Norm of the operator $A\colon L^2[0,1] \to L^2[0,1]$, $x(t) \mapsto (t-0.5)\cdot x(t)$ I'm trying to calculate a norm of the operator $$A\colon L^2[0,1] \to L^2[0,1],\qquad  x(t) \mapsto (t-0.5)\cdot x(t).$$
I started finding it as follows:
$$\|Ax(t)\|^2 = \int_{0}^1 |(t-0.5)\cdot x(t)|^2\,dt$$
Then I've tried to apply Cauchy-Schwarz inequality, but I can't get $\int_{0}^1 |x(t)|^2\,dt$ (that would be equal to $\|x\|^2$) in my expression, so I can't proceed further.
P.S. In addition to norm itself, I have to specify some value or sequence on which the norm is reached
Please, tell me about the right way to find $\|A\|$ here
 A: The operator you are describing falls in the class of what is known as "multiplication operators".
Let $(X,\mu)$ be a finite measure space (in your case $X=[0,1]$ and the measure is just the regular Lebesgue measure). Define a map $M:L^\infty(X)\to\mathbb{B}(L^2(X))$ by taking a function $\phi\in L^\infty(X)$ and mapping it to the operator $M_\phi$ defined by $M_\phi(f):=\phi\cdot f$. It can be very easily shown that $M$ is an injective $*$-homomorphism and thus isometric. If you are unfamiliar with elementary $C^*$-theory, you can prove that $\|M_\phi\|=\|\phi\|_\infty$ by an elementary argument; I will provide the details for this elementary argument only if OP asks for them.
Now in your case your operator $A$ is the multiplication operator $M_\phi$ for the map $\phi:[0,1]\to\mathbb{C}$, $\phi(t)=t-0.5$. So $$\|A\|=\sup_{t\in[0,1]}|t-0.5|=0.5$$
Edit: As OP requested, we prove that $\|M_\phi\|=\|\phi\|$. First note that, if $f\in L^2(X)$ with $\|f\|=1$, then $\|M_\phi(f)\|^2=\int_X|f(x)|^2|\phi(x)|^2d\mu(x)\leq\|\phi\|_\infty^2\cdot\int_X|f(x)|^2d\mu(x)=\|\phi\|_\infty^2$. This shows that $\|M_\phi\|\leq\|\phi\|_\infty$. Now we show the other inequality. Let $\varepsilon>0$. We find a measurable set $E\subset X$ with $\mu(E)>0$ such that $\phi(x)>\|\phi\|_\infty-\varepsilon$ for all $x\in E$, this is by the way that the essential supremum norm is defined. Now for $f=1_E$, the indicator function (which satisfies $\|f\|^2=\int_X|f(x)|^2d\mu(x)=\mu(E)>0$ gives us
$$\|M_\phi(f)\|^2=\int_X1_E\cdot|\phi(x)|^2d\mu=\int_E|\phi(x)|^2d\mu(x)\geq\mu(E)\cdot(\|\phi\|_\infty-\varepsilon)^2=$$ $$=\|f\|^2\cdot(\|\phi\|_\infty-\varepsilon)^2 $$
so
$$\|M_\phi\|\ge\frac{\|M_\phi(f)\|}{\|f\|}\ge\|\phi\|_\infty-\varepsilon$$
and letting $\varepsilon\to0^+$ yields $\|M_\phi\|\ge\|\phi\|_\infty$.
2nd Edit: I just noticed that OP also asks for a function or sequence of functions that give us the norm of the operator. The details of my previous edit actually provide us with a way to find such a sequence of functions: the supremum norm of $\phi(t)=t-0.5$ is equal to $0.5$, so, for $\varepsilon>0$ very small we consider the function $f_{\varepsilon,0}$ that is the indicator function of the interval $[1-\varepsilon,1]$. These have norm $\int_0^1f_{\varepsilon,0}(t)^2dt=\varepsilon$, so we normalize them: we consider $f_\varepsilon=\frac{f_{\varepsilon,0}}{\sqrt{\varepsilon}}$. These functions have norm 1 and note that
$$\|M_\phi(f_\varepsilon)\|^2=\frac{1}{\varepsilon}\int_{1-\varepsilon}^{1}(t-0.5)^2dt=\frac{1}{3\varepsilon}[(t-0.5)^3]_{1-\varepsilon}^{1}=$$ $$=\frac{0.5^3-(0.5-\varepsilon)^3}{3\varepsilon}=\frac{3\cdot0.25\cdot\varepsilon-3\varepsilon^2\cdot0.5+\varepsilon^3}{3\varepsilon}=0.25-0.5\varepsilon-\frac{\varepsilon^2}{3}=$$ $$=\|M_\phi\|^2-0.5\varepsilon-\frac{\varepsilon^2}{3} $$
So by taking $\varepsilon_n=\frac{1}{n}$, the corresponding functions give you a sequence $(f_n)$ of norm 1 functions in $L^2$ so that $\|Af_n\|\to\|A\|$.
