Let $\{\phi_n\}$ be an orthonormal set in $L^2(\mu)$, $g\in L^2(\mu)$ such that $|\phi_n(x)|\le |g(x)|$ and $\sum a_n\phi_n(x)$ converges a.e.

Question

Let $\mu$ be a positive measure on measurable space $(X,\mathfrak{M})$. Let $\{\phi_n\}$ be an orthonormal set in $L^2(\mu)$ and $g\in L^2(\mu)$ such that $$|\phi_n(x)|\le |g(x)|\text{ for a.e. }x\ \forall n$$ and $\sum\limits_{n=1}^\infty a_n\phi_n(x)$ converges a.e., then prove that $\lim a_n=0$

I define $f(x)=\sum\limits_{n=1}^\infty a_n\phi_n(x)$. I want to prove that $f\in L^2(\mu)$. If I canprove so, as $\{\phi_n\}$ is orthonormal set, $\lVert f\rVert_2=\sum|a_n|^2<\infty$, therefore $\lim a_n=0$.

By Minkowski integral inequality, we have-

$\lVert f\rVert_2$

$=\left[\int|\sum a_n\phi_n(x)|^2\ dx\right]^{1/2}$

$\le \sum\left[\int|a_n\phi_n(x)|^2 dx\right]^{1/2}$

$\le \sum|a_n|\left[\int|g(x)|^2\ dx\right]^{1/2}$

$=\sum |a_n|\lVert g\rVert_2$

But this says nothing. Can anyone suggest me a wayout to finish the proof? Thanks for your help in advance.

Answer 1

The series $\sum_{n=1}^\infty a_n\phi_n(x)$ converges (a.e.) so that $\lim_{n\rightarrow\infty}\left\vert a_n\phi_n\left(x\right)\right\vert=0$ almost everywhere.

If the sequence $\left\{a_{n}\right\}$ is unbounded there is a subsequence $\left\{a_{n_k}\right\}$ such that $1\leq\left\vert a_{n_k}\right\vert$ for all $k$. Then $\left\vert\phi_{n_k}\left(x\right)\right\vert \leq \left\vert a_{n_k}\phi_{n_k}\left(x\right)\right\vert\rightarrow0$ so that $\left\vert\phi_{n_k}\right\vert\rightarrow0$ almost everywhere in this case and $$ 1 = \lim_{k\rightarrow\infty}\left(\int_X\left\vert\phi_{n_k}\right\vert^2d\mu\right)^{1/2} = \left(\int_X\left(\lim_{k\rightarrow\infty}\left\vert\phi_{n_k}\right\vert\right)^2d\mu\right)^{1/2} = 0 $$ by the Dominated Converge Theorem applied to the sequence $\left\{\phi_{n_k}\right\}$. A contradiction.

Therefore $\left\{a_{n}\right\}$ is bounded, say by $M$. In this case $\left\vert a_n \phi_n \right\vert\leq \left\vert M g \right\vert$ so that $$ \begin{align} \lim_{n\rightarrow\infty}\left\vert a_n \right\vert &= \lim_{n\rightarrow\infty}\left\vert a_n \right\vert\left(\int_X\left\vert\phi_n\right\vert^2d\mu\right)^{1/2} \\ &= \lim_{n\rightarrow\infty}\left(\int_X\left\vert a_n \phi_n\right\vert^2d\mu\right)^{1/2} \\ &= \left(\int_X\left(\lim_{n\rightarrow\infty}\left\vert a_n \phi_n\right\vert\right)^2d\mu\right)^{1/2} \\ &=0 \end{align} $$ again by the Dominated Converge Theorem.

Hence $a_n\rightarrow0$.

Answer 2

Petronella's answer says it all but personally I would like to address more on the case that when $\{a_{n}\}$ is bounded:

Take an arbitrary subsequence $\{a_{n_{k}}\}$ of $\{a_{n}\}$. Since $\{a_{n_{k}}\}$ is bounded, a standard analysis fact leads to a convergence (further) subsequence $\{a_{n_{k_{l}}}\}$ of $\{a_{n_{k}}\}$. Just as before, we have by Lebesgue's theorem that \begin{align*} \lim_{l}a_{n_{k_{l}}}=\left(\int|\lim_{l}a_{n_{k_{l}}}\phi_{n_{k_{l}}}(x)|^{2}\right)^{\frac{1}{2}}=0. \end{align*} Therefore, we conclude that, every subsequence of $\{a_{n}\}$ has a further subsequence which converges to zero, a standard analysis fact then yields $a_{n}\rightarrow 0$.