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Let $f:[0,1]\times[0,1]\to\mathbb R$ be continuous. For each $y\in[0,1]$ define $f_y:[0,1]\to\mathbb R$ by $f_y(x)= f(x,y)$. Show that the set $A=\big\{ f_y\,\big|\, y\in[0,1]\big\}$ is compact in ${\cal C}[0,1]$.

I tried to use the Arzela-Ascoli Theorem, that is $A$ is comapct if and only if $A$ is closed, pointwise bounded and equicontinuous.

I managed to show that $A$ is pointwise bounded by Extreme Value Theorem. I am not sure how to prove that $A$ is closed and equicontinuous.

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Hint: Take a sequence of point $y_n$, the set of functions $f_{y_n}$ is uniformly bounded since $f$ is continuous on $[0,1]\times [0,1]$ and its image is compact since $[0,1]\times [0,1]$ is compact.

It is uniformly equicontinuous continuous since $f$ is uniformly continuous as a continuous function defined on a compact.

Ascoli implies you can extract a subsequence which converges.

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You can do this directly. Each sequence in $A$ has the form $f_{y(1)},f_{y(2)},\ldots$ for some sequence $y(1),y(2),\ldots$ in $[0,1]$. By compactness some subsequence of $y(n)$ tends to some $y \in [0,1]$. For ease of notation assume $y(n) \to y$. Then we claim $f_{y(n)} \to f_y$. This can be proved using uniform continuity of $f$.

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Since the domain of $f$ is compact and $f$ is continuous, it is uniformly continuous. Therefore, one can show that the function $y \mapsto f_y : [0, 1] \to C[0, 1]$ is continuous using a straightforward $\delta$-$\epsilon$ argument. The image of a compact set under a continuous map is compact; thus, $\{f_y : y \in [0, 1]\}$ is compact.

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