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Look at $(\mathbb{R},\mathcal{B}(\mathbb{R}),\frac{1}{2\pi}e^{-x^2/2}d\lambda(x)$), where $\lambda$ denotes the Lebesgue measure.

The $L^2$-space of this measure space is a Hilbert-space, and consists of those real functions such that: $\int_\mathbb{R}f^2(x)\frac{1}{2\pi}e^{-x^2/2}d\lambda(x)<\infty$.

It can be shown that there exists an orthonormal sequence of functions in this space that is also a basis, call the basis-elements $H_i(x), i=0,2,3...$.

Now assume that $f$ is in this Hilbert-space, that is assume that $\int_\mathbb{R}f^2(x)\frac{1}{2\pi}e^{-x^2/2}d\lambda(x)<\infty$. Then we have that $f(x)=\sum\limits_{i=0}^\infty c_iH_i(x)$. Where $c_i=\int_\mathbb{R}f(x)H_i(x)\frac{1}{2\pi}e^{-x^2/2}d\lambda(x)$.

What I am having problems with is the last part $f(x)=\sum\limits_{i=0}^\infty c_iH_i(x)$. This convergence is in the space above, that is we have that $\int_R [f(x)-\sum\limits_{i=0}^nc_iH_i(x)]^2\frac{1}{2\pi}e^{-x^2/2}d\lambda(x)$ goes to 0 as n goes to infinity.

But what can we say about the pointwise convergence of $\sum\limits_{i=0}^\infty c_iH_i(x)$ as a sequence of real functions? It can be shown that $\sum\limits_{i=0}^\infty c_i^2<\infty$, this follow from the Hilbert-space theory. But what about the sequence $\sum\limits_{i=0}^\infty| c_iH_i(x)|$, does it converge for almost every $x$?, and if so, how can I show it?

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    I would not expect absolute convergence a.e. without additional assumptions on $f$. There is a closely related MO question; [this answer](http://mathoverflow.net/a/145235) may be useful. – 2017-01-09
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    Perhaps I'm missing something, but for instance if you replace your measure space with $[0,1)$ and Lebesgue measure. Then for $H_k=e^{2 \pi i k} $, $k\in \mathbb{Z} $, the question becomes the pointwise convergence of Fourier series, a notoriously hard question in real analysis. So I don't expect an easy answer, but again I may be overlooking something here. – 2017-01-09
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    Why do you want the pointwise convergence ? Many elements in $L^2$ are not well-defined pointwise, for example $f(x) = x^{-1/4} \in L^2([0,1])$ and $g(x) = \sum_{k=1}^\infty \sum_{m=0}^{k} \frac{1}{2^{2k}} f(x-m/k) \in L^2([0,1])$ but $g(x)$ is not well-defined at any $x \in \mathbb{Q} \cap [0,1]$ – 2017-01-09
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    @user1952009 The reason I need it is this: The basis-elements are the Hermite polynomials, and we have that $H_0=1$, and if $G$ is a standard normal variable, we have that $E[H_i(G)]=0, i >1$. The author then says that $c_0=E[f(G)]$, and the reasoning for the last equation is that in the equation $f(x)=\sum\limits_{i=0}^\infty c_iH_i(x)$, we put $G$ in for $x$ and take expecation. But I do not quite how he can just take the expectation, we need to use the dominated convergence theorem or something?, and that is why I need some sort of pointwise convergence. – 2017-01-09
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    If you know that $f \in L^2(\frac{e^{-x^2/2}}{\sqrt{2\pi}})$ then $E[f(G)] = \int_{-\infty}^\infty f(x)\frac{e^{-x^2/2}}{\sqrt{2\pi}}dx = \langle f,H_0 \rangle$ which is well-defined, and for any sequence $f_n \to f$ in $L^2(\frac{e^{-x^2/2}}{\sqrt{2\pi}})$ then $\langle f_n,H_0 \rangle \to \langle f,H_0 \rangle $ (this is the Cauchy-Schwarz inequality : $f \mapsto \langle f,g \rangle$ is a bounded operator if $f,g$ are in the Hilbert space) – 2017-01-09
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    @user1952009 Thank you very much! So you have shown what is needed(the value of $c_0$). But did the author then make a mistake by taking the expectation of $\sum\limits_{i=0}^\infty c_i H_i(G)$, we do not know if this is meaningful? – 2017-01-09
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    You can prove pointwise convergence under classical Fourier conditions such as where there are left- and right- limits and first derivatives. Showing pointwise a.e. convergence is difficult enough for the ordinary Fourier series that I doubt this has been shown for the Hermite Fourier series. – 2017-01-09

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