$\newcommand{\R}{\mathbb{R}} \newcommand{\eps}{\epsilon} \newcommand{\norm}[1]{\Vert #1 \Vert}$ Let $A$ and $B$ be subsets of the Euclidean space. Consider the continuous function $f:A\times B \rightarrow \R$. Let $h:B \rightarrow \R$ defined by $$ h(y) = \sup_{x\in A} f(x,y) $$ Then, it is known that $h$ is lower semicontinuous. That is, $$ \forall y \in B:\forall \eps > 0:\exists \delta > 0:\forall z \in B: \norm{y - z} < \delta \Rightarrow h(z) > h(y) - \eps $$
My question is, on what condition, can we say that the supremum is continuous. I guess that $f$ being Lipschitz on $A\times B$ is sufficient, inspired by Supremum of continuous functions is continuous?
My attempt: First, I show the lower semicontinuity of the supremum $h$ of the continuous function $f$.
Let $y \in B$. Then, by the definition of supremum, $$ \begin{align} P_1)&~\forall x \in A: f(x,y) \leq h(y)\\ P_2)&~\forall \eps >0:\exists x \in A:f(x,y) > h(y) - \eps \end{align} $$ Let us fix $\eps$ and choose $x$ so that $f(x,y) > h(y) - \eps/2$. Now, let $z \in B$ and $\norm{z - y} < \delta$ for some $\delta > 0$ to be chosen. Because $\norm{z-y} = \norm{(x,z) - (x,y)}$, we have $\norm{(x,z) - (x,y)} < \delta$ and by the continuity of $f$, we can choose $\delta > 0$ so that $\norm{f(x,z) - f(x,y)} < \eps/2$. Hence, from $f(x,z) \leq h(z)$ $$ h(y)-h(z) < f(x,y) + \eps/2 - f(x,z) \leq \eps/2 + |f(x,y) - f(x,z)| < \eps $$ which shows that $h$ is lower semicontinuous.
To show that $h$ is also continuous, we have to additionally show that we can choose $\delta > 0$ such that $ h(z) - h(y) < \eps$ whenever $\norm{z-y} < \delta$. We can choose $\delta$ so that $|f(x,z) - f(x,y)| < \eps/3$. Since $h(z)$ is the supremum of $f(\cdot,z)$, there is $p \in A$ such that $f(p,z) > h(z) - \eps/3$.
$$ \begin{align} h(z) - h(y) &< f(p,z) + \eps/3 - f(x,y)\\ &= f(p,z) - f(x,z) +f(x,z) - f(x,y) + \eps/3\\ &\leq f(p,z) - f(x,z) + |f(x,z) - f(x,y)| + \eps/3\\ &< f(p,z) - f(x,z) + 2\eps/3 \end{align} $$ But I could not find a clue that $|f(p,z) - f(x,z)|$ can be made arbitrarily small by the choice of $\delta$.
