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For which one $p \ge 1$ is with $$A(x_n)_{n=1}^{\infty}=\left(\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right)_{m=1}^{\infty}$$ defined bounded linear operator $A:\ell^p \to \ell^p$? Find norm $\| A\|.$

First, I have to show that if $x = (x_n)_{n=1}^{\infty} \in \ell^p$ then $\displaystyle y=(y_m)_{m=1}^{\infty} \stackrel{\text{def}}{=}\left(\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right)_{m=1}^{\infty} \in \ell^p$, that is $\displaystyle\sqrt[p]{\sum_{m=1}^{\infty}|y_m|^p} < +\infty.$

All right, $$\begin{align*} |y_m|=\left|\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right|&\leq\frac{1}{m} \sum_{n=1}^{m} \frac{|x_n|}{\sqrt{n}}\\ &\stackrel{\star}{\leq}\frac{1}{m} \sqrt[p]{\sum_{n=1}^{m}|x_n|^p} \sqrt[q]{\sum_{n=1}^{m} n^{-q/2}} \\ &\leq \frac{\| x \|_p}{m}\sqrt[q]{\sum_{n=1}^{m} n^{-q/2}}.\end{align*}$$

So $$\begin{align*} \sum_{m=1}^{\infty} |y_m|^p &\leq \sum_{m=1}^{\infty} \left|\frac{\| x \|_p}{m}\sqrt[q]{\sum_{n=1}^{m} n^{-q/2}}\right|^p\\&=\sum_{m=1}^{\infty} \frac{1}{m^p} \| x\|^p_p \left(\sum_{n=1}^{m} n^{-q/2}\right)^{p/q}\\ &\leq \|x\|^p_p \sum_{m=1}^{\infty} \frac{1}{m^p} \left(\sum_{n=1}^{\infty} n^{-q/2}\right)^{p/q} \end{align*}$$

If I see well, last sum is convergent for $q/2 >1 \iff q>2$ and $p>1$ (but with condition $1/p + 1/q =1$).

I think my estimate is too sharp.

And we find that $$\| A \| \leq \sqrt[p]{\sum_{m=1}^{\infty} \frac{1}{m^p} \left(\sum_{n=1}^{\infty} n^{-q/2}\right)^{p/q}}$$

But for part (if my approximation is good enough) with $\geq$ I don't have idea (to be honest, I used Hölder inequality, so we want $x_n = c \frac{1}{\sqrt{n}}$, that will give us something divergent I think so).

$\star$ - one doubt also: I used here Hölder inequality, but that works only for $p,q >1$, but in my question we have $p \geq 1$, so basically I have to work case $p=1$ as separate?

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  • $\begingroup$ I think Hölder inequality works for all $1\leq p,q\leq +\infty$ such that $\frac{1}{p}+\frac{1}{q}=1$. $\endgroup$
    – Neph
    May 31 '13 at 22:02
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As mention by robjohn, only the case $p>1$ allows $A$ to be bounded.

In your last estimate, you changed $m$ to $\infty$, and there was no reason for that step, as you already had an upper bound.

One way to get a lower bound for the norm of $A$ is as follows. Fix $k\in\mathbb N$. Let $$ x_n=n^{-q/2p}. $$ Then, since $-q/2p-1/2=-q/2$, \begin{align} \|Ax\|_p^p&=\sum_{m=1}^\infty|(Ax)_m|^p =\sum_{m=1}^\infty \left|\frac1m\,\sum_{n=1}^m\frac{n^{-q/2p}}{n^{1/2} }\right|^p\\ \ \\ &=\sum_{m=1}^\infty \left(\frac1m\,\sum_{n=1}^m{n^{-q/2}}\right)^p\\ \ \\ &=\sum_{m=1}^\infty \frac1{m^p}\left[\left(\sum_{n=1}^m{|n^{-q/2p}|^p}\right)^{1/p}\left(\sum_{n=1}^m{n^{-q/2}}\right)^{1/q}\right]^p\\ \ \\ &=\|x\|_p^p\,\sum_{m=1}^\infty\frac1{m^p}\left(\sum_{n=1}^m{n^{-q/2}}\right)^{p/q}. \end{align} So $$ \|A\|\geq \left(\sum_{m=1}^\infty\frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}\right)^{1/p}, $$ as you already had the upper bound.

The series converges for any $p>1$. Indeed, for $m$ large enough we have $$ \sum_{n=1}^mn^{-q/2}\leq 1+\int_1^mt^{-q/2}=1+\frac{m^{1-q/2}-1}{1-q/2}\leq c\,m^{1/2} $$ for some $c>0$. So $$ \frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}\leq c^{1/q}\,m^{p/2q-p} =c^{1/q}\,m^{-p/2-1/2}. $$ Since $p>1$, the series $\sum_{m=1}^\infty\frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}$ converges.

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  • $\begingroup$ Thank you Martin for this. Could you just put a line or two of explanation for $\| Ax \|_p^p \geq \ldots$, because I am having trouble to understand it well. $\endgroup$
    – Cortizol
    Jun 1 '13 at 10:24
  • $\begingroup$ I have added a few lines. I don't know if that is what you were looking for. $\endgroup$ Jun 1 '13 at 14:45
  • $\begingroup$ Do we need here $\sum_{m=1}^k \left| \frac 1m \sum_{n=1}^m \frac {n^{-q/2}}{\left(\sum_{n=1}^kn^{-q/2} \right)^{1/p}} \right|^p = \sum_{m=1}^k \frac 1{m^p} \, \left( \sum_{n=1}^m n^{-q/2} \right)^{p-1}$ sign $\leq$ instead $=$, because, if I see good, $m \leq k$, so $\sum_{n=1}^k n^{-q/2} \geq \sum_{n=1}^m n^{-q/2}$. And that might be problem. I am probably wrong, I am very tired. $\endgroup$
    – Cortizol
    Jun 1 '13 at 15:40
  • $\begingroup$ I am sorry, but something is wrong with Comment, and I can't edit him because it's too long $\endgroup$
    – Cortizol
    Jun 1 '13 at 15:43
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    $\begingroup$ @ArthurFischer Thank you very much (at least, I voted for you, so I have some use of it :-)) $\endgroup$
    – Cortizol
    Jun 1 '13 at 16:14
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Brute force and Hölder give us, with $\frac1p+\frac1q=1$, $$ \begin{align} \sum_{m=1}^\infty\left|\frac1m\sum_{n=1}^m\frac{x_n}{\sqrt{n}}\right|^p &\le\sum_{m=1}^\infty\frac1{m^p}\left(\sum_{n=1}^m\frac{|x_n|}{\sqrt{n}}\right)^p\\ &\le\sum_{m=1}^\infty\frac1{m^p}\|x_n\|_p^p\left(\sum_{n=1}^m\frac1{n^{q/2}}\right)^{p/q}\\ &\le\|x_n\|_p^p\sum_{m=1}^\infty\frac1{m^p}\left(2\sqrt{m}\right)^{p-1}\tag{$q\ge1$}\\ &=2^{p-1}\|x_n\|_p^p\zeta\left(\frac{p+1}{2}\right) \end{align} $$ Thus, $A$ is bounded for $p\gt1$ with norm no greater than $2^{1-1/p}\zeta\left(\frac{p+1}{2}\right)^{1/p}\le2\,\zeta\left(\frac{p+1}{2}\right)$.

For $(x_n)=(1,0,0,\dots)\in\ell^1$, $$ \frac1m\sum_{n=1}^m\frac{x_n}{\sqrt{n}}=\frac1m $$ Thus, we cannot have that $A$ is bounded on $\ell^1$.

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  • $\begingroup$ How did you find that $\displaystyle\sum_{n=1}^{m} n^{-q/2} \leq 2 \sqrt{m}$ (if I see good, because $p/q=p-1$, so you didn't change that). It's probably something obvious, but I can't see it. And thanks for example that $A$ is unbounded on $\ell^1$. $\endgroup$
    – Cortizol
    Jun 1 '13 at 7:47
  • $\begingroup$ @Cortizol: since $q\ge1$, we have that $$ \sum_{n=1}^m n^{-q/2}\le\sum_{n=1}^m n^{-1/2}\le\int_0^m\frac{\mathrm{d}x}{\sqrt{x}}=2\sqrt{m} $$ $\endgroup$
    – robjohn
    Jun 1 '13 at 11:27
  • $\begingroup$ Now everything is clear. Thank you Rob for your help. I accepted Martin answer because he found sequence $(x_n)$ such that $\| A \| \geq \ldots$, I hope that is all right with you. $\endgroup$
    – Cortizol
    Jun 1 '13 at 16:36

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