I know it should be true, but for some reason I can't get the calculations to work out in order to show that if $f$ is smooth and compactly supported, the power series $\sum_{j=0}^\infty \frac{(\partial_x^2)^j}{j!} f $ converges in $L^2$.
I hope someone can help!
This a potential answer to the question. It is not clear whether its proof technique can be completed or not.
We can obtain the result by considering convergence of the Fourier transform of the partial sums. Since $f$ is smooth and compactly supported it is in $L^2(\mathbb{R})$ as are all of its derivatives.
Let $f_n$ be the $n$th partial sum in the series: $f_n = \sum_{j=0}^n \frac{(\partial_x^2)^j}{j!} f$. We can compute the Fourier transform of $f_n$ as: $$\mathcal{F}(f_n)(k) = \mathcal{F}\left(\sum_{j=0}^n \frac{(\partial_x^2)^j}{j!} f\right)(k) = \sum_{j=0}^n \frac{(2\pi i k)^{2j}}{j!} \hat{f}(k) = \hat{f}(k) \sum_{j=0}^n \frac{(-4\pi^2 k^2)^j}{j!}$$
Therefore (quesionably, see below) $\hat{f}_n \to \hat{f}e^{-4\pi^2 k^2}$ in $L^2(\mathbb{R})$. Then by Plancherel's theorem, $f_n \to \mathcal{F}^{-1}(\hat{f}e^{-4\pi^2 k^2})$ in $L^2(\mathbb{R})$.
$\hat{f}_n \to \hat{f}e^{-4\pi^2 k^2}$ in $L^2(\mathbb{R})$
There seems to be some disagreement over whether this statement is true so I'll add details (and am unable to complete the proof). As demonstrated above, $$\hat{f}_n(k) = \hat{f}(k) \sum_{j=0}^n \frac{(-4\pi^2 k^2)^j}{j!}$$ 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 have asked the community for assistance with this at this thread.
This should not be true because every term in the series has the same compact support. Since this is a solution to the heat equation, it should spread out.
Edit: Another way of looking at this follows from the "uncertainty principle" : the Fourier transform of a "bump function" cannot decay exponentially fast.
Thus we expect $||(\frac {d}{dx})^n f|| = ||k^n \tilde f||$ to grow more rapidly than $\int{_0 ^\infty}k^n e^{-kx}dx \approx n! . $
Yet another way is that the "bump functions" are not a domain of essential self-adjointness. This is explained in the case of the operator $i\frac{d}{dx}$ on pp. 97-98 of An Introduction to the Mathematical Structure of Quantum Mechanics.
