# Essential self-adjointness of the Laplace operator via the Fourier transform

I'm working through some notes on showing the essential self-adjointness of the Laplace operator on $\Bbb R$ via the Fourier transform (see here) but there seems to be a little bit of liberty taken at one point which I can't quite justify. It is shown that $-\Delta$ is symmetric on $C_c^{\infty}(\Bbb R)$ and is also positive. These are pretty straightforward as is often the case with differential operators. The part which bothers me is showing that it is in fact essentially self-adjoint.

The easiest way (I think) to do it is via the Fourier transform since the Fourier transform turns the Laplace operator into a multiplication operator. The author wants to show that $\operatorname{Im}(-\Delta\pm i)$ is dense in $L^2(\Bbb R)$, or equivalently $\operatorname{Im}(-\Delta \pm i)^{\perp} = \{0\}$, which is of course one of the equivalent conditions for a symmetric operator to be essentially self-adjoint. I'll replicate the argument here and explain where I think some rigor is missing.

Suppose $g\in L^2(\Bbb R)$ is such that

$$0 = \langle (-\Delta \pm i)f,g\rangle \tag{1}$$

for all $f\in C_c^{\infty}(\Bbb R)$. Since the Fourier transform is unitary, we have

$$0 = \langle \mathcal{F}(-\Delta\pm i)f,\mathcal{F}g\rangle.$$

For $f\in C_c^{\infty}(\Bbb R)$, $-\mathcal{F}\Delta f(x) = 4\pi^2|x|^2\mathcal{F}f(x)$ so this becomes

$$0 = \langle (4\pi^2 |x|^2\pm i)\mathcal{F}f,\mathcal{F}g\rangle.$$

We want to show that $g$ is zero, so it stands to reason that we would push $4\pi^2|x|^2\pm i$ onto $\mathcal{F}g$ to then get that

$$0 = \langle \mathcal{F}f, (4\pi^2|x|^2\mp i)\mathcal{F}g\rangle.$$

Since the Fourier transform is unitary, $\mathcal{F}(C_c^{\infty}(\Bbb R))$ is dense in $L^2(\Bbb R)$ so this would say that $(4\pi^2|x|^2\mp i)\mathcal{F}g = 0$, giving that $\mathcal{F}g=0$ and so $g=0$ since the Fourier transform is unitary.

Herein lies my problem. There is no guarantee a priori that $(4\pi^2|x|^2\mp i)\mathcal{F}g$ is in $L^2(\Bbb R)$ so we cannot apply this reasoning. The obvious way around this is to require that $g$ be such that $g\in L^2(\Bbb R)$ and $(4\pi^2|x|^2\mp i)\mathcal{F}g\in L^2(\Bbb R)$. The set of $g$ satisfying this is dense since it contains $C_c^{\infty}(\Bbb R)$ by Paley-Wiener (or, better yet, contains the Schwartz space). For such $g$, the above argument would show that $g$ must indeed be the zero function. However it isn't clear to me that this is sufficient to say that any such $g$ satisfying $(1)$ must be zero. Is there something I am missing? Can density arguments be used somehow or is this a fundamentally flawed argument?

• Related – Giuseppe Negro Feb 4 '15 at 20:45
• I agree with your doubts. Moreover this proof does not use in any way the symbol of $-\Delta$, it could in principle work with any differential operator. Something is wrong – Giuseppe Negro Feb 4 '15 at 20:49
• @GiuseppeNegro I think the only way the symbol really comes about is insofar that $|x|^2$ doesn't have any eigenfunctions in $L^2(\Bbb R)$ (giving that $\mathcal{F}g$ must be zero), which is quite elegant but there is definitely something missing. This domain mismatch issue is of course closely related to the domain of $(-\Delta)^*$ but I was hoping that somehow it could be skirted via the Fourier transform. I've seen other texts that go through the details surrounding $(-\Delta)^*$ and do it without the Fourier transform but it's not very clean by any means. – Cameron Williams Feb 4 '15 at 20:54

## 2 Answers

It is actually not too difficult to repair the argument: You know $0=\left\langle (4\pi^2 |x|^2\pm i)\hat{f},\hat{g}\right\rangle$; now the set $\{ (4\pi^2 |x|^2\pm i)\hat{f} \ | \ f\in C^\infty_0\}$ is dense in $L^2$, whence $\hat{g}=0$ and $g=0$. Why is this set dense? Adapt the proof (last paragraph) of Proposition 1.1.15 (page 5) here.

[It is not too difficult to find other references which carry out a correct proof using the Fourier transform, for instance http://people.math.gatech.edu/~loss/14SPRINGTEA/laplacian.pdf or Davies' "Spectral theory and differential operators" (Theorem 3.5.3). In Davies' book the result is generalized to a certain class of constant coefficient differential operators (depending on the symbol, as it has been mentioned in the comments by G.N.).]

• I saw the notes by Loss but it dealt a little too heavily with some of the technical details so I was hoping there was a way to repair this argument. The proposition you pointed out is exactly what I was going for. I had a sneaking suspicion that that set was dense but I assumed it was a much lengthier proof. Thanks a bunch! – Cameron Williams Feb 7 '15 at 14:39

This started as a comment but soon became too big, so I am making it an answer.

The Fourier transform makes it clear that $-\Delta$ is self-adjoint on its natural domain in free space $H^2(\mathbb{R}^n)$ because it unitarily diagonalizes it. That is, the Fourier transform is unitary and it intertwines $\mathcal{F}\circ(-\Delta) = a(\xi)\circ\mathcal{F}$, where $a$ is a real valued function acting as a multiplication. (Precisely $a(\xi)=\lvert \xi\rvert^2$, but only the fact that it is real valued matters here).

Therefore, $-\Delta$ is self-adjoint on its natural domain iff $a$ is. The natural domain of this last operator is

$$\text{Dom}(a)=\left\{ \phi\in L^2(\mathbb{R}^n)\ :\ a(\xi)\phi(\xi)\in L^2(\mathbb{R}^n)\right\}.$$ It is clear from realness of $a(\xi)$ that $\text{Dom}(a)\subset \text{Dom}(a^\star)$, that is, that $a$ is symmetric. The other inclusion is Riesz's representation theorem: if $\psi \in \text{Dom}(a^\star)$, then by definition $$\frac{\left\lvert \int \phi(\xi)\overline{a(\xi)\psi(\xi)}\, d\xi\right\rvert}{\lVert \phi\rVert_2}\le C$$ for all $\phi\in \text{Dom}(a)$, which surely is a dense subset of $L^2$ space. By Riesz's representation theorem the supremum of this ratio is exactly $\lVert a\psi\rVert_2$, so this norm is finite and $\psi\in \text{Dom}(a)$.

You probably already knew this stuff, but I am posting here anyway hoping that it might help.