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!

  • $\begingroup$ -Do you mean $\exp(\Delta)$ in the title? As DavidC.Ullrich noticed, that doesn't seem to match with your power series $(\sum_{j=0}^\infty \frac{(\partial_x^2)^j}{j!})$ (or is yours the negative of the usual Laplacian). $\endgroup$ – muaddib Aug 15 '15 at 15:16

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.

| cite | improve this answer | |
  • $\begingroup$ Excellent proof! +1 $\endgroup$ – Nicolas Aug 7 '15 at 18:31
  • $\begingroup$ One of us is missing something. I don't see what that sum has to so with $\exp(-\Delta)$ in the first place! In one variable: $\exp(-\Delta)=\sum(-\partial_x^2)^j/j!$; now on the Fourier transform side this becomes $F(\exp(-\Delta)f)(k)=e^{k^2}\hat f(k),$ not $e^{-k^2}$. ??? $\endgroup$ – David C. Ullrich Aug 15 '15 at 14:53
  • $\begingroup$ @DavidC.Ullrich - That's a good point! Maybe the OP miswrote the title. His question asks about $\sum_{j=0}^\infty \frac{(\partial_x^2)^j}{j!} f $ converges in $L^2$ so I'll work with that. $\endgroup$ – muaddib Aug 15 '15 at 15:02

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.

| cite | improve this answer | |
  • $\begingroup$ In Answer #1 I question this step: "Therefore $\hat{f}_n \to \hat{f}e^{-4\pi^2 k^2}$ ...". This would hold if the remainder of the series converged uniformly to zero (as a function of $k$), but I don't think it is true in general. $\endgroup$ – Keith McClary Aug 8 '15 at 2:58
  • $\begingroup$ $|| f_n - f_{n-1} ||^2$ is like $\int k^{4n} |f(k)|^2 dk / (n!)^2$. There are limitations on how slowly Fourier transforms of "bump functions" can drop off. math.stackexchange.com/questions/138302/… $\endgroup$ – Keith McClary Aug 9 '15 at 17:40
  • $\begingroup$ Are you sure about this? As far as i can see, the $n$'th partial sum is perfectly well-defined in the time domain, and the limit is well-defined in the frequency domain (as it is merely an exponential). $\endgroup$ – Ukhrir Aug 10 '15 at 20:54
  • $\begingroup$ The "Therefore ..." would work if $\sum_{j=0}^n \frac{(-4\pi^2 k^2)^j}{j!}$ converged uniformly to $e^{-4\pi^2 k^2}$ but it is far from uniform. There is classical theory relating to the Question: encyclopediaofmath.org/index.php/Quasi-analytic_class $\endgroup$ – Keith McClary Aug 12 '15 at 0:37
  • $\begingroup$ That makes sense, I guess I have to use the solution i ended up with which is $H_0=-\Delta$ that we see that as $F(H_0 f)=\xi^2Ff$ we have that $F^{-1}\exp(-t\xi^2)Ff=\exp(-tH_0)f$ by the unbounded functional calculus, since $\exp(-x)$ is bounded on the spectrum of $H_0$. Therefore we may calculate for $t\geq 0$, and $f\in C_c^{\infty}$. $(\exp(-tH_0)f)(x)=F^{-1}\exp(-t\xi^2)Ff$ $F^{-1} ( \exp(-t\xi) \hat{f})$ $\frac{1}{(2\pi t)^m}\exp(-\frac{-x^2}{2t})*f(y)$ which is what we desired. We may extend this to the entirety of $L^2$ via. standard methods of approximating $L^2$ functions $\endgroup$ – Ukhrir Aug 12 '15 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.