1. ## Uniform convergence

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.

2. Originally Posted by 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.
I don't see where you have produced an actual sequence. You start with some point $\displaystyle x_n$, and go from there. You need to produce a full-fledged sequence.

3. For any $\displaystyle N \in \mathbb{Z}$, there is an $\displaystyle n > N$ and an $\displaystyle x_n$ so that $\displaystyle |f(x_n) - f_n(x_n)| \ge 2\epsilon$, which is where I get the sequence from. Although I guess it would've been more clear if I had labeled the $\displaystyle x_n's$ as $\displaystyle x_N$?