Question: Let $\Omega$ be a compact subset of the function space $C([0,1])$, the set of continuous functions on $[0,1]$.
Define $A:= \sup \{ \int ^1_0 |f(x)|dx: f \in \Omega\}$.
Prove $\exists$ a function $f_0 \in \Omega$ such that $A= \int^1_0 |f_0(x)|dx$.
Recall that the norm in $C([0,1])$ is
$||f-g||=\sup\{|f(x)-g(x)|: x \in [0,1]\}$.
I am stuck. How do I prove this? Note that this exercise is from an introductory real analysis over metric spaces, not functional analyis.
What I have tried:
After quite some time, I have only been able to show that $\{ \int ^1_0 |f(x)|dx: f \in \Omega\}$ is bounded and therefor $A$ is finite. Let me know if you spot any errors.
Consider $\bigcup_{f \in \Omega}\{g \in\Omega: ||f-g||<1\}$.
It is clear that this is an open cover of $\Omega$.
Because $\Omega$ is compact, there are finitely many
$f_1,..,f_k$ such that $\Omega \subset \{g \in\Omega: ||f_1-g||<1\} \cup…\cup\{g \in\Omega: ||f_k-g||<1\}$.
Now, each $f_i$ is continuous on $[0,1]$, so by the Extreme Value Theorem, $f_i$ is bounded on $[0,1]$. So $\exists M_i \in \mathbb R$ such that $\sup_{x \in [0,1]}|f_i(x)|<M_i$.
Set $M=\max\{M_1,..,M_k\}$.
Let $g \in \Omega$. then, $\exists f_i$ such that $\sup_{x \in [0,1]}|f_i(x)-g(x)|<1$.
We also have that $\sup_{x \in [0,1]}|f_i(x)|<M$.
In other words,
$|g(x)-f_i(x)|+|f_i(x)|<M+1$ for all $x \in [0,1]$.
By triangle inequality,
$|g(x)|<M+1$ for all $x \in [0,1]$.
And so, $\int^1_0 |g(x)|dx < \int^1_0(M+1)dx=M+1$.
This shows that $\{ \int ^1_0 |f(x)|dx: f \in \Omega\}$ is bounded above and therefore $A$ is a finite real number.
Best Answer
Let $T:\Omega\rightarrow\mathbb{R}$ be defined by $T(f)=\|f\|_{L^{1}[0,1]}$. We claim that $T$ is continuous. Indeed, we have \begin{align*} |T(f)-T(g)|\leq\|f-g\|_{L^{1}[0,1]}\leq\|f-g\|_{L^{\infty}[0,1]}. \end{align*} Now $T$ is continuous on a compact set, and hence it attains its maximum.