# 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 $C$?

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Have you tried computing $\|Ax\|_2$? – EPS May 13 '14 at 5:33
@Sam It's $\sqrt{2} |t \operatorname{sgn}(t-1)x(t)|$ – michaeluskov May 13 '14 at 5:51
Where did you get the $\sqrt 2$? Also, the $L_2$ norm is given by $\Vert x(t) \Vert^2 = \int_0^2 (x(t))^2 dt$; it doesn't explicitly depend on $t$. – Robert Lewis May 13 '14 at 6:00

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!!!

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Thanks! But how to find a norm (supremum of C)? – michaeluskov May 13 '14 at 6:45
@user23791: I guess you saw my edits! You're most welcome, and thanks for the "acceptance"! – Robert Lewis May 13 '14 at 7:02