,s $_{)}$z ( f$_{π}$ ) ] . ≤ [ sup π ∈ Π$_{r}$$_{m}$ z ( f$_{π}$ ) − lim sup s ↓ 0 sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,s $_{)}$z ( f$_{π}$ ) ] + o (1) . (A.4) sup π ∈ Π$_{r}$ z ( f$_{π}$ ) ≤ z ( f$_{π}$$_{r}$ ) + r. (A.5) ‖ π ⋆ r − π$_{r}$ ‖ ≤ inf π $^{⋆}$∈ Π ⋆ ‖ π ⋆ − π$_{r}$ ‖ + r ≤ sup π ∈ Π$_{r}$ inf π $^{⋆}$∈ Π ⋆ ‖ π ⋆ − π ‖ + r. (A.6) 26