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