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This question was asked in my assignment of Functional analysis and I was not able to solve this particular problem.

Question:(a) Show that the formula $Af(x) = \frac{1} {x} \int_{0}^x f(y) dy$ defines a bounded operator $A: L^2 (( 0,\infty)) \to L^2 ((0,\infty))$ with $||A||\leq 2$.

(b)Show that in fact $||A||=2.$

(c) Determine the adjoint operator of A.

(d) Show that $A^*A= A+A^* =AA^* .$ Deduce that the operator $U= I-A $ is unitary.

Attempt: (a) $ ||A|| \leq || \frac{1} {X} \int^x_0 f(y) dy ||\leq \frac{1} {X} \int_{0}^{x}||f(y)|| dy$ but I am not able to show that $||A||\leq 2$ because I don't understand how to find the minimum value of integral of $f(y)$.

(b) Due to this I am also not able to show how ||A||=2.

(c) Using the definition , I got that $A^* = {x} ({\int^{x}_{0} f(y) dy})^{-1}$ but using this definition I am not able to show that $AA^* = A+ A^*$.

Can you please tell is my $A^*$ wrong and how to rightly calculate it?

  • First rework your answer to (a) but before that, write down the definitions of the norm of an operator and of the $L^2$-norm of an $f.$ – Anne Bauval Oct 29 '22 at 13:13
  • @3ibfwcbi: you already got an answer. But note that the expression you wrote for $A^*$ is not even linear. – Martin Argerami Oct 30 '22 at 14:32

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The bound of the Hardy operator $Hf(x)=\frac1x\int^x_0f(t)\,dt$ has been discussed in several posting in MSE, for example posting 1 and posting 2

As for the adjoint: for $f,g\in L_2(0,\infty)$, an application of Fubini's theorem yields

\begin{align} \int^\infty_0\Big(\frac1x\int^x_0f(y)\,dy\Big)\overline{g}(x)\,dx&=\int^\infty_0\Big(\frac1x\int^\infty_0\mathbb{1}_{(0,x]}(y)f(y)\,dy\Big)\overline{g}(x)\,dx\\ &=\int^\infty_0f(y)\int^\infty_0\mathbb{1}_{[y,\infty)}(x)\frac{\overline{g}(x)}{x}\,dx\Big)\,dy\\ &=\int^\infty_0f(y)\Big(\int^\infty_y\frac{\overline{g}(x)}{x}\,dx\Big)\,dy \end{align} Hence, $$A^*g(x)=\int^\infty_x\frac{g(y)}{y}\,dy$$

For part (c), another application of Funibi's theorem yields \begin{align} AA^*f(x)&=\frac1x\int^x_0A^*f(y)\,dy=\frac1x\int^x_0\Big(\int^\infty_y\frac{f(t)}{t}\,dt\Big)\,dy\\ &=\frac1x\left(\int^x_0\frac{f(t)}{t}\Big(\int^t_0\, dy\Big)\, dt +\int^\infty_x\frac{f(t)}{t}\Big(\int^x_0\, dy\Big)\,dt\right)\\ &=Af(x) + A^*f(x) \end{align} Hr=er th region of integration $R:=\{(y, t):0<y\leq x, y\leq t\}$ is expressed as $R:=\{(y, t): 0<t\leq y, 0<y\leq t\}\cup\{(y, t): x<t,0<y\leq x\}$.

Similarly \begin{align} A^*Af(x)&=\int^\infty_x\frac{Af(y)}{y}\,dy=\int^\infty_x\frac1{y^2}\Big(\int^y_0 f(t)\,dt\Big)\,dy\\ &=\int^x_0f(t)\Big(\int^\infty_x \frac{dy}{y^2}\big)\,dt+\int^\infty_xf(t)\Big(\int^\infty_t\frac{dt}{y^2}\Big)\,dt\\ &=Af(x)+A^*f(x) \end{align} Here, the region of integration $Q:=\{(y, t): x<y, 0<t\leq y\}$ is split into $\{(y, t): x<y, 0<t\leq x\}\cup\{(y, t): x<t, t<y\}$.

Part (d) of the OP follows from (c).

Mittens
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  • Diaz Solutions of (c) and (d) I completely understood. Can you please give solution of (a) and (b) also? I am not able to understand them using the links mentioned . I am very new in this topic. –  Oct 30 '22 at 13:31
  • @3ibfwcbi: The first one is not complicated if one uses the generalized Minkowski inequality. On of the postings does that. The other part is more delicate. The second posting I refer to has several good solutions which are very interesting on their own. I suggest you take a look at them. – Mittens Oct 30 '22 at 14:20
  • I looked at them but I couldn't understand. Kindly help if you have some time to spare. –  Oct 31 '22 at 01:38
  • Are you familiar with the generalized Minkowski inequality? If not, here is a different proof of the Hardy inequality (boundedness of $A$) that is based on (a) showing the statement for continuous functions of compact support then, and then by density arguments to extend result for all $f\in L_p$ (in your case $p=2$. That $p/(p-1)=2$ is the best constant, is more involved and I don't have the energy to rewrite my previous solution. The idea is to get a sequence of functions $f_n$ with $|f_n|_2=1$ and such that $|Af_n|_2\xrightarrow{n\rightarrow\infty}\frac{p}{p-1}=2$ (for your case $p=2$) – Mittens Oct 31 '22 at 01:52
  • See this solution that does not use Minkowski's inequality – Mittens Oct 31 '22 at 01:53
  • @OlivierDiaz How should I prove the equality of part (b)? I have done (a), thanks to your references! –  Oct 31 '22 at 12:52
  • @3ibfwcbi: part b is requires a little finesse. In my previous message I told you what the idea was, posting 2 in my answer has a couple of solutions. – Mittens Oct 31 '22 at 14:21