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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?

yo1995
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1 Answers1

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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 $s<s_{0}+1$, and $\displaystyle\int_{0}^{\infty}\chi_{(0,s_{0}+1)}(t)|f(t)|dt\leq\left(\int_{0}^{s_{0}+1}|f(t)|^{2}dt\right)^{1/2}(s_{0}+1)^{1/2}<\infty$. Now $\chi_{(0,s)}(t)f(t)\rightarrow\chi_{(0,s_{0})}(t)f(t)$ a.e., so the rest is by Lebesgue Dominated Convergence Theorem.

user284331
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  • Thanks. I have on question. Would it be more clearer if I'd define the sequence $s_n := s - \frac{1}{n}$ instead of working with $s_0+1$.

    Apart from that, I think that proof definetly makes sense and is really helpful.

     – yo1995 Jan 17 '18 at 17:21
  • If you take $s_{n}=s-1/n$, then this is just a sequence which converges to $s$, but by the very definition of continuity, if you want to switch to sequential characterisation, the sequence should be arbitrary, not a fixed sequence like $s_{n}=s-1/n$. – user284331 Jan 17 '18 at 19:09
  • The virtue of setting $s_{0}+1$ is that, for arbitrary sequence ${s_{n}}$ that $s_{n}\rightarrow s_{0}$, then for large $n$, $s_{n}<s_{0}+1$, so the sequence of functions $\chi_{(0,s_{n})}(\cdot)|f(\cdot)|$ can be controlled by the independent integrable function $\chi_{(0,s_{0}+1)}(\cdot)|f(\cdot)|$ of $n$, then Lebesgue Dominated Convergence Theorem goes through. – user284331 Jan 17 '18 at 19:44