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Let $u$ be a measurable function in $[0,1]$ and define $T:L^p(0,1)\to L^p(0,1)$ by $Tu(x)=\frac{1}{x}\int_0^x u(t)dt\quad\forall x \in [0,1]$. Let $1<p<\infty$. Prove that $T$ is bounded, non-compact. Determine the spectral radius of $T$ and prove that in the case $p=2$ the operator $TT^*-T^*T$ has range $1$. Can anyone help me? Thank you.

mechanodroid
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user62138
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1 Answers1

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Here is the proof that $T$ is bounded:

Hardy's Inequality for Integrals

Here is the exact calculation of its norm:

Computing the best constant in classical Hardy's inequality

To find its spectral radius, use the formula $\text{radius}(T) = \lim\limits_{n\to\infty}\|T^n\|^{1/n}$

To compute $T^n$ and $T^*$, look at http://faculty.missouri.edu/~stephen/preprints/hardy.html

Here is compactness https://math.stackexchange.com/questions/262221/is-hf-1-over-x-int-0x-ftdt-compact?rq=1

Probably this will get marked as a duplicate, but I don't see anywhere spectral radius was asked before.

Another way to find a lower bound for the spectral radius is to consider $u(t) = t^{-1/r}$ for $r>p$. This will give eigenfunctions.

Stephen Montgomery-Smith
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    You beat me to this! Ok, i'll just add that $$T^(u)(x)=\int_x^1 t^{-1}u(t)dt$$$$(TT^-T^*T)(u)=\left(\int_0^1 u(t)dt\right) \chi_{(0,1)}$$ – Norbert Nov 30 '13 at 20:49
  • I had the advantage that I had it all written up beforehand! – Stephen Montgomery-Smith Nov 30 '13 at 20:51
  • So what is the final formula for $T^n$ ? I can't understand from that paper. – Norbert Nov 30 '13 at 20:54
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    It's on page 5. In this special case $$ T^{n+1} f(x) =\frac1x \int_0^x \frac{(\log(x/t))^n}{n!} u(t) , dt $$ – Stephen Montgomery-Smith Nov 30 '13 at 21:01
  • And to find a lower bound on $|T^{n+1}|$, assume $u(t)$ is decreasing (because you only need to consider specific functions to find the lower bounds). Then $$T^{n+1}f(x) \ge \frac1x \int_0^{ax} \frac{(\log(x/t)^n}{n!} u(t) dt \ge \frac1a \frac{|\log(a)|^n}{n!} u(ax) $$ for $0<a<1$. Then use $|u(a\cdot)|_p = a^{-1/p} |u|_p$. Choose $a$ and $n$ suitably to show the spectral radius of $T$ is bounded below by $|T|$. – Stephen Montgomery-Smith Nov 30 '13 at 21:29
  • Maybe there is an easier way, though. – Stephen Montgomery-Smith Nov 30 '13 at 21:31
  • I think I found the easier way and I added it to the end of the solution. – Stephen Montgomery-Smith Nov 30 '13 at 22:50
  • is there a problem writing $\int^{x}_{1} \frac {g (t)}{t} $ ? thats what I get as adjont – user123124 Oct 25 '15 at 18:02
  • @User2313 That is exactly the negative of what I obtained. Maybe you made a sign error somewhere, because I am rather certain I am correct. – Stephen Montgomery-Smith Oct 25 '15 at 18:08
  • @StephenMontgomery-Smith Im sure you are I just cant get where I go wrong, flipping $\int^{x}{0} f = - \int^{x}{1} f$ right? – user123124 Oct 25 '15 at 18:18
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    @User2313 No, $\int_1^x = - \int_x^1$. – Stephen Montgomery-Smith Oct 25 '15 at 18:40
  • @Stephen Montgomery-Smith I cant see how this appear..we are integrating g(t)/t from zero to x in the partial integration how do we get to start from 1? – user123124 Nov 17 '15 at 16:24
  • @User2313 $\langle g,Tf\rangle = \int_{y=0}^1 g(y) \frac1y \int_{x=0}^y f(x) , dx , dy = \int_{x=0}^1 f(x) \int_{y=x}^1 \frac{g(y)} y , dy , dx = \langle T^* g,f \rangle$. – Stephen Montgomery-Smith Nov 17 '15 at 16:30
  • @Stephen Montgomery-Smith donst that suggest we define a primitve for g(t)/t beyond 1? – user123124 Nov 17 '15 at 18:58
  • @User2313 How can $g(t)/t$ have a primitive when $t>1$? We don't even have that $g(t)$ is defined there, since $g:[0,1] \to \mathbb C$. – Stephen Montgomery-Smith Nov 17 '15 at 19:10
  • @StephenMontgomery-Smith I was doing integration by parts, considering $\int_{0}^{y} f(y) dy$ as one function and $g(t)/t$ as the other, integrating the latter but from $0$ to $x$, but this donst add up to what you have here. It looks like you have been integrating $g(t)/t$ from $x$ to $1$, then made the flipp suggested a few comments up. – user123124 Nov 17 '15 at 19:36
  • Correction it look like you been integrating $g(t)/t$ from $1$ to $x$ – user123124 Nov 17 '15 at 19:43
  • @User2313 But I meant for $0 \le x \le 1$, and I used $\int_a^b = -\int_b^a$. – Stephen Montgomery-Smith Nov 17 '15 at 19:48
  • @StephenMontgomery-Smith are we defining a primitve from $1$ to $x$ for $x \le 1$ , is that possible? – user123124 Nov 17 '15 at 20:08
  • @User2313 Yes. It's just a matter of notation. – Stephen Montgomery-Smith Nov 17 '15 at 20:38
  • @StephenMontgomery-Smith right, this is kinda of topic but could one define a primivite for $f$ being any continous function from an arbitrary point $a$ as $F(x)=\int_a^x f(y) dy$ ? ofc this would only be the primitve for $f$ on this partricular intevall. It seems reasonable..but still kind of wierd. – user123124 Nov 18 '15 at 08:26
  • A primitive for $f$ would be any function of the form you gave plus a constant. Making $a$ arbitrary isn't quite enough, e.g. consider the case $f = 0$. – Stephen Montgomery-Smith Nov 18 '15 at 22:12
  • @StephenMontgomery-Smith how can one just choose this primitive to be $0$ at $1$ ? $g$ is predetermined right? We need $g(1)=0$ I think, otherwise we get boundary terms. And since g is predet. we cant just choose – user123124 Dec 13 '15 at 10:10
  • @User1 You can choose any primitive you like, as long as you are consistent (use the same one throughout). – Stephen Montgomery-Smith Dec 13 '15 at 16:36
  • @StephenMontgomery-Smith let $g$ be cts(given that we later want to extend the adjont to $L^{p}$ later) and let $G(x)= g(1)+ \int_{1}^{x} g(t) $. Im sorry for being persistant but at what point do we have the freedom to choose here? Are you suggesting we take the constant $C=-g(1)$? I thought the "constant of integraton" was the number $g(1)$ – user123124 Dec 13 '15 at 18:34
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    I don't see what your problem is. – Stephen Montgomery-Smith Dec 13 '15 at 18:58
  • @StephenMontgomery-Smith http://math.stackexchange.com/questions/156376/definite-integral-and-constant-of-integration , there it was :) – user123124 Dec 13 '15 at 19:55
  • I'm afraid that the compactness link is broken? – TrostAft Oct 19 '20 at 14:31
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    @TrostAft I fixed the $T^n$ and $T^*$ link. – Stephen Montgomery-Smith Oct 19 '20 at 23:15
  • @StephenMontgomery-Smith It's good that that one got fixed too, but I'm referring to the link after "Here is compactness [...]". I get redirected to a Page not Found screen, on which it states that the question had been removed for reasons of moderation. Perhaps it was removed as a duplicate? – TrostAft Oct 21 '20 at 04:33
  • @TrostAft Yes, I don't have any control over that. If you can find a better reference, please let me know. – Stephen Montgomery-Smith Oct 21 '20 at 05:54