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

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


and considering its induced metric


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


which completes the proof.


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