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.


1 Answer 1


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.

  • $\begingroup$ What is $L^1[0,1]$ and $L^{\infty}[0,1]$ $\endgroup$
    – Mike L
    Commented Mar 21, 2023 at 5:51
  • $\begingroup$ In this case, $\|f\|_{L^{1}[0,1]}=\int_{0}^{1}|f(x)|dx$ and $\|f\|_{L^{\infty}[0,1]}=\sup_{x\in[0,1]}|f(x)|$. $\endgroup$
    – user284331
    Commented Mar 21, 2023 at 5:52
  • $\begingroup$ @MichaelLooper What all you need is $|T(f)-T(g)| \leq \sup \{|f(x)-g(x)|:0\leq x \leq 1\}$. $\endgroup$ Commented Mar 21, 2023 at 5:58
  • $\begingroup$ @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? $\endgroup$
    – Mike L
    Commented Mar 21, 2023 at 6:41
  • $\begingroup$ Yes, it is in fact uniform continuous. $\endgroup$
    – user284331
    Commented Mar 21, 2023 at 6:46

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