Linear map on $L^{2}$ For $a\in L^{\infty}([0,1],\mathbb{K})$, we define
$$
M_{a}:L^{2}([0,1],\mathbb{K})\mapsto L^2([0,1],\mathbb{K})
$$
by
$$M_a(f)=x\to a(x)f(x)$$
($\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$)

*

*Show that it is a linear map and that it is continuous.

*Show that $\Vert M\Vert_{\mathcal{L}(L^{2}([0,1],\mathbb{K}))}=\Vert a\Vert_{L^\infty([0,1],\mathbb{K})}$

*Look for a necessary and sufficient condition on which there exists a function $f\in L^2([0,1],\mathbb{K})$ with $\Vert f\Vert_{L^2([0,1],\mathbb{K})}=1$ such that
$$
\Vert M_af\Vert_{L^2([0,1],\mathbb{K})}=\Vert a\Vert_{L^\infty([0,1],\mathbb{K})}
$$
First, take $a(x)=x$, then generalize.
My work:

*

*[Same proof for $a(x)=x$] $M_a$ is well defined as if $f\in L^2([0,1],\mathbb{K})$, we have
$$\int_{[0,1]}\vert a(x)f(x)\vert^2dx\leqslant \int_{[0,1]}\Vert a\Vert_{L^\infty}^2\vert f(x)\vert^2dx=\Vert a\Vert_{L^\infty}^2\Vert f\Vert_{L^2}^2<\infty$$
It is linear as the product of two numbers of $\mathbb{K}$ is a bilinear map on $\mathbb{K}\times\mathbb{K}$.

Let $f\in L^2([0,1],\mathbb{K})$ such that $\Vert f\Vert_{L^2}=1$. The inequality above show that
$$\int_{[0,1]}\vert M_a(f)(x)\vert^2dx\leqslant \Vert a\Vert_{L^\infty}^2$$
so $M$ is continuous and we have
$$\Vert M\Vert_{\mathcal{L}(L^{2}([0,1],\mathbb{K}))}\leqslant\Vert a\Vert_{L^\infty([0,1],\mathbb{K})}$$


*For $a(x)=x$, let $\varepsilon>0$ and let $f\in L^2([0,1],\mathbb{K})$ such that $\Vert f\Vert_{L^2}=1$ and $Supp(F)\subset[1-\varepsilon,1]$. Then we have
$$
\begin{align*}
\Vert M_a\Vert
&\geqslant \Vert M_af\Vert_{L^2}\\
&=\int_{[1-\varepsilon,1]}\vert xf(x)\vert^2dx\\
&\geqslant 1-2\varepsilon+\varepsilon^2
\end{align*}
$$
When $\varepsilon\to0$, we have $\Vert M\Vert=1$.

In the general case, can we also say that for a fixed $\varepsilon>0$, there exists $\delta>0$ and $x_0\in[0,1]$ such that for almost all $x$ in $B(x_0,\delta)$, we have $\vert a(x)\vert\geqslant\Vert a\Vert_{L^\infty}-\varepsilon$ ? If that is the case, then the same proof as above also works.


*We need to make sure that we get $f(x)a(x)=\Vert a\Vert_{L^\infty}$ for x such that $a(x)\neq 0$, and to compensate for the loss when $a(x)=0$. A natural way to do it is to define $f(x)=\frac{\Vert a\Vert_{L^\infty}}{a(x)\sqrt{\lambda({a(x)\neq 0})}}$ if $a(x)\neq 0$ and 0 elsewhere (where $\lambda$ denotes the Lebesgue measure).
However, we get that for this particular $f$
$$\Vert f\Vert_{L^2}=\frac{\Vert a\Vert_{L^\infty}}{\lambda({a(x)\neq 0})}\int \frac{1}{a(x)^2}dx$$
I do not know how to proceed next.

Edit: Question number 3) is still there to be done.
 A: Answer of $(2)$ In this answer, I will use $L^2,L^\infty$ for simplicity instead of $L^2\big([0,1],\Bbb K\big)$ and $L^\infty\big([0,1],\Bbb K\big)$. Also, these are sets of equivalence classes, but I will not distinguish in between equivalence classes and their representatives, as any two representatives are almost everywhere equals.
From the above, we have $$\int|\alpha\cdot f|^2\leq||\alpha||_\infty^2\int|f|^2,\text{ hence }||M_\alpha||\leq||\alpha||_\infty.$$ To prove the reverse direction, assume at first $\alpha$ is simple function with $\alpha\not\equiv 0$ and consider the measurable set $E=\alpha^{-1}(c)$, where $|c|=||\alpha||_\infty$. Since, $\alpha$ is non-trivial simple function we have $m(E)>0$, as $|c|$ is the maximum value of the simple function $|\alpha|$. Now, letting $$f=\frac{\chi_E}{\sqrt{m(E)}}\text{ we have }||f||_2=1 \text{ and }$$$$||M_\alpha(f)||_2=\frac{1}{\sqrt{m(E)}}\bigg(\int_E|\alpha|^2\bigg)^{1/2}=\frac{1}{\sqrt{m(E)}}\bigg(m(E)||\alpha||_\infty^2\bigg)^{1/2}=||\alpha||_\infty.$$
So, we have $||M_\alpha||=||\alpha||_\infty$, in this case. But, $\alpha\equiv 0$ implies $||M_\alpha||=0=||\alpha||_\infty$, trivially. So, we are done for simple functions.
Now, consider arbitrary essentially bounded function $\psi$, and since simple functions are dense in $L^\infty$ we have a sequence of simple functions $\{\varphi_n\}$ with $\lim||\psi-\varphi_n||_\infty=0$. Now, $$||M_\psi f-M_{\varphi_n}f||_2=||M_{\psi-\varphi_n}f||_2\leq ||\psi-\varphi_n||_\infty||f||_2\text{ for all }f\in L^2$$$$\implies \lim||M_\psi -M_{\varphi_n}||\leq\lim||\psi-\varphi_n||_\infty=0 $$$$\implies ||M_\psi||=\lim ||M_{\varphi_n}||=\lim||\varphi_n||_\infty=||\psi||_\infty.$$
So, we are done.
