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Let $f:\left[0,1\right]\times\left[0,1\right]\rightarrow\mathbb{R}$ be continuous and let $g:\left[0,1\right]\rightarrow\left[0,1\right]$ be also a continuous function. Is the following function continuous? \begin{eqnarray*} h:\left[0,1\right] & \rightarrow & \mathbb{R}\\ x & \rightarrow & \sup_{g\left(x\right)\leq t\leq1}f\left(t,x\right) \end{eqnarray*}

Thank you.

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    I would not exactly label this problem as "functional analysis" – 2017-01-06
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    Surely the range of $h$ is $\mathbb{R}$? – 2017-01-06

2 Answers 2

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Define $\phi(x,s) = f(g(x)+s(1-g(x)), x)$, then $h(x) = \max_{t \in [0,1]} \phi(x,t)$.

Suppose $x_n \to x$, then $\phi(x_n,t) \to \phi(x,t)$ for all $t$ and since $h(x_n)\ge \phi(x_n,t)$, we see $\liminf_n h(x_n) \ge \phi(x,t)$ for all $t$ and so $\liminf_n h(x_n) \ge h(x)$.

Let $t_n$ be such that $h(x_n) = \phi(x_n,t_n)$, then let $x_{n_k}, t_{n_k}$ be such that $\phi(x_{n_k}, t_{n_k}) \to \limsup_n h(x_n)$ and $ t_{n_k} \to t^*$, then $\phi(x_{n_k}, t_{n_k}) \to \phi(x,t^*) \le h(x)$ and so $\limsup_n h(x_n) \le h(x)$.

Hence $\lim_n h(x_n) = h(x)$ and so $h$ is continuous.

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    Your idea is good but the function h(x) is not equal to the function as you wrote. – 2017-01-06
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    I inadvertently switched the $x,t$ parameters. The functions $h$ in the question and above are equal now. The idea is an old & standard one, not mine! – 2017-01-06
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    @copper.hat Great answer! Very slick! – 2017-01-06
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    @Nobody: Actually, I prefer solutions such as yours, sequence arguments tend to be slick in the sense of quick but opaque whereas the uniform continuity arguments tend to be more revealing. Complete non sequitur; one of my favourite films is "My Name is Nobody". – 2017-01-06
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    @copper.hat Thank you, but those horrid estimates take a while for me to cook up—I much prefer an answer such as yours! Small comment: I believe Binjiu's concern is that you meant $\phi(x,s)=f(1-sg(x),x)$, although it doesn't end up mattering because $g$ was an arbitrary continuous function. I actually just find it funny when other users refer to "Nobody" a la Odysseus and Polyphemus :) – 2017-01-06
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    @Nobody: Thanks for catching my error, I didn't read carefully enough and thought it was $0 \le t \le g(x)$. A slightly less classical nobody is in Beckett's Godot: "nothing happens, nobody comes, nobody goes, it's awful." A new interpretation, maybe? – 2017-01-06
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    @copper.hat No problem. You know, I've never had the chance to see/read `Waiting for Godot', but I've always wanted to. – 2017-01-06
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Note that $f$ and $g$ are uniformly continuous. Let the metric $d$ on $[0,1]\times[0,1]$ be $d((a,b),(x,y))=|x-a|+|y-b|$. This is equivalent to the usual metric. Fix $\varepsilon>0$ and pick $\delta_1,\delta_2>0$ so small that if $|a-b|<\delta_1$, then $|g(a)-g(b)|<\dfrac{\delta_2}{2}$ where if $d((a,b),(x,y))<\delta_2$, then $|f(a,b)-f(x,y)|<\varepsilon$. Let $x_0\in[0,1]$ and $\delta=\dfrac{1}{2}\min\{\delta_1,\delta_2\}$.

If $|x-x_0|<\delta$, then $|h(x)-h(x_0)|=\left|\sup_{g(x)\le t\le 1}f(t,x)-\sup_{g(x_0)\le t\le 1}f(t,x_0)\right|$. The sups are attained because of compactness, so we may as well let $f(a,x)=\sup_{g(x)\le t\le 1}f(t,x)$ and $f(b,x_0)=\sup_{g(x_0)\le t\le 1}f(t,x_0)$. Note that $|f(g(x),x)-f(g(x_0),x_0)|<\varepsilon$ as we contrived that $|g(x)-g(x_0)|+|x-x_0|<\delta_2$.

First suppose that $a$ and $b$ do not lie between $g(x)$ and $g(x_0)$. Then $f(a,x)

Now suppose WLOG that $b$ is between $g(x)$ and $g(x_0)$ (the case of $a$ is identical). We have that $|f(a,x)-f(a,x_0)|<\varepsilon$ and $|f(g(x),x)-f(b,x_0)|<\varepsilon$. Hence, $f(a,x)<\varepsilon+f(a,x_0)\le \varepsilon+f(b,x_0)$ and $f(b,x_0)<\varepsilon +f(g(x),x)\le \varepsilon +f(a,x)\ldotp$ Hence, $|f(a,x)-f(b,x_0)|<\varepsilon$.

Hence, $h$ is continuous.

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    So good, thanks. – 2017-01-06
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    @Binjiu Thank you, but copper.hat's answer (I think we actually posted simultaneously) is *much* slicker than mine, and you should accept his! – 2017-01-06