# $\Omega \subset C([0,1])$, $A:= \sup \{ \int ^1_0 |f(x)|dx: f \in \Omega\}$. Prove $\exists$ a $f_0 \in \Omega$ such that $A= \int^1_0 |f_0(x)|dx$.

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 $$\sup_{x \in [0,1]}|f_i(x)|. Set $$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 $$\sup_{x \in [0,1]}|f_i(x)|. In other words, $$|g(x)-f_i(x)|+|f_i(x)| for all $$x \in [0,1]$$. By triangle inequality, $$|g(x)| for all $$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.

## 1 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.

• What is $L^1[0,1]$ and $L^{\infty}[0,1]$ Commented Mar 21, 2023 at 5:51
• In this case, $\|f\|_{L^{1}[0,1]}=\int_{0}^{1}|f(x)|dx$ and $\|f\|_{L^{\infty}[0,1]}=\sup_{x\in[0,1]}|f(x)|$. Commented Mar 21, 2023 at 5:52
• @MichaelLooper What all you need is $|T(f)-T(g)| \leq \sup \{|f(x)-g(x)|:0\leq x \leq 1\}$. Commented Mar 21, 2023 at 5:58
• @user284331 Two questions: (1) you are letting $\delta = \epsilon$ if we fill in the proof of continuity? (2)And you are actually showing uniform continuity here? Commented Mar 21, 2023 at 6:41
• Yes, it is in fact uniform continuous. Commented Mar 21, 2023 at 6:46