Consider a continuous function $\phi: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}_{\geq 0}$ and a locally bounded function $\psi:\mathbb{R}^n \rightarrow \mathbb{R}^m$.

So we study functions of the kind $\phi(x,\psi(y))$ for $x,y \in \mathbb{R}^n$.

Prove the following proposition.

Fixed $\bar x \in \mathbb{R}^n$, for any $\epsilon>0$ there exists $\delta > 0$ such that

$$ \max_{(x,y) \in X \times X} \{ \phi(x,\psi(y)) - \phi(y,\psi(y)) \} < \epsilon $$

where $X = {\bar x} + \delta \mathbb{B}$ is the closed ball around $\bar x$.

  • 2
    $\begingroup$ Hint: Use uniform continuity of $\phi$ on compact sets. $\endgroup$ – Francesco Sica May 14 '12 at 3:31
  • $\begingroup$ I know. I'm just wondering if the discontinuity of $\psi(\cdot)$ can cause problems. $\endgroup$ – user693 May 14 '12 at 5:17
  • $\begingroup$ So maybe we can write $$\max_{(x,y) \in X^2} \{ \phi(x,\psi(y))-\phi(y,\psi(y)) \} \leq \max_{(x,y) \in X^2, \ z \in \overline{\psi(\{X\})}} \{ \phi(x,z)-\phi(y,z) \}. $$ Then we notice that $(x,y,z) \mapsto \phi(x,z)-\phi(y,z)$ is continuous and the result follows. $\endgroup$ – user693 May 14 '12 at 5:33

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