# For $f \in C_c^\infty(\mathbb{R})$, does $\hat{f}(k)\sum_{j=0}^n \frac{(-k^2)^j}{j!}$ converge to $\hat{f}(k)e^{-k^2}$ in $L^2(\mathbb{R})$.

As the title states: For $f \in C_c^\infty(\mathbb{R})$, does $\hat{f}(k)\sum_{j=0}^n \frac{(-k^2)^j}{j!}$ converge to $\hat{f}(k)e^{-k^2}$ in $L^2(\mathbb{R})$ where $C_c^\infty(\mathbb{R})$ is the space of smooth compactly supported functions. (and to be extra clear, I mean convergence in the $L^2(\mathbb{R})$ norm, not e.g. pointwise).

While attempting to answer Power Series expansion for $\exp(-\Delta)$ I used this step. Only later, under the objections of Keith McClary (thank you) have I found it increasingly difficult to prove. I would like help finding a proof of this statement, either by finishing my partial attempt, or using a different technique altogether. I'm just not sure if it's even true at this point.

Here is my attempt at a proof from that thread: Suppose $f \in C_c^\infty(\mathbb{R})$ and define $$\hat{f}_n(k) = \hat{f}(k) \sum_{j=0}^n \frac{(-4\pi^2 k^2)^j}{j!}$$ Observe that $\hat{f}$ is in $L^2(\mathbb{R})$ by Plancharel's theorem. Likewise, $\hat{f}(k)e^{-4\pi^2 k^2}$ is in $L^2(\mathbb{R})$ (because $e^x \leq 1$ for all $x \leq 0$).

We want to show that $$||\hat{f}(k)e^{-4\pi^2 k^2} - \hat{f}_n(k)||_2 \to 0$$ i.e. for all $\epsilon > 0$, there exists an $N$ such that for all $n\geq N$ $$||\hat{f}(k) \sum_{j=n+1}^\infty \frac{(-4\pi^2 k^2)^j}{j!}||_2 \leq \epsilon$$

To that end, we use the bound for alternating series that $|\sum_{j=n+1}^\infty a_j| \leq |a_{n+1}|$. Hence, $$|\hat{f}(k) \sum_{j=n+1}^\infty \frac{(-4\pi^2 k^2)^j}{j!}| \leq |\hat{f}(k)| \frac{(4\pi^2 k^2)^{n+1}}{(n+1)!}$$

It is well known that if a function $f$ is smooth then its Fourier transform satisfies $$|\hat{f}(k)| \leq \frac{C_m}{k^m}$$ for all $m > 0$ (see e.g. this thread). Using this estimate we can make the tail of the series arbitrarily small. $$\int_T^\infty |\hat{f}(k)|^2 \left|\frac{(4\pi^2 k^2)^{n+1}}{(n+1)!}\right|^2 dk \leq C_m^2 \int_T^\infty \left|\frac{(4\pi^2)^{2n+2} (k^{2n + 2 - 2m})}{(n+1)!}\right| dk$$

(This is the false step as is:) Using this, we can choose $N, T$ sufficiently large that $$\int_T^\infty |\hat{f}(k)|^2 \left|\frac{(4\pi^2 k^2)^{n+1}}{(n+1)!}\right|^2 dk \leq \frac{\epsilon}{4}$$ for all $n \geq N$.

To finish the proof, we choose $N_1, T$ so that $\forall n \geq N_1$: $$||(\hat{f}e^{-4\pi^2 k^2} - \hat{f}_n)\chi_{[-T, T]}||_2 < \frac{\epsilon}{2}$$

Using this and the bounds for the tail of the integral establish above we can show there exists $N_2 \geq N_1$ such that for all $n \geq N_2$ $$||\hat{f}e^{-4\pi^2 k^2} - \hat{f}_n||_2 \leq \epsilon$$

• I doubt that this works. I also doubt that you really care. Seems to me that what's relevant to $\exp(-\Delta)$ is $\hat f(k)e^{k^2}$, not $e^{-k^2}$. And of course you're not going to get the corresponding result for $e^{k^2}$. One of us dropped a minus sign - could be me, I don't think so... – David C. Ullrich Aug 15 '15 at 14:56
• Did you read the rest of my comment? I doubt that you really care about this convergence to $e^{-k^2}\hat f(k)$ because it seems to me that this has nothing to do with the original question about $\exp(-\Delta)$. – David C. Ullrich Aug 15 '15 at 15:02
• @DavidC.Ullrich - Aha I see. Well I am more interested in the convergence to $e^{-k^2}\hat{f}(k)$ than to $e^{k^2}\hat{f}(k)$. I think he miswrote the title. This would be something like the solution to the heat equation at time $1$. (or maybe his Laplacian is the negative of the usual one). But nice catch! – muaddib Aug 15 '15 at 15:05
• You should use $C_c$ not $C_0$ for compactly supported continuous functions. $C_0$ is the set of functions which merely decay at infinity. – Cameron Williams Aug 15 '15 at 16:11
• I know it is. For whatever reason, some authors love using $C_0$ for this purpose when $C_0$ for most analysts means something else. Of course they're the same in a finite space, but still. – Cameron Williams Aug 15 '15 at 16:19