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How can I show the following known fact (fact not known to me)

$$\text{let}\quad\{y_n\}_{n=1}^{\infty} \in l^2\quad\text{then}\quad\left\{ \frac{1}{n}\sum_{j=1}^{n} y_j\right\}_{n=1}^{\infty} \in l^2\,?$$

I tried Cauchy-Schwarz but that does not look like a right way to start

EDIT: I noticed that in a place where this fact is used is $y_n \ge 0$ (maybe its easier to show for that case). But $y_n \ge 0$ is not mentioned when they mentioned fact from above (I doubt that they forgot to say and $y_n$ must be $\ge 0$).

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  • $\begingroup$ if you want to show $\{ \frac{1}{n}\sum_{j=1}^{n} y_j\}_{n=1}^{\infty} \in l^2$ is bounded, just use triangle inequality for $l_2$ if $\{y_n\}_{n=1}^{\infty}$ is bounded. $\endgroup$ – runaround Feb 19 '16 at 15:47
  • $\begingroup$ I need $\sum_{n=1}^{\infty}( \frac{1}{n}\sum_{j=1}^{n} y_j )^2 < \infty$ $\endgroup$ – jack Feb 19 '16 at 15:50
  • $\begingroup$ Have you tried to use completeness of $\ell^2$? $\endgroup$ – noctusraid Feb 19 '16 at 17:01
  • $\begingroup$ No, what exactly should I try to do ? $\endgroup$ – jack Feb 19 '16 at 17:05
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This follows from Hardy's inequality. Looking at its proof, you see that Cauchy-Schwarz IS the right way to go, applied to $$ \sum_{j=1}^n y_j = \sum_{j=1}^n j^{-1/4} j^{1/4}y_j. $$

To give intuition why you first write the sum like this: Cauchy-Schwarz is sharp when two functions are multiples of each other. If you apply it simply to $fg$, with $f(j)=1, g(j)=y_j$ then they are less and less multiples of each others as $n\to \infty$ because $y_j$ converges to zero and $1$ does not. With the choice above, $f(j)=j^{-1/4}$, $g(j)=j^{1/4}y_j$, both are appropriately multiples, at least at infinity. This is of course very handwaving, but if you consider $y_j=j^{-1/2}$, which is at the very edge of being (not) in $\ell^2$, then $f$ and $g$ above are actually the same

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    $\begingroup$ In the Q,it is not necessary for $y_n\geq 0$ . $\endgroup$ – DanielWainfleet Feb 19 '16 at 20:44
  • $\begingroup$ @user254665 How does this work when $y_1,y_2,\ldots$ are not necessarily non-negative? $\endgroup$ – Cm7F7Bb Feb 19 '16 at 22:45
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    $\begingroup$ @user254665 when $y_n$ are not all nonnegative, note that $(1/n(\sum y_n))^2\leq (1/n \sum |y_n|)^2$ and that $\{|y_n|\}\in \ell^2$ $\endgroup$ – Bananach Feb 20 '16 at 5:21
  • $\begingroup$ @Bananach (I checked statement of inequality and I got everything I need) but about edit and proof, I looked at proof (found it in book Inequalities from Hardy) and its a long one. but he used some other inequality (it is C.S. for p=2 which is my case) but I dont see anything about $j^{-1/4}$. And I tought maybe I am not looking at same proof as the one you mentioned. $\endgroup$ – jack Feb 20 '16 at 19:04
  • $\begingroup$ math.stackexchange.com/questions/99728/… $\endgroup$ – Bananach Feb 21 '16 at 5:00
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The general statement of Hardy's inequality is stated as:

Suppose $1<p<\infty$, $f\in L^p\left((0,\infty)\right)$, and $F : (0,\infty) \to \mathbb{C}$ is defined by $$ F(x) = \frac{1}{x} \int_0^x fdm. $$ Then $$ \|F\|_p \leq \frac{p}{p-1} \|f\|_p. $$

To prove it, assume first that $f \geq 0$ and $f\in C_c\left((0,\infty)\right)$. Integration by parts gives $$ \int_0^\infty F^p(x) dx = -p \int_0^\infty F^{p-1}(x)xF'(x)dx. $$ Not that $xF'(x) = f(x) - F(x)$. Apply Hölder's inequality to $\int F^{p-1}fdm$ and rearrange: $$ (p-1)\int_0^\infty F^pdm \leq p \left(\int_0^\infty F^p dm\right)^{1-1/p} \left(\int_0^\infty f^pdm\right)^{1/p}, $$ which is $$ \|F\|_p \leq \frac{p}{p-1} \|f\|_p. $$

It is not difficult to generalize this to all $f\in C_c\left((0,\infty)\right)$, noting that $$ |F(x)| \leq \frac{1}{x}\int_0^x |f|dm. $$

Now suppose $f \in L^p\left((0,\infty)\right)$, and choose a sequence $\{f_n\}$ in $C_c\left((0,\infty)\right)$ such that $\|f_n-f\|_p \to 0$ as $n \to \infty$, and $\{f_n(x)\}$ converges to $f(x)$ for a.e $x$. This is possible because $C_c\left((0,\infty)\right)$ is dense in $L^p\left((0,\infty)\right)$, and because every $L^p$ convergent sequence has a pointwise convergent subsequence. By Fatou's Lemma and Jensen's inequality, \begin{align} |F(x)-F_n(x)|^p &\leq \frac{1}{x^p} \int_0^x \lim_{j\to \infty}|f_j-f_n|^pdm \\ &\leq \frac{1}{x^p} \liminf_{j\to \infty} \int_0^x |f_j-f_n|^pdm \\ &\leq \frac{1}{x^p} \liminf_{j\to \infty} \int_0^\infty |f_j-f_n|^pdm \\ &= \frac{1}{x^p} \|f-f_n\|_p^p \end{align} for all $x\in (0,\infty)$. Hence $F$ is the pointwise limit of $\{F_n\}$. Again, by Fatou's lemma, \begin{align} \int_0^\infty |F-F_n|^pdm &\leq \liminf_{j \to \infty}\int_0^\infty |F_j-F_n|^pdm \\ &\leq \liminf_{j\to \infty} \left(\frac{p}{p-1}\|f_j-f_n\|_p\right)^p \\ &= \left(\frac{p}{p-1}\|f-f_n\|_p\right)^p \end{align} Therefore, $\{F_n\}$ converges to $F$ in $L^p\left((0,\infty)\right)$, and we have $$ \|F\|_p = \lim_{n\to \infty} \|F_n\|_p \leq \lim_{n\to \infty} \frac{p}{p-1} \|f_n\|_p = \frac{p}{p-1} \|f\|_p. $$

To obtain the series inequality for $\{a_n\} \in \ell^2$, first assume that $|a_{n+1}| \geq |a_n|$. Then it follows from the integral inequality that $$ \sum_{N=1}^{\infty}\left(\frac{1}{N}\sum_{n=1}^N |a_n|\right)^p \leq \left(\frac{p}{p-1}\right)^p \sum_{n=1}^\infty |a_n|^p. $$ In the general case, since $|a_n|\to 0$ as $n\to \infty$, we can arrange the sequence so that $|a_n| \geq |a_{n+1}|$. Absolute convergence ensures that such rearrangements would not change the series.

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