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I'm hoping someone can help me with this integration problem I've been struggling with.

Let $\{f_n\}$ be a sequence in $L^2(\mathbb{R})$ such that $\sum_{n=1}^\infty \lVert f_n\rVert^2_2<\infty$ and $\sum_{n=1}^\infty f_n (x)=0$ for a.e. $x\in\mathbb{R}$. I need to show that for every $g\in L^2(\mathbb{R})$, $$\sum_{n=1}^\infty \int_\mathbb{R} f_ng\,d\mu$$ exists and is equal to zero, where $\mu$ is Lebesgue measure.

The assumptions of the problem lead me to believe that some sort of Cauchy-Schwartz Inequality result can be used, maybe along with Dominated Convergence Theorem. However, every attempt I've made at using CSI has led to unwanted square roots.

I was, however, able to prove with Fatou's Lemma that $\sum_{n=1}^\infty |f_n|^2$ is integrable. I'm not sure if it helps, but I thought I'd share it in case it does.

I'd appreciate any help for this problem.

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    $\begingroup$ Are you sure the condition is $\sum_{n=1}^{\infty}\|f_n\|_2^2<\infty$, and not $\sum_{n=1}^{\infty}\|f_n\|_2<\infty$? $\endgroup$
    – triple_sec
    Aug 12, 2015 at 8:32
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    $\begingroup$ I believe that would make the problem nicer. Unfortunately, what I've typed is what I have. I'm not ruling out the possibility that the problem statement has a typo, though. I'd be interested in finding a counterexample of the statement then. $\endgroup$
    – srnoren
    Aug 12, 2015 at 20:14
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    $\begingroup$ I did try providing a proof or a counterexample myself, to no avail yet. I guess your instructor either made a typo or set the bar high for students... $\endgroup$
    – triple_sec
    Aug 13, 2015 at 4:54

1 Answer 1

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The claim is not true (we will even show that it is very false :).

To see this, let $\left(g_{n}\right)_{n\in\mathbb{N}}$ be any sequence in $L^{2}\left(\mu\right)$ which satisfies $\sum_{n=1}^{\infty}g_{n}=0$ pointwise almost everywhere. To avoid trivialities, also assume $\left\Vert g_{n}\right\Vert _{L^{2}}>0$ for all $n$.

Now, for each $n\in\mathbb{N}$ let $$ k_{n}:=\left\lceil n^{2}\cdot\left\Vert g_{n}\right\Vert _{L^{2}}^{2}\right\rceil \in\mathbb{N}, $$ so that we have $$ \sum_{\ell=1}^{k_{n}}\left\Vert \frac{g_{n}}{k_{n}}\right\Vert _{L^{2}}^{2}=\frac{1}{k_{n}^{2}}\cdot\sum_{\ell=1}^{k_{n}}\left\Vert g_{n}\right\Vert _{L^{2}}^{2}=\frac{\left\Vert g_{n}\right\Vert _{L^{2}}^{2}}{k_{n}}\leq\frac{1}{n^{2}}. $$ Finally, define a new sequence $\left(f_{n}\right)_{n\in\mathbb{N}}$ of $L^{2}$ functions so that we have $$ \left(f_{n}\right)_{n}=\left(\underbrace{\frac{g_{1}}{k_{1}},\dots,\frac{g_{1}}{k_{1}}}_{k_{1}\text{ terms}},\frac{g_{2}}{k_{2}},\dots,\frac{g_{2}}{k_{2}},\frac{g_{3}}{k_{3}},\dots,\frac{g_{3}}{k_{3}},\dots\right). $$ It is then not hard to see that we have $$ \sum_{n=1}^{\infty}f_{n}=\sum_{n=1}^{\infty}\sum_{\ell=1}^{k_{n}}\frac{g_{n}}{k_{n}}=\sum_{n=1}^{\infty}g_{n}=0\text{ almost everywhere}. $$ This is surely justified if the series $\sum_{n=1}^{\infty}g_{n}$ converges absolutely almost everywhere, but even if this is not the case, it should hold. For simplicity, let us assume that the convergence is absolute.

By our choice of $k_{n}$, we have (as seen above) $$ \sum_{n=1}^{\infty}\left\Vert f_{n}\right\Vert _{L^{2}}^{2}=\sum_{n=1}^{\infty}\sum_{\ell=1}^{k_{n}}\left\Vert \frac{g_{n}}{k_{n}}\right\Vert _{L^{2}}^{2}\leq\sum_{n=1}^{\infty}\frac{1}{n^{2}}<\infty. $$ If the claim was true, this would imply $$ \int\left(\sum_{n=1}^{N}f_{n}\right)\cdot g\,{\rm d}\mu\xrightarrow[N\to\infty]{}\sum_{n=1}^{\infty}\int f_{n}\cdot g\,{\rm d}\mu $$ for all $g\in L^{2}\left(\mu\right)$, i.e. weak convergence of $F_{N}:=\sum_{n=1}^{N}f_{n}$ to $0$ in $L^{2}\left(\mu\right)$.

But the uniform boundedness principle implies that weakly convergent sequences are bounded (in $L^{2}$), so that we get $$ \left\Vert \sum_{n=1}^{N}f_{n}\right\Vert _{L^{2}}\leq C $$ for all $N\in\mathbb{N}$ and some constant $C>0$. By taking $N=k_{1}+\dots+k_{\ell}$ for arbitrary $\ell\in\mathbb{N}$, this implies in particular $$ \left\Vert \sum_{n=1}^{\ell}g_{n}\right\Vert _{L^{2}}\leq C $$ for all $\ell$ and thus also $$ \left\Vert g_{\ell+1}\right\Vert _{L^{2}}=\left\Vert \left(\sum_{n=1}^{\ell+1}g_{n}\right)-\left(\sum_{n=1}^{\ell}g_{n}\right)\right\Vert _{L^{2}}\leq2C $$ for all $\ell\in\mathbb{N}$.

But it is easy to see that we can choose the $\left(g_{n}\right)_{n\in\mathbb{N}}$ from the beginning with $\left\Vert g_{n}\right\Vert _{L^{2}}\xrightarrow[n\to\infty]{}\infty$, for example $$ g_{2n-1}=\chi_{\left[n,2n\right]}\text{ and }g_{2n}=-g_{2n-1}\text{ for all }n\in\mathbb{N}. $$ Note that the convergence of the (pointwise) series $\sum_{n}g_{n}=0$ is absolute in this case, since only finitely many terms are nonzero for each fixed $x\in\mathbb{R}$.

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  • $\begingroup$ This looks good. I wonder if the assumption is supposed to be $\sum_{n=1}^\infty \lVert{f_n\rVert}_2<\infty$, or if this result had an incorrect proof originally. I might create a new problem on MSE with the other assumption instead. Anyways, thanks for your time and effort! $\endgroup$
    – srnoren
    Aug 15, 2015 at 19:41
  • $\begingroup$ @srnoren: If you assume $\sum \Vert f_n \Vert_2 < \infty$, the proof is very easy. You then know that $\sum_{n=1}^N f_n$ converges for $N \to \infty$ to some $L^2$ function (by completeness, use the Cauchy criterion). Since you also have pointwise convergence, the limit must be zero. $\endgroup$
    – PhoemueX
    Aug 15, 2015 at 20:28

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