This is not my own question, but while answering a part of a question of this link Linear map on $L^{2}$, I failed. Since no one tried to answer this question, I think that it is fair to re-ask a part of the question. If it is not fair, let me know, I will remove my question.
Let $\alpha\in L^\infty[0,1]$ and $M_\alpha:L^2[0,1]\ni f\longmapsto a\cdot f\in L^2[0,1]$. Then, in the above link I have shown in my answer that $||M_\alpha||=||\alpha||_\infty$. Now, as in the link, OP asked a question, which is given below.
Does there exist $f\in L^2[0,1]$ with $||f||_2=1$ such that $||M_\alpha||=\big|\big|M_\alpha(f)\big|\big|_2=||\alpha||_\infty$?
I believe that it is not always true, certainly true if $\alpha$ is simple function, as I have shown in my answer. So, I am looking for a counterexample.
Any help will be appreciated, thanks in advance.
Pick $\alpha(x)=x$. Then we have $\Vert \alpha \Vert_\infty = 1$. Assume that there exists $f\in L^2$ with $\Vert f \Vert_2=1$ such that $1=\Vert M_\alpha \Vert = \Vert M_\alpha (f) \Vert_2$. For $0<\varepsilon <1$ we have $$ 1^2=\Vert M_\alpha(f) \Vert_2^2 = \int_0^1 x^2 f(x)^2 dx = \int_0^{1-\varepsilon} x^2 f(x)^2 dx + \int_{1-\varepsilon}^1 x^2 f(x)^2 dx \leq \int_0^{1-\varepsilon} (1-\varepsilon)^2 f(x)^2 dx + \int_{1-\varepsilon}^1 x^2 f(x)^2 dx \leq \int_0^1 f(x)^2 dx - (2\varepsilon - \varepsilon^2)\int_0^{1-\varepsilon} f(x)^2 dx = 1 - (2\varepsilon - \varepsilon^2)\int_0^{1-\varepsilon} f(x)^2 dx$$ As $2\varepsilon - \varepsilon^2>0$ we get that $$ \int_0^{1-\varepsilon} f(x)^2 dx =0. $$ However, we have that for every $\varepsilon \in (0,1)$, thus, $f=0$. This contradics $\Vert f \Vert_2=1$.
You can take anything that does not have a "plateau on top", i.e. such that $\{ x \in [0,1] \ : \ \alpha(x) = \Vert \alpha \Vert_\infty \}$ is a nullset (the proof is then the same, but we restrict to $\{ x\in [0,1] \ : \ \vert \alpha(x) \vert \leq \Vert \alpha \Vert_\infty - \varepsilon\}$).