# Sup Norm and Uniform Convergence

My book says

Convergence in sup norm $||f_n-f||\to 0$ is equivalent to uniform convergence and this follows immediately from definitions.

but I just want to check:

$\Rightarrow$ If lim$_{n\to\infty}||f_n-f||=0$, then sup$\{|f_n(x)-f(x)|:x\in[a,b]\}\to 0\Rightarrow |f_n(x)-f(x)|\to 0 \forall x\in[a,b]\Rightarrow (f_n)\to f$ uniformly.

And then running in reverse:

$\Leftarrow$ If $f_n\to f$ uniformly, then $|f_n(x)-f(x)|\to 0 \forall x\in[a,b]\Rightarrow$sup$\{|f_n(x)-f(x)|:x\in[a,b]\}\to 0\Rightarrow ||f_n-f||\Rightarrow 0$.

My question is, why $|f_n(x)-f(x)|\to 0 \forall x\in[a,b]\Rightarrow$sup$\{|f_n(x)-f(x)|:x\in[a,b]\}\to 0$. I think it's because sup is really max here because functions are continuous and $[a,b]$ is compact. Is this right? Does it hold in general if the functions aren't continuous?

The implication in "My question is ... $\to 0$" is in fact not true; "for every $x \in [a,b]$ we have $|f_{n}(x) - f(x)| \to 0$" says only the pointwise convergence of $(f_{n})$.

A reason why the two statements is equivalent is the equivalence of the following two statements:

(i) For every $\varepsilon > 0$ there is some $N \geq 1$ such that we have $n \geq N$ only if $|f_{n}(x) - f(x)| \leq \varepsilon$ for all $x \in [a,b]$.

(ii) For every $\varepsilon > 0$ there is some $N \geq 1$ such that we have $n \geq N$ only if $\sup_{x \in [a,b]}|f_{n}(x) - f(x)| \leq \varepsilon$.

• Can I ask if the equality between (i) and (ii) is true in general, or only for continuous functions on compact space? – user153582 Nov 12 '15 at 12:31
• The equivalence is true in general; nothing to do with continuity. :) – Megadeth Nov 12 '15 at 12:34

Well that implication doesn't hold at all. In fact the statement that $$|f_n(x)-f(x)|\rightarrow 0\quad \forall x\in[a,b]$$ is not equivalent to saying that $f_n\rightarrow f$ uniformly. In fact it is the statement that $f_n\rightarrow f$ pointwise, which is weaker.

Uniform convergence means that for every $\varepsilon$ we can find an $N$ such that for all $n > N$ we have $|f_n(x)-f(x)|<\varepsilon$ for all $x$.

The difference with pointwise convergence being that for pointwise we can choose a different $N$ for every $x$, but for uniform convergence we must find an $N$ that works for all $x$ at once.

Consider the set of all bounded functions from $$[a,b]$$ to $$\mathbb{R}$$ which we denote by $$B([a,b],\mathbb{R})$$. Defining a norm for this set as the usual Sup norm

$$\left\|f\right\|=\text{Sup}\{|f(x)|:x\in[a,b]\}$$

and considering its induced metric

$$d(f,g)=\text{Sup}\{|f(x)-g(x)|:x\in[a,b]\}$$

one can show that this will be a metric space (check this as an exercise). Now, we claim that convergence in this metric space is equivalent to uniform convergence on $$[a,b]$$. To check this out, let us write the definitions as follows

\begin{align*} &\text{Convergent in B([a,b],\mathbb{R})} \iff \\ &\forall\epsilon_1\gt0,\,\,\exists N_1\gt0,\,\,n\ge N_1 \implies \text{Sup}\{|f_n(x)-f(x)|,x\in[a,b]\}<\epsilon_1 \\ \\ &\text{Uniformly Convergent on [a,b]} \iff \\ &\forall\epsilon_2\gt0,\,\,\exists N_2\gt0,\,\, \forall x\in [a,b],\,\, n\ge N_2 \implies |f_n(x)-f(x)|<\epsilon_2. \end{align*}

First assume convergence in $$B([a,b],\mathbb{R})$$. Consequently, $$\epsilon_2$$ is given and we should find $$N_2$$ such that the desired result holds. Choose $$\epsilon_1:=\epsilon_2$$ and take $$N_2:=N_1$$. Then according to

$$|f_n(x)-f(x)|\le\text{Sup}\{|f_n(x)-f(x)|,x\in[a,b]\}<\epsilon_2,\qquad \forall x\in [a,b]$$

uniform convergence on $$[a,b]$$ follows immediately. Next, assume uniform convergence on $$[a,b]$$. So $$\epsilon_1$$ is given and we should find $$N_1$$ such that the result holds. Choose $$\epsilon_2:=\frac{\epsilon_1}{2}$$ and take $$N_1:=N_2$$. For fixed $$n$$, $$\frac{\epsilon_1}{2}$$ is an upper bound for $$\{|f_n(x)-f(x)|\,\big|x\in[a,b]\}$$ and we have

$$\text{Sup}\{|f_n(x)-f(x)|,x\in[a,b]\}\le\frac{\epsilon_1}{2}\lt\epsilon_1$$

which completes the proof.