a) Compute $T(1_{[\alpha,\beta]})$ for all $0<\alpha <\beta<0$

Question

Let H be a Hilbert space $H=L_2(0,\infty), dt) $ with $dt$ the Lebegues measure.

For each $ f \epsilon \ H $, define the function $ Tf:(0,\infty)\rightarrow \mathbb{C} $ by $(Tf) (s) = \frac{1}{s}\int_{(0,s)}f(t) dt , s>0 $

a) Compute $T(1_{[\alpha,\beta]})$ for all $0<\alpha <\beta<0$

b) Show that $\|T(1_{[\alpha,\beta]}\|_2 \ge \frac{\beta-\alpha}{\sqrt\beta}$

My work:

a) We have to differ between three cases

Case (i) $s \in [\alpha, \beta]$

$T(1_{[\alpha,\beta]}) = \frac{s- \alpha}{s} $

Case (ii) $s < \alpha$ $T(1_{[\alpha,\beta]}) = 0 $

Case (iii) $s > \beta$ $T(1_{[\alpha,\beta]}) = \frac{\beta- \alpha}{s} $

b) $\|T(1_{[\alpha,\beta]}\|_2 =(\int_{(0,\infty)}(\frac{1}{s}\int_{(0,s)}f(t) dt)^2 ds)^\frac{1}{2} $ And with this part I'm stuck.

For case ii) i have a contradiction. since I'm integrating over zero. For case i) I have $ (\int_{(0,\infty}(\frac{s- \alpha}{s})^2)^{\frac{1}{2}}$ which has the value $\infty$ Similar for Case iii) Where is my mistake?

I just also found here that someone has the same Problem. However I don't understand how they found their solution?

Any help is much appreciated.

Answer 1

You don't need to find the full $L^2$ norm here. Just note

$$\|T(1_{[\alpha,\beta]})\|_2 > \left (\int_b^\infty \left(\frac{\beta-\alpha}{s}\right)^2\, ds\right)^{1/2} =\frac{\beta-\alpha}{\sqrt \beta}.$$