functional analysis - Show that $ Tf $ is continuous and measurable on a Hilbert space $H=L_2((0,infty))$

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 $

Show that each $f \epsilon H $ is continuous and measurable.

Can someone help me how I can show this?

Answer

The following argument is for the continuity of the linear operator $T$, I misunderstood the question.
\begin{align*}

\|Tf\|_{L^{2}}&=\left(\int_{0}^{\infty}\dfrac{1}{s^{2}}\left|\int_{0}^{s}f(t)dt\right|^{2}ds\right)^{1/2}\\
&\leq 2\left(\int_{0}^{\infty}|f(t)|^{2}dt\right)^{1/2}\\
&=2\|f\|_{L^{2}},
\end{align*}
so $\|T\|_{L^{2}\rightarrow L^{2}}\leq 2<\infty$, hence $T$ is bounded and hence continuous. Note that Hardy's inequality is used for the inequality. Look up this one for a proof.

The following argument is for the continuity of the function $Tf$.

Consider a fixed $s_{0}\in(0,\infty)$, then $\displaystyle\int_{0}^{s}f(t)dt=\int_{0}^{\infty}\chi_{(0,s)}(t)f(t)dt$, then $\chi_{(0,s)}(t)|f(t)|\leq\chi_{(0,s_{0}+1)}(t)|f(t)|$ for all $s