Find the norm of the operator $A:L_2[0,2] \rightarrow L_2[0,2]$ defined by $(Ax)(t) = t \, \operatorname{sgn}(t-1) x(t)$ I have operator:
$\boldsymbol{L}_2[0,2]  \to \boldsymbol{L}_2[0,2], ( Ax)( t ) =  \boldsymbol{t} \operatorname{sgn}(t-1)x(t)$
I need to find operator norm or say that operator isn't bounded.
Operator is bounded if $\exists C \left\| Ax \right\|  \leqslant C\left\| x \right\|$
How to find this $C$?
 A: I think the operator $A$ has a bound $C$, $\Vert Ax(t) \Vert \le C \Vert x(t) \Vert$, no greater than $2$, to wit:
$\Vert Ax(t) \Vert^2 = \int_0^2 (t \text{sgn}(t - 1) x(t))^2 dt = \int_0^2 t^2 (x(t))^2 dt \tag{1}$
since $(\text{sgn}(t - 1))^2 = 1$ for $0 \le t \le 2$, except perhaps at $t = 1$, but the measure of the set $\{ 1 \}$ is zero; it won't affect the integrals.  Since $0 \le t^2 \le 4$ on $[0, 2]$, we have
$\int_0^2 t^2 (x(t))^2 dt \le 4 \int_0^2 (x(t))^2 dt = 4\Vert x(t) \Vert^2, \tag{2}$
and combining this with (1) yields
$\Vert Ax(t) \Vert^2 \le 4 \Vert x(t) \Vert^2, \tag{3}$
or
$\Vert Ax(t) \Vert \le 2 \Vert x(t) \Vert; \tag{4}$
which shows we must have the bound $C \le 2$.  Thus the operator norm $\Vert A \Vert$ of $A$ must satisfy $\Vert A \Vert \le 2$, taking $\Vert A \Vert$ to be the infimum of all such bounds.
We can in fact show that $\Vert A \Vert = 2$ by, for any small $\epsilon > 0$, exhibiting a function $y(t) \in L_2[0, 2]$ such that $\Vert Ay(t)\Vert \ge (2 - \epsilon)\Vert y(t) \Vert$.  This may be done as follows: set
$y(t) = 0, 0 \le t < 2 - \epsilon, \tag{5}$
$y(t) = \dfrac{1}{\sqrt{\epsilon}}, 2 - \epsilon \le t \le 2; \tag{6}$
then
$\Vert y(t) \Vert^2 = \int_0^2 (y(t))^2 dt = \int_{2 - \epsilon}^2 \epsilon^{-1} dt = 1, \tag{7}$
so that $\Vert y(t) \Vert = 1$; but
$\Vert Ay(t) \Vert^2 = \int_{2 - \epsilon}^2 \epsilon^{-1} t^2 dt \ge (2 - \epsilon)^2\int_{2 - \epsilon}^2 \epsilon^{-1}dt= (2 - \epsilon)^2 \Vert y(t) \Vert^2, \tag{8}$
that is,
$\Vert Ay(t) \Vert \ge (2 - \epsilon) \Vert y(t) \Vert; \tag{9}$
thus we must have $\Vert A \Vert = 2$.
Hope this helps!  Cheers,
and as always,
Fiat Lux!!!
