Suppose $f(x , y)$ is continuous in both variables. For any $\epsilon > 0$ and some $y_0$, let $h_{\epsilon}(x) = \max_{y^{'}: \| y^{'} - y_0 \| \leq \epsilon} f(x , y^{'})$. It seems to me that $h_{\epsilon}(x)$ is continuous in $\epsilon$ on $(0 , \infty)$, that is, for any $\epsilon_n \rightarrow \epsilon_0 > 0$, one has $h_{\epsilon_n}(x) \rightarrow h_{\epsilon_0}(x)$.
Is this true? How should I prove it?
Yes, the function is continuous. Let $R$ be a large parameter; then $f$ is uniformly continuous in $B(y_0,R)$. Suppose $\epsilon<\epsilon'\leq R$. Then $h_\epsilon(x)\leq h_{\epsilon'(x)}$ (supremum over a larger set is no larger). Now let $y'$ be a point such that $h_{\epsilon'}(x) = f(x,y')$. Then there is a point $y\in B(y_0,\epsilon)$ with $\Vert y'-y\Vert\leq \delta = \epsilon'-\epsilon$. We then have
$$ 0 \leq h_{\epsilon'}(x) - h_\epsilon(x) \leq f(x,y')-f(x,y) $$
and this can be made arbitrarily small if $\delta$ is small enough.
