Sup Norm and Uniform Convergence

Question

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?

Answer 1

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 $\sup_{x \in [a,b]}|f_{n}(x) - f(x)| \leq \varepsilon$.

Answer 2

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.

Answer 3

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

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

and considering its induced metric

$$d(f,g)=\left\|f-g\right\|_{\infty}=\text{sup}\{|f(x)-g(x)|:x\in[a,b]\}$$

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. For convergence in $B([a,b],\mathbb{R})$ we have

\begin{equation} \forall\epsilon_1\gt0,\,\,\exists N_1\gt0,\,\,n\ge N_1 \implies \left\|f_n-f\right\|_{\infty}<\epsilon_1, \end{equation}

which can be rewritten as follows

\begin{equation} \bbox[10px,border:2px solid red]{ \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.} \tag{1} \end{equation}

For uniform convergence on $[a, b]$, let us start with the definition of pointwise convergence on $[a, b]$ which reads as

\begin{equation} \forall x\in [a,b], \,\, \forall\epsilon_2\gt0,\,\,\exists N_2\gt0,\,\, n\ge N_2 \implies |f_n(x)-f(x)|<\epsilon_2. \end{equation}

Note that the choice of $N_2$ depends on $x$. To convert this into uniform convergence, we desire some $N_2$ that works for every $x\in[a, b]$ and hence the name uniform. Consequently, the previous definition converts to

\begin{equation} \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{equation}

which is logically equivalent to

\begin{equation} \bbox[10px,border:2px solid red]{ \forall\epsilon_2\gt0,\,\,\exists N_2\gt0,\,\, n\ge N_2 \implies (\forall x\in [a,b], \,\, |f_n(x)-f(x)|<\epsilon_2),} \tag{2} \end{equation}

where we used the logical fact that $\forall x, \big( p \implies q(x)\big)$ is equivalent to $p \implies (\forall x, \,\, q(x))$, where the delicate point is that $p$ does not depend on $x$. Now that we have understood the definitions precisely, the proof is easy to carry out. Just keep (1) and (2) in mind.

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

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

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 $\{|f_n(x)-f(x)|: x \in[a,b]\}$ and due to the completeness of $\mathbb{R}$ this set must have a supremum, which is smaller than all of its upper bounds. This leads us to

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

which completes the proof.