Computing the Fourier transform of $H_k(x)e^{-x^2/2}$, where $H_k$ is the Hermite polynomial. [Notations] The definition of Fourier transform of a $L^1$ function $f$ is given by the formula $\int f(x)e^{-ix\cdot\xi}dx$, with no normalizing factors; similarly for the Fourier-Plancherel transform on $L^2$. The Hermite polynomial of order $k$ is given by $H_k(x)=(-1)^ke^{x^2}\left(\frac{d}{dx}\right)^k(e^{-x^2})$. This is  "physicists' Hermite polynomials", I believe. I will use $\mathscr{F}$ to denote the Fourier transform.

I'm trying to understand the proof of the following theorem.

For any $k=0,1,2,\ldots,$ $$\mathscr{F}(H_k(x)e^{-x^2/2})=\sqrt{2\pi}(-i)^kH_k(\xi)e^{-\xi^2/2}.$$

The proof starts by showing that for any fixed $t\in\mathbb{C}$, the series $\sum H_k(x)e^{-x^2/2}\frac{t^k}{k!}$ is (absolutely) convergent in $L^2$ with its limit being $e^{-x^2/2+2xt-t^2}$. This requires the generating-function form of the Hermite polynomials, namely $\sum H_k(x)\frac{t^k}{k!}=e^{2xt-t^2}$, and some orthogonality relations. Then, by continuity of $\mathscr{F}$ on $L^2$, we obtain for any fixed $t\in\mathbb{C}$, $$\sum\mathscr{F}(H_k(x)e^{-x^2/2})\frac{t^k}{k!}=\mathscr{F}(e^{-x^2+2xt-t^2})=\sqrt{2\pi}e^{-\xi^2/2}e^{t^2-2it\xi}$$ and by writing $e^{t^2-2it\xi}=e^{-(-it)^2+2(-it)\xi}$, we get $$\sqrt{2\pi}e^{-\xi^2/2}e^{t^2-2it\xi}=\sum \sqrt{2\pi}(-i)^kH_k(\xi)e^{-\xi^2/2}\frac{t^k}{k!}$$ with convergence in $L^2$. Thus we have something like 

For all $t\in\mathbb{C}$, $$\sum\mathscr{F}(H_k(x)e^{-x^2/2})\frac{t^k}{k!}=\sum \sqrt{2\pi}(-i)^kH_k(\xi)e^{-\xi^2/2}\frac{t^k}{k!}.$$


[Question] Can we conclude from here that $\mathscr{F}(H_k(x)e^{-x^2/2})=\sqrt{2\pi}(-i)^kH_k(\xi)e^{-\xi^2/2}$ ?
My book just says "equating the coefficients of $t^k$ yields the result", but the situation here differs from the typical ones in complex analysis, since the 'coefficients' are elements of $L^2$, not just numbers. I want to say something like "Plugging in $t=0$ gives the result for $k=0$. Differentiating once and plugging in $t=0$ gives the result for $k=1$, and so on.", but I'm not familiar with derivatives of a $L^2$ valued function, and I don't think that is a valid argument.
How can we conclude $\mathscr{F}(H_k(x)e^{-x^2/2})=\sqrt{2\pi}(-i)^kH_k(\xi)e^{-\xi^2/2}$ from the above power-series looking equation in $L^2$? Please help me!
 A: Yes, once you have the equation
$$
           \sum_{n}\mathscr{F}(e^{-x^2/2}H_n(x))\frac{t^n}{n!}
           = \sum_{n}\sqrt{2\pi}(-i)^n H_n(\xi)e^{-\xi^2/2}\frac{t^n}{n!},
$$
then you can fix $\xi$, and view this as a power series equation that holds for all $t$. Therefore, for this fixed $\xi$, the power series coefficients must be identical, which leads to
$$
           \mathscr{F}(e^{-x^2}H_n(x))=\sqrt{2\pi}(-i)^n H_n(\xi)
$$
The left side implicitly depends on $\xi$, because it is the transform variable. The transform is continuous in $\xi$ because of the exponentially decaying nature of the function being transformed. So
$$
         \mathscr{F}(e^{-x^2}H_n(x))(\xi)=\sqrt{2\pi}(-i)^n H_n(\xi),\;\;\xi\in\mathbb{R}.
$$
Omitting the arguments of the functions, these functions are equal:
$$          \mathscr{F}(e^{-x^2}H_n(x)) = \sqrt{2\pi}(-i)^n H_n
$$
A: Once you know that the series in question converges absolutely in $L^2$ 
then the identification becomes automatic.
Writing $\psi_k(x) = \frac{1}{k!} H_k(x) e^{-x^2/2}$, $k\geq 0$
then each
$\psi_k$ is an element in the Hilbert space $L^2({\Bbb R})$ 
having norm
$\frac{2^{k/2} } {\sqrt{k!}} \pi^{1/4}$.
One has $\lim_k \|\psi_k\|_{L^2}^{1/k} = 0 $,  so
the series
$$ \Psi_t(x) = 
\lim_{N\rightarrow \infty}
     \sum_{k=0}^{N-1} t^k \psi_k(x)$$
is norm convergent in $L^2$ for all complex $t$.
$\Psi_t$ is therefore an analytic $L^2$ valued function of $t\in {\Bbb C}$.
Coefficients in an analytic series are uniquely determined by the function and may be recovered e.g. by a Cauchy integral (which works fine in Hilbert or Banach spaces). You may also recover them simply from looking at successive derivatives at $0$, as you mention.
Since Fourier transformation (denoted  by a 'hat') preserves the $L^2$ norm 
 we automatically get:
 $$ \widehat{\Psi}_t(\xi) = 
\lim_{N\rightarrow \infty}
 \sum_{k=0}^{N-1} t^k \widehat{\psi}_k(\xi),$$
again an analytic $L^2$ valued function.
An issue is how to calculate the two sums (as $L^2$ limits) but
as you say, in your book they prove that
$\Psi_t(x)$ (in $L^2$) is identical to 
$$ \Phi_t= 
\exp \left( 2xt-t^2-x^2/2 \right) $$
This then resolves the problem since as elements of $L^2$:
$$\widehat{\Psi}_t(\xi)=
\widehat{\Phi}_t(\xi) = \exp \left( -2it\xi+t^2-\xi^2/2 \right)
  = \Phi_{-it}(\xi) =\Psi_{-it}(\xi)$$
Identifying the analytic expansions for 
$\widehat{\Psi}_t(\xi)=\Psi_{-it}(\xi)$
you must have:
$\widehat{\psi_k}(\xi) = (-i)^k \psi_k(\xi)$ in $L^2$ (where possibly a factor of $\sqrt{2\pi}$ should be included).
As both are actually smooth functions,  you get the wanted identity.
Returning to the limit in $L^2$, let me provide a relatively simple proof: The defining equation,
$H_k(x)=(-1)^k e^{x^2} \partial_x^k e^{-x^2}$ shows that $\psi_k=\frac{1}{k!}H_k(x) e^{-x^2/2}$ itself may be expressed through $\Phi_t$
as a Cauchy integral as follows:
$$ \psi_k(x)= 
  (-1)^k e^{x^2/2} \oint_{\partial B(x,R)}
    \frac{e^{-z^2}}{(z-x)^{k+1}} \frac{dz}{2\pi i} = 
  \oint_{\partial B(0,R)}
    \frac{e^{-x^2/2+2xu-u^2}}{u^{k+1}} \frac{du}{2\pi i}$$
where $R>0$ is arbitrary. From this we get the bound $|\psi_k(x)| \leq R^{-(k+1)} e^{-x^2/2+2|x|R +R^2}$ which is clearly square integrable
(in $x$) so that
$\psi_k$ is an element in $L^2$.
Now fix a value of $t$ and let $R>|t|$. Then by summing a geometric series:
 $$
     \sum_{k=0}^{N-1} t^k \psi_k(x) = 
  \oint_{\partial B(0,R)}
    (1 - (t/u)^N)\frac{e^{-x^2/2+2xu-u^2}}{u-t} \frac{du}{2\pi i}=
\Phi_t(x) + {\cal R}^{N}(t,x)$$
where we have the following bound for the remainder:
 $$ |{\cal R}^{N}(t,x)| \leq \frac{(|t|/R)^N}{R-|t|} 
 e^{-x^2/2+2R|x|+R^2}.$$
This is indeed a function in $L^2({\Bbb R})$ and goes to zero in norm when
 $N\rightarrow \infty$, thus justifying the above identifications.
