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Let $K$ compact in a metric space $M$, $N$ is a metric space, $\mathcal{C}(K,N)$ the space of continuous functions $f: K \longrightarrow N$ and $E \subset \mathcal{C}(K,N)$ a family of functions such that $\overline{E}$ is compact in $\mathcal{C}(K,N)$ with respect to the topology generated by the metric of uniform convergence (see the the post scriptum for the definition). Fix $a \in K$ and define $\varphi: \overline{E} \times K \longrightarrow \mathbb{R}$, $\varphi(f,x) := d(f(x),f(a))$. Show that $\varphi$ is continuous.

Initially, I was stuck because I don't know what is the metric defined on $\overline{E} \times K$ (observe that I can't take any metric since that I don't know if this space is a finite-dimensional normed space), but I tried prove the continuity of $\varphi$ via sequence and I would like to know if what I did is right.

$\textbf{My attempt:}$

Let be $(f_n,x_n)$ a sequence in $\overline{E} \times K$ such that $\lim\limits_{n \rightarrow \infty} (f_n,x_n) = (f,x)$. By continuity of $f$ and the metric $d$,

$$\lim\limits_{n \rightarrow \infty} \varphi(f_n,x_n) = \lim\limits_{n \rightarrow \infty} d(f_n(x_n),f_n(a)) = \lim\limits_{n \rightarrow \infty} d(f_n(x_n),f_n(a)) = d(\lim\limits_{n \rightarrow \infty} f_n(x_n), \lim\limits_{n \rightarrow \infty} f_n(a)) = d(\lim\limits_{n \rightarrow \infty} f_n(x_n), f(a)),$$

so the continuity of $\varphi$ will follow if I'm be able to show that $\lim\limits_{n \rightarrow \infty} f_n(x_n) = f(x)$. By the convergence of $(f_n)$ to $f$ and by the continuity of $f$ (observe that $f$ is continuous because $\overline{E}$ is compact by hypothesis, therefore is closed), we know that, given $\varepsilon > 0$, exist $N \in \mathbb{N}$ such that

$$n \geq N \Longrightarrow d(f_n(x),f(x)) < \frac{\varepsilon}{3}$$

and

$$n \geq N \Longrightarrow d(f(x_n),f(x_n)) < \frac{\varepsilon}{3}$$

Thus,

$$d(f_n(x_n),f(x)) \leq d(f_n(x_n),f_N(x_n)) + d(f_N(x_n),f_N(x)) + d(f_N(x),f(x)) < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon,$$

whenever $n \geq N$, therefore $\lim\limits_{n \rightarrow \infty} f_n(x_n) = f(x)$ and the continuity of $\varphi$ is proved. $\square$

Thanks in advance!

$\textbf{EDIT:}$

$\textbf{Definition:}$ if $X \subset M$ is arbitrary and $f,g: X \longrightarrow M$ are bounded functions, then the metric of uniform convergence is defined by

$$d(f,g) := \sup_{x \in X} d(f(x),g(x)).$$

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I assume, that you have a norm $\lVert \cdot\rVert_N$ on $N$; otherwise, you cannot define the sup norm $\lVert \cdot\rVert_{\infty}$ on $\mathcal{C}(K,N)$, right?

Let $(f_n)_{n \geq 0}$ be a sequence in $\bar{E}$ and $f \in \bar{E}$ with $$ \lim_{n \rightarrow \infty} \lVert f_n - f \rVert_{\infty} = 0 .$$ Furthermore, let $(x_n)_{n \geq 0}$ be a sequence in $K$ and $x \in K$ with $$ \lim_{n \rightarrow \infty} d_M( x_n, x) = 0 .$$

If we can show, that this is sufficient to conclude, that $$ \lim_{n \rightarrow \infty} \lvert \varphi(f_n, x_n) - \varphi(f, x) \rvert = 0 ,$$ then we can conclude, that $\varphi$ is continuous.

To see, that this is the case, we calculate (using the reverse triangle inequality for our first upper bound) $$ \lvert \varphi(f_n, x_n) - \varphi(f, x) \rvert = $$ $$ \lvert \lVert f_n(x_n) - f_n(a) \rVert_N - \lVert f(x) - f(a) \rVert_N \lvert \leq $$ $$ \lVert (f_n(x_n) - f_n(a)) - (f(x) - f(a)) \rVert_N \leq $$ $$ \lVert f_n(x_n) - f(x) \rVert_N + \lVert f_n(a) - f(a)) \rVert_N \leq $$ $$ \lVert f_n(x_n) - f(x) \rVert_N + \lVert f_n - f \rVert_{\infty} \leq $$ $$ \lVert f_n(x_n) - f(x_n) \rVert_N + \lVert f(x_n) - f(x) \rVert_N + \lVert f_n - f \rVert_{\infty} \leq $$ $$ \lVert f_n - f \rVert_{\infty} + \lVert f(x_n) - f(x) \rVert_N + \lVert f_n - f \rVert_{\infty} $$

Taking the limit $n \rightarrow \infty$ in this chain of inequalities completes the proof.

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  • $\begingroup$ Sorry, it was a typo, $\mathcal{C}(K,N)$ is not equipped with the sup norm, I edited my post and put the definition of the metric of uniform convergence $\endgroup$ – George Apr 22 at 23:11
  • $\begingroup$ I see! I think you can still use my exact proof and just replace the metric induced by the sup norm with the metric of uniform convergence. Everything will fall into place just like before. $\endgroup$ – Joker123 Apr 22 at 23:23

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