[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!
Best Answer
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: $$ \mathscr{F}(e^{-x^2}H_n(x)) = \sqrt{2\pi}(-i)^n H_n $$