4

I am trying to prove that the given operator is bounded in the $L_2[0, 1]$ space: $$ (Ax)(t) = t^{r-1} \int\limits^{t}_{0} \frac{x(s)}{s^r} ds $$

The way I'm trying to do it is via a Hölder's inequality, Fubini's theorem and also using Hardy's inequality for integrals. I know that $r$ must be less than $\frac{1}{2}$ in order for the operator above to be bounded. However, I am unable to prove that. Can anyone help me? Thank you.

wxist
  • 481

2 Answers2

4

$$\frac{1}{t}\int^t_0\big(\tfrac{t}{s}\big)^rx(s)\,ds=\int^1_0\frac{x(ut)}{u^r}\,du$$

Then, using Minkowski's inequality (the generalized version rather) \begin{align} \Big(\int^1_0\Big|\int^1_0\frac{x(ut)}{u^r} \,du\Big|^2\,dt\Big)^{1/2}&\leq \int^1_0\frac{1}{u^r}\Big(\int^1_0|x(ut)|^2\,dt\Big)^{1/2}\,du\\ &=\int^1_0\frac{1}{u^r}\frac{1}{u^{1/2}}\Big(\int^u_0|x(t)|^2\,dt\Big)^{1/2}\,du\\ &\leq \|x\|_2\int^1_0\frac{1}{u^{r+1/2}}\,du \end{align}

For the last integral to converge, it is necessary that $r+\tfrac12<1$.

Mittens
  • 39,145
  • As I understand it, to prove that the operator is unbounded for $r < \frac{1}{2}$, I need to give an example of a function on which the operator is unbounded, I tried to give some standard functions, but nothing happened, could not you help with that? – wxist Nov 12 '22 at 07:55
  • Try $x(t)=\frac{1}{t^p}$ with $2p<1$. See what happens to $Ax$. The operator may not even be define din all off $L – Mittens Nov 12 '22 at 13:18
  • I made a mistake in the comment above, you proved that operat is bounded for $r < \frac{1}{2}$. To prove that the operator is unbounded for $r \geq \frac{1}{2} $, I need to give an example of a function, I tried to consider the functions $x(t) = t^p$, but nothing happened, could you help? – wxist Nov 12 '22 at 13:31
  • @wxist: I just wrote a rather long comment in the answer section to address the case$r=1/2$. I left the other cases to you. – Mittens Nov 12 '22 at 14:35
1

This is to address a comment of the OP which is not part of the original posting.

Denote $A_r$ the operator $A_rx(t)=\int^1_0\frac{x(tu)}{u^r}\,du$ on $L_2(0,1)$.

Suppose $r=1/2$. If $x_p(s)=\frac{1}{s^p}$ with $p<1/2$, then $\|x\|_2=\frac{1}{(1-2p)^{1/2}}<\infty$. Let $X_p(t)=(1-2p)^{1/2}x_p(t)$. Then, $$A_{1/2}X_p(t):=\frac{\sqrt{1-2p}}{t^p}\int^1_0\frac{1}{u^{p+1/2}}\,du=\frac{1}{t^p}\frac{2}{\sqrt{1-2p}}=\frac{2}{1-2p}X_p(t)$$ Hence $$\|A_{1/2}X_p\|_2=\frac{2}{1-2p}$$

Notice that $\|X_p\|_2=1$ for all $p<1/2$, and that $\lim_{p\rightarrow1/2-}\|AX_p\|_2=\infty$, that is $A_{1/2}$ is unbounded.

Similar examples can be used to show that $A_r$ is unbounded of $r>1/2$.

Mittens
  • 39,145