# Prove that $\{f_n\}$ converges to $f$ with respect to the uniform norm if and only if it converges uniformly to $f$.

## Problem

Let $$f$$ and $$f_1,f_2,\dots$$ be continuous functions on $$[a,b]$$. Prove that $$\{f_n\}$$ converges to $$f$$ with respect to the uniform norm if and only if it converges uniformly to $$f$$.

Definition$$\quad$$ The function $$\|\cdot\|_{\infty}:C[a,b]\to\mathbb{R}$$ defined by the formula \begin{align*} \|f\|_{\infty} = \sup\left\{|f(x)|:x\in[a,b]\right\} \end{align*} is a norm (the continuity of $$f$$ and the compactness of $$[a,b]$$ imply that $$\|f\|_{\infty}$$ is finite; this is important because the codomain of a norm is $$\mathbb{R}$$). This is norm is called the uniform norm on $$C[a,b]$$.

## My Attempt

Let $$\epsilon>0$$. Suppose first that $$\{f_n\}$$ converges to $$f$$ with respect to $$\|\cdot\|_{\infty}$$. This implies that $$\lim_{n\to\infty}\|f_n(x)-f(x)\|_{\infty} = \lim_{n\to\infty}\sup\{|f_n(x)-f(x)|:x\in[a,b]\}=0$$. Then there is an integer $$N$$ such that $$n\geq N$$ implies $$\sup\{|f_n(x)-f(x)|:x\in[a,b]\}<\epsilon$$, and so $$|f_n(x)-f(x)|<\epsilon$$ for all $$x\in[a,b]$$. So $$\{f_n\}$$ converges uniformly to $$f$$ on $$[a,b]$$.

Conversely, suppose that $$\{f_n\}$$ converges uniformly to $$f$$ on $$[a,b]$$. Then there is an integer $$N$$ such that $$n\geq N$$ implies $$|f_n(x)-f(x)|\leq\epsilon$$ for all $$x\in[a,b]$$ and so $$\sup\{|f_n(x)-f(x)|:x\in[a,b]\}\leq\epsilon$$. Since $$\epsilon$$ is arbitrary, it follows that $$\lim_{n\to\infty}\sup\{|f_n(x)-f(x)|:x\in[a,b]\} = 0$$, and thus $$\{f_n\}$$ converges to $$f$$ with respect to $$\|\cdot\|_{\infty}$$.

## My Question

I am not sure whether my attempt is correct. I would really appreciate it if someone could help me check! Especially, in the second paragraph, I claimed that $$\lim_{n\to\infty}\sup\{|f_n(x)-f(x)|:x\in[a,b]\} = 0$$. However, by the definition of a convergent sequence, we have to have strict inequality to conclude the limit exists (see below).

The following definition is taken from Baby Rudin:

Definition$$\quad$$ A sequence $$\{p_n\}$$ in a metric space $$X$$ is said to converge if there is a point $$p$$ in $$X$$ with the following property: For every $$\epsilon>0$$ there is an integer $$N$$ such that $$n\geq N$$ implies that $$d(p_n,p)<\epsilon$$. In this case, we wrtie $$\lim_{n\to\infty}p_n=p$$.

• Note that $\|f\|_\infty < \epsilon$ implies $|f(x)| < \epsilon$ for all $x$. In the reverse direction, if $|f(x)| < \epsilon$ for all $x$ then $\|f\|_\infty \le \epsilon$ (not $<$!). Commented May 9 at 2:58
• @copper.hat Thanks for your comment! Quick question: Does $|f(x)|\leq\epsilon$ for all $x$ imply $\|f\|_{\infty}\leq\epsilon$? Commented May 9 at 10:35

Your attempt is correct. Don't worry about the strict inequality in the definition. You can show that: For a sequence $$(x_n)$$ and an element $$x$$,
$$\forall \epsilon > 0, \exists N \in \mathbb{N}, d(x_n, x) < \epsilon \ \forall n \ge N \Longleftrightarrow \forall \epsilon > 0, \exists N \in \mathbb{N}, d(x_n, x) \le \epsilon \ \forall n \ge N$$
$$(\Longrightarrow)$$ This is straightforward. As, $$a < b$$ implies $$a \le b$$
$$(\Longleftarrow)$$ Fix $$\epsilon > 0$$. There exists $$N \in \mathbb{N}, d(x_n, x) \le \dfrac{\epsilon}{2} \ \forall n \ge N$$. Thus, $$d(x_n, x) < \epsilon \ \forall n \ge N$$