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?



Is the pointwise supremum of a continuous function continuous?
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.


