Series of $L^2$ functions converges pointwise almost everywhere

Question

Let $\{f_n\}_{n=1}^\infty$ be a sequence of functions $f_n : \mathbb{R}^d \to \mathbb{C}$ in $L^2(\mathbb{R}^d)$ such that, for all $N \in \mathbb{N}$

$$\sum_{n = 1}^N \| f_n \|_{L^2}^2 \le C,$$

where $C$ is some positive constant independent of $N$. Furthermore, suppose there is some function $F \in L^2(\mathbb{R}^d)$ such that

$$\| F - \sum_{n=1}^N f_n \|_{L^2}^2 \to 0$$

I would like to know if it is true that $\sum_{n=1}^\infty f_n$ converges pointwise almost everywhere to the function $F$.

If we are dealing with functions in $L^1(\mathbb{R}^d)$ rather than $L^2(\mathbb{R}^d)$, then the answer to this question would be yes, and seen in this question here. However, in the $L^2$ case we are seemingly missing a step. We can of course start by applying monotone covergence:

$$\int_{\mathbb{R}^n} \sum_{n=1}^\infty |f_n|^2 dx = \sum_{n =1}^\infty \int_{\mathbb{R}^n} |f_n|^2 dx \le C.$$

Hence $\sum_{n =1}^\infty |f_n|^2$ converges to a finite number almost everywhere. But this does not then imply that $\sum_{n=1}^\infty f_n$ converges almost everywhere (and we would have this implication in the $L^1$ case).

Does the proof break down from here, or is there a way around this difficulty? Any hints or solutions are greatly appreciated.

Answer 1

Fix a positive $R$. Then we have by Cauchy-Schwarz inequality $\lVert f_n\rVert_2^2\geqslant\int_{[-R,R]^d} \left|f_n(x)\right|^2\mathrm dx \geqslant\int_{\left[-R,R\right]^d} \left|f_n(x)\right|\mathrm dx /\lambda_d\left(\left[-R,R\right]^d\right)$. Therefore, defining $g_n(x):=f_n(x)\mathbf 1_{\left[-R,R\right]^d}(x)$, we have:

1. a constant $C'$ depending on $R$ such that $\sum_{n=1}^{ N}\left\lVert g_n\right\rVert_1\leqslant C'$ for each $N$;
2. the convergence $$\lim_{N\to +\infty} \left\lVert F\mathbf 1_{\left[-R,R\right]^d} -\sum_{n=1}^Ng_n \right\rVert_1=0.$$

Consequently, using the result in the linked thread, we deduce that $\left(\sum_{n=1}^Nf_n -F\right)\mathbf 1_{\left[-R,R\right]^d}\to 0$ almost everywhere. Since $R$ is arbitrary and a countabe union of sets of measurable zero still have measure $0$, it follows that $\sum_{n=1}^Nf_n \to F$ almost everywhere.