Power series expression for $\exp(-\Delta)$

Question

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!

Answer 1

This a potential answer to the question. It is not clear whether its proof technique can be completed or not.

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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 $L^2(\mathbb{R})$.

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Update:

$\hat{f}_n \to \hat{f}e^{-4\pi^2 k^2}$ 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 $n \geq N_2$ $$||\hat{f}e^{-4\pi^2 k^2} - \hat{f}_n||_2 \leq \epsilon$$

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I have asked the community for assistance with this at this thread.

Answer 2

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.