Uniform convergence

**Math Major** · February 1st 2011, 05:35 PM

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

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

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

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

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

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

But then, if $n' > N$ , we have that
$2\epsilon \le |f(x_{n'}) - f_{n'}(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + |f(x) - f(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + \epsilon$

or

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

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

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

**Drexel28** · February 1st 2011, 06:10 PM

Originally Posted by Math Major

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

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

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

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

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

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

But then, if $n' > N$ , we have that
$2\epsilon \le |f(x_{n'}) - f_{n'}(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + |f(x) - f(x_{n'})| \le |f_{n'}(x_{n'}) - f(x)| + \epsilon$

or

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

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

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 $x_n$ , and go from there. You need to produce a full-fledged sequence.

**Math Major** · February 1st 2011, 06:33 PM

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

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