real analysis - Let $f:[a,b]toBbb{R}$ be continuous. Does $max{|f(x)|:aleq xleq b}$ exist?

Let $f:[a,b]\to\Bbb{R}$ be continuous. Does

$$\begin{align}\max\{|f(x)|:a\leq x\leq b\} \end{align}$$ exist?

MY WORK

I believe it does and I want to prove it.

Since $f:[a,b]\to\Bbb{R}$ is continuous, then $f$ is uniformly continuous. Let $\epsilon> 0$ be given, then $\exists\, \delta>$ such that $\forall x,y\in [a,b]$ with $|x-y|<\delta,$ it implies $|f(x)-f(y)|<\epsilon.$

Then, for $a\leq x\leq b,$

$$\begin{align} f(x)=f(b)+[f(x)-f(b)]\end{align}$$
$$\begin{align} |f(x)|\leq |f(b)|+|f(x)-f(b)|\end{align}$$
$$\begin{align} \max\limits_{a\leq x\leq b}|f(x)|\leq |f(b)|+\max\limits_{a\leq x\leq b}|f(x)-f(b)|\end{align}$$

I am stuck at this point. Please, can anyone show me how to continue from here?

Answer

Because $[a,b]$ is compact, every sequence in $[a,b]$ has a subsequence that limits to a point in $[a,b]$. Pick a sequence $x_n \in [a,b]$ such that $\lim_{n \rightarrow \infty} |f(x_n)| = \sup_{[a,b]}|f|$. Now get a convergent subsequence $x_{n_k}$ that converges to some $x \in [a,b]$.

By continuity of $|f|$,

$|f(x)| = |f( \lim_{k\rightarrow \infty}x_{n_k})| = \lim_{k\rightarrow \infty} |f(x_{n_k})| = \lim_{n \rightarrow \infty} |f(x_n)| = \sup_{[a,b]}|f|$.