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Let $f_n \to f$ uniformly, $f$ be continuous, and $x_n \to z$. Prove that $f_n(x_n) \to f(z)$.

My attempt:

Pick $\epsilon > 0$.

By continuity of $f$, we have $|f(x_n) - f(z)| \to 0$. So, pick $n_0$ such that $\forall n \geq n_0: |f(x_n) - f(z)| < \epsilon/2$.

Since $f_n \to f$ uniformly, we can pick $n_1$ such that $\forall n \geq n_1: \forall x: |f_n(x) - f(x)| < \epsilon/2$.

Then, for $n \geq \max\{n_0,n_1\}$ $$|f_n(x_n) - f(z)| \leq |f_n(x_n) - f(x_n)| + |f(x_n) - f(z)| < \epsilon/2 + \epsilon/2 = \epsilon$$

Is this correct?

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    $\begingroup$ Yes, it is correct! $\endgroup$ – eddie May 28 '18 at 8:12
  • $\begingroup$ Thanks for the verification. $\endgroup$ – user370967 May 28 '18 at 8:12
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Your attempt is correct. I would like to bring to your attention a partial converse to this statement. I will mention the hypotheses.

Let $X \subset \mathbb R^n$. If $f_n : X \to \mathbb R$ is a sequence of functions and $f : X \to \mathbb R$ is a continuous function such that $f_n(x_n) \to f(x)$ for every $x_n \to x$ in the domain, then $f_n \to f$ uniformly.

For if not, then(by definition of being "not uniformly convergent") there is an $\epsilon > 0$, and for each $n \geq 1$ an $x_n$ in the domain, such that $|f_n(x_n) - f(x_n)| > \epsilon$.

Now, assume that $X$ is compact, then $x_n$ has a convergent subsequence $x_{n_k} \to x \in X$. By hypothesis, $f_{n_k}(x_{n_k}) \to f(x)$, and $f(x_{n_k}) \to f(x)$ by continuity of $f$. Subtracting, $f_{n_k}(x_{n_k}) - f(x_{n_k}) \to 0$, a contradiction to the fact that it is always greater than $\epsilon$.

So, if $X$ is compact, then conditions given are equivalent, something that is not obvious and is a useful proposition.

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  • $\begingroup$ Nice statement. Another one I didn't know about! Thanks! This proofs generalises to arbitrary metric spaces it seems. $\endgroup$ – user370967 May 28 '18 at 8:28
  • $\begingroup$ You are welcome! Turns out this came from a competitive exam that was held recently. $\endgroup$ – астон вілла олоф мэллбэрг May 28 '18 at 8:29
  • $\begingroup$ Yes, this would generalize to arbitrary metric spaces as it stands. $\endgroup$ – астон вілла олоф мэллбэрг May 29 '18 at 4:13

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