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**Math Major** Suppose that $\displaystyle f:[0,1] \rightarrow \mathbb{R} $ is continuous and that $\displaystyle {f_n} $ is such that $\displaystyle f_n \rightarrow f $ pointwise.

Show that $\displaystyle f_n \rightarrow f $ uniformly if and only if for every sequence $\displaystyle {x_n}$ such that $\displaystyle x_n \rightarrow x $, we have that $\displaystyle f_n(x_n) \rightarrow f(x) $.

My work: I've finished the forward implication. This is my work on the reverse. We'll prove the contrapositive.

Suppose that $\displaystyle f_n $ does not converge to $\displaystyle f$ uniformly. So, there is some $\displaystyle \epsilon > 0 $ so that for any $\displaystyle N \in \mathbb{Z} $, there is an $\displaystyle n \ge N$ and an $\displaystyle x_n $ so that $\displaystyle |f(x_n) - f_n(x_n)| \ge 2\epsilon $.

So, $\displaystyle {x_n} $ is a bounded sequence of reals, and so has a convergent subsequence, say $\displaystyle {x_{n'}}$ such that $\displaystyle x_{n'} \rightarrow x \in [0,1] $

Since $\displaystyle f$ is continuous at $\displaystyle x$, we can find an $\displaystyle N \in \mathbb{Z}$ such that for any $\displaystyle n > N$, $\displaystyle |f(x) - f(x_n)| < \epsilon$

But then, if $\displaystyle n' > N$, we have that

$\displaystyle 2\epsilon \le |f(x_{n'}) - f_{n'}(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + |f(x) - f(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + \epsilon $

or

$\displaystyle \epsilon \le |f_{n'}(x_{n'}) - f(x)| $

and so $\displaystyle f_{n'}(x_{n'}) $ does not converge to $\displaystyle f(x)$.

Something about this argument feels incredibly wrong, and I was wondering if anyone could point out where I went awry.