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Consider a function $f:\mathcal{X}\times \mathcal{Y}\rightarrow \mathbb{R}$ with $\mathcal{X}\subseteq \mathbb{R}^k$ and $\mathcal{Y}\subseteq \mathbb{R}^p$.

Under which sets of conditions is $\sup_{x\in \mathcal{X}} f(x,y)$ continuous?

Similar questions are asked here and here (among the others) but I can't summarise the main findings.

In particular, is having $f(x,y)$ jointly continuous in $x$ and $y$ plus $\mathcal{X}$ compact sufficient?

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  • The same proof as in http://math.stackexchange.com/questions/609012/how-prove-this-gx-sup-fx-y0-le-y-le-1-is-continuous-on-0-1 applies. – Jochen Apr 21 '16 at 11:28
  • I don't understand whether they need $\mathcal{Y}$ compact as well. Moreover, do they need compactness or closeness? – Star Apr 21 '16 at 11:41
  • You need compactness of $Y$ to have $f$ uniformly continuous. – Jochen Apr 22 '16 at 07:43

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I add an answer in case others are confused on this point, as I was, after reading the comments above.

Write $g(y) = \sup_{x \in \mathcal{X}} f(x,y)$; the claim that $f$ continuous and $\mathcal{X}$ compact implies that $g$ continuous is true, as long as $\mathcal{Y}$ is somewhat reasonable (say, closed; this can probably be weakened).

The intuition is that continuity of $g$ at a point $y_0 \in \mathcal{Y}$ depends only on local information at $y_0$; so it is without loss of generality to consider some compact $\mathcal{Y}' \subset \mathcal{Y}$ (for example, a closed ball around $y_0$ intersected with $\mathcal{Y}$), pass to an auxiliary $\tilde{f} : \mathcal{X} \times \mathcal{Y}' \to \mathbb{R}$ which agrees with $f$ locally at $\mathcal{X} \times \{ y_0 \}$, and run the proof linked by Jochen on this function.

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