prove that there is no $\varphi \in C([0,1], \mathbb{R})$ such that $||\varphi|| = |L(\varphi)|$

Question

Define the functional $L:C([0,1], \mathbb{R}) \rightarrow \mathbb{R}$ by $L(\varphi) = \displaystyle\int_0^{1/2}\varphi(t)dt - \displaystyle\int_{1/2}^{1}\varphi(t)dt$. we consider in $C([0,1], \mathbb{R})$ the supremum norm. I have showed that $||L|| = 1$, however I couldn't show that there is no $\varphi \in C([0,1], \mathbb{R})$ such that $||\varphi|| = |L(\varphi)|$.

This exercise was proposed before the Hanh-Banach Theorem, so I don't know if you need this Theorem. Does anyone have any idea how to do?

Answer 1

Assume $\varphi$ such that $\|\varphi\|_{\infty}=1$. Then, we see that $$ \left|\int_0^{\frac{1}{2}} \varphi(t)\textrm{d}t\right|\leq \frac{1}{2} $$ with equality if and only if $\varphi$ is constant and $|\varphi|\equiv 1$. Similarly for the other integral. Since $\varphi$ is continuous, if $\|\varphi\|_{\infty}=1$ and $|L(\varphi)|=1$ we get that there exists $e^{i\theta}\in S^1$ such that $\varphi(t)=e^{i\theta}$.

Then, however, we get that $L(\varphi)=0,$ which is a contradiction.

Answer 2

The only $L^\infty$ function with $|L(\varphi)|=||\varphi||$ is $$ \varphi_1(x)= \begin{cases} ||\varphi_1||,& 0\leq x \leq 1/2 \\ -||\varphi_1||,& 1/2<x\leq 1 \end{cases} $$ almost everywhere (and all constant multiples). Obviously, this function is not continuous.