A quick heuristic.
Just to be clear, we recall that for all $u\in\mathcal{S}'(\mathbb{R}^n)$, up to some abuse of notations : $$\mathcal{F}(-\Delta u)(\xi) = |\xi|^2 \mathcal{F}u(\xi)\text{.}$$
Here, I define the Fourier transform as
$$ \mathcal{F}f(\xi) := \int_{\mathbb{R}^n} f(x) e^{-i\,x\cdot\xi} \mathrm{~d}x\,\text{, (}f\in\mathrm{L}^1(\mathbb{R}^n)\text{, } \xi\in\mathbb{R}^n\text{).}$$
Therefore, if one wants to solve
$$-\Delta u =f\,\text{, in } \mathbb{R}^n.$$
One should have then, at least morally,
$$u(x) = \int_{\mathbb{R}^n} \mathcal{F}f(\xi) \cdot \frac{1}{|\xi|^2} e^{i\,x\cdot\xi} \mathrm{~d}\xi = \mathcal{F}^{-1}\left( \xi\mapsto\mathcal{F}f(\xi) \cdot \frac{1}{|\xi|^2} \right) = [f \ast \mathcal{F}^{-1}(\xi\mapsto|\xi|^{-2})](x). $$
Thus, the question reduce to know if one can prove (and give a sense to) the equality
$$\mathcal{F}^{-1}(\xi\mapsto|\xi|^{-2})(x) = C_n \frac{1}{|x|^{n-2}}.$$
Notice that, at the first glance, such representation formula involving such kernel makes sense only when $n\geqslant 3$. For $n=2$, one has to replace $\frac{1}{|x|^{n-2}}$ by $\log(|x|)$.
Answers to the post.
Now, I am starting by answering (partially) to the first part of your second question.
- For what function spaces for u
is it valid? I assume that u
must belong to a space such that the Laplacian is invertible (i.e. a bijection), but for which spaces is this true?
You can make sense of the right hand-side integral for Schwartz functions, $u\in\mathcal{S}(\mathbb{R}^n)$, or non-negative measurable functions, $u\in\mathrm{L}^1_{\text{loc}}(\mathbb{R}^n)$ (allowing the integral to take infinite values).
The question of actual function spaces on which it is well defined and such that it induces a bijection is much more delicate. I will adress this issue a bit later.
- How is it derived and is there a reason it is identical to the fundamental solution of the Laplacian?
It is actually the same as the fundamental solution. In fact, we can go a bit further and also find a similar way to identify the fractional Laplacian as such kind of convolution operator.
We define the following subspace of Schwartz functions for convenience
$$\mathcal{S}_0(\mathbb{R}^n) := \{\,u\in\mathcal{S}(\mathbb{R}^n)\,|\, 0\notin \mathrm{supp}\,\mathcal{F}u\,\}$$
- Definition: For $s\in\mathbb{R}$, we define the fractional Laplacian of order $\frac{s}{2}$, for any suitable $u$, as
$$ (-\Delta)^\frac{s}{2} u := \mathcal{F}^{-1}(\xi\mapsto |\xi|^{s} \mathcal{F}u(\xi)). $$
In particular, the definition makes sense for any $u\in \mathcal{S}_0(\mathbb{R}^n)$ and one even has in this case,
$$ (-\Delta)^\frac{s}{2} u \in \mathcal{S}_0(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n). $$
This is left as a tedious (but not hard) exercise.
We are particularly interested here by the behavior of the fractional Laplacian
$$ (-\Delta)^{-\frac{s}{2}}\,\text{, } s>0\text{.} $$
Here is the main result you are looking for:
- Proposition: Let $n\geqslant 2$, for any $s\in(0,n)$, we have the following equality in $\mathcal{S}'(\mathbb{R}^n)$
\begin{align*}
\int_{\mathbb{R}^n} \frac{1}{\left\lvert\xi\right\rvert^s} e^{-ix\cdot\xi}\,\mathrm{d}\xi = (2\pi)^\frac{n}{2}\frac{2^\frac{n-s}{2}\Gamma\left(\frac{n-s}{2}\right)}{2^\frac{s}{2}\Gamma\left(\frac{s}{2}\right)}\cdot \frac{1}{\left\lvert x\right\rvert^{n-s}}\text{ ,}
\end{align*}
i.e., for all $\varphi\in\mathcal{S}(\mathbb{R}^n)$,
$$ \left\langle \mathcal{F}\left[|\cdot|^{-s}\right], \varphi\right\rangle = (2\pi)^\frac{n}{2}\frac{2^\frac{n-s}{2}\Gamma\left(\frac{n-s}{2}\right)}{2^\frac{s}{2}\Gamma\left(\frac{s}{2}\right)} \cdot \left\langle |\cdot|^{-(n-s)}, \varphi\right\rangle\text{.}$$
I propose here a very elementary proof with a lot of details, for other proofs one may consult [1, Proposition 1.29] or [2, Section 2.4.3, Theorem 2.4.6]. Setting $s=2$ gives the result you are looking for.
- Proof of the proposition: Let $s \in(0, n)$, and $\xi \in \mathbb{R}^{n} \setminus\{0\}$. We denote by $\Gamma$ the usual Gamma function defined for any $x>0$ by
\begin{align*}
\Gamma(x)=\int_{0}^{+\infty} t^{x-1} e^{-t} \mathrm{~d} t\text{ .}
\end{align*}
We clearly have
\begin{align*}
\Gamma\left(\frac{s}{2}\right)=\int_{0}^{+\infty} t^{\frac{s}{2}-1} e^{-t} \mathrm{~d} t\text{ .}
\end{align*}
Since $\xi \neq 0, t \longmapsto t\left\lvert\xi\right\rvert^{2}$ is a diffeomorphism of $\mathbb{R}_{+}$, of class $C^{1}$, we deduce
\begin{align*}
\Gamma\left(\frac{s}{2}\right)=\int_{0}^{+\infty}\left\lvert\xi\right\rvert^{s} t^{\frac{s}{2}-1} e^{-t\left\lvert\xi\right\rvert^{2}} \mathrm{~d} t\text{ ,}
\end{align*}
so that
\begin{align*}
\frac{1}{\left\lvert\xi\right\rvert^s}=\frac{1}{\Gamma\left(\frac{s}{2}\right)} \int_{0}^{+\infty} t^{\frac{\pi}{2}-1} e^{-t\left\lvert\xi\right\rvert^{2}} \mathrm{~d} t \text{ . }
\end{align*}
Now, let $\phi \in \mathcal{S}\left(\mathbb{R}^{n}\right)$. First we consider the function
\begin{align*}
F: \mathbb{R}_{+}^{*} \times \mathbb{R}^{n} &\longrightarrow \mathbb{C} \\
(t, \xi) &\longmapsto \frac{1}{\Gamma\left(\frac{s}{2}\right)} t^{\frac{s}{2}-1} e^{-t\left\lvert\xi\right\rvert^{2}} \phi(\xi)
\end{align*}
so that $F$ satisfies near the origin
\begin{align*}
F(t, \xi) \underset{(0,0)}{\sim} \frac{1}{\Gamma\left(\frac{s}{2}\right)} t^{\frac{s}{2}-1} \phi(0)
\end{align*}
And for any $\alpha \geqslant 0, \beta \in \mathbb{R}_{+}^{n},$ writing $\xi^{\beta}=\xi_{1}^{\beta_{1}} \ldots \xi_{n}^{\beta_{n}}$, we see from the fact that $\phi\in \mathcal{S}\left(\mathbb{R}^{n}\right)$ that
\begin{align*}
t^{\alpha} \xi^{\beta} F(t, \xi) \underset{t,\left\lvert\xi\right\rvert \rightarrow+\infty}{\longrightarrow} 0 \text{ .}
\end{align*}
We deduce that $F \in \mathrm{L}^{1}\left(\mathbb{R}_{+} \times \mathbb{R}^{n}\right)$. Hence, applying Fubini-Lebesgue's Theorem, one has :
\begin{align*}
\int_{\mathbb{R}_{+}} \int_{\mathbb{R}^{n}} \frac{1}{\Gamma\left(\frac{s}{2}\right)} t^{\frac{s}{2}-1} e^{-t\left\lvert\xi\right\rvert^{2}} \phi(\xi) \mathrm{d} \xi \mathrm{d} t=\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}_{+}} \frac{1}{\Gamma\left(\frac{s}{2}\right)} t^{\frac{s}{2}-1} e^{-t\left\lvert\xi\right\rvert^{2}} \mathrm{~d} t\right) \phi(\xi) \mathrm{~d} \xi \text{, }
\end{align*}
thus, one can write
\begin{align*}
\frac{1}{\Gamma\left(\frac{s}{2}\right)} \int_{0}^{+\infty} t^{\frac{s}{2}-1}\left\langle e^{-t\left\lvert\cdot\right\rvert^{2}}, \phi\right\rangle \mathrm{d} t=\int_{\mathbb{R}^{n}} \frac{\phi(\xi)}{\left\lvert\xi\right\rvert^{s}} \mathrm{~d} \xi\text{ .}
\end{align*}
Note that the latter integral is absolutely convergent (one may use polar coordinates and the decreasing properties of $\phi\in\mathcal{S}(\mathbb{R}^n)$). Now, we introduce the distribution $T_s \in \mathcal{S}^{\prime}\left(\mathbb{R}^{n}\right)$ which is defined for any $\phi \in \mathcal{S}\left(\mathbb{R}^{n}\right)$ by
\begin{align*}
\langle T_s, \phi\rangle &:=\int_{\mathbb{R}^{n}} \frac{\phi(\xi)}{\left\lvert\xi\right\rvert^{s}} \mathrm{~d} \xi \\
&=\frac{1}{\Gamma\left(\frac{s}{2}\right)} \int_{0}^{+\infty} t^{\frac{s}{2}-1}\left\langle e^{-t\left\lvert\cdot\right\rvert^{2}}, \phi\right\rangle \mathrm{~d} t\text{ .}
\end{align*}
We want to compute $\mathcal{F}T_s$, for $\phi\in\mathcal{S}(\mathbb{R}^n)$, we use the definition of the Fourier transform on $\mathcal{S}'$ which gives
\begin{align*}
\langle \mathcal{F}T_s, \phi\rangle &:= \langle T_s, \mathcal{F}\phi\rangle\\
&=\frac{1}{\Gamma\left(\frac{s}{2}\right)} \int_{0}^{+\infty} t^{\frac{s}{2}-1}\left\langle e^{-t\left\lvert\cdot\right\rvert^{2}}, \mathcal{F}\phi\right\rangle \mathrm{~d} t\\
&=\frac{1}{\Gamma\left(\frac{s}{2}\right)} \int_{0}^{+\infty} t^{\frac{s}{2}-1}\left\langle \mathcal{F}\left[ e^{-t\left\lvert\cdot\right\rvert^{2}}\right], \phi\right\rangle \mathrm{~d} t\\
&=\frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{s}{2}\right)} \int_{0}^{+\infty} t^{\frac{s}{2}-1}\left(\int_{\mathbb{R}^n} \frac{1}{t^\frac{n}{2}}e^{-\frac{\left\lvert x \right\rvert^2}{4t}} \phi(x)\mathrm{~d}x \right) \mathrm{~d} t\text{ ,}
\end{align*}
where we used the well known fact $\mathcal{F}\left[ e^{-t\left\lvert\cdot\right\rvert^{2}}\right] (x) = \frac{\pi^\frac{n}{2}}{t^\frac{n}{2}}e^{-\frac{\left\lvert x \right\rvert^2}{4t}}$, for all $t>0$, $x\in\mathbb{R}^n$. Now, we consider the function
\begin{align*}
G: \mathbb{R}_{+}^{*} \times \mathbb{R}^{n} & \longrightarrow \mathbb{C} \\
(t, x) & \longmapsto t^{\frac{s}{2}-\frac{n}{2}-1} e^{-\frac{|x|^{2}}{4 t}} \phi(x)
\end{align*}
using the change of variable $(t, x) \longmapsto(t, \sqrt{t} x)$ which is a diffeomorphism of class $C^{1}$ over $\mathbb{R}_{+}^{*} \times \mathbb{R}^{n}$, with jacobian $(t, x) \longmapsto t^{\frac{n}{2}},$ we have
\begin{align*}
G(t, \sqrt{t} x) t^{\frac{n}{2}} \underset{(0,0)}{\sim} t^{\frac{s}{2}-1} \phi(0)
\end{align*}
And for any $\alpha \geqslant 0, \beta \in \mathbb{N}^{n}$, writing $x^{\beta}=x_{1}^{\beta_{1}} \ldots x_{n}^{\beta_{n}},$ we see that
\begin{align*}
t^{\alpha} x^{\beta} G(t, \sqrt{t} x) t^{\frac{n}{2}} \underset{t,|x| \rightarrow+\infty}{\longrightarrow} 0
\end{align*}
Since the composition with a diffeomorphism does not change the integrability, we deduce that $G \in \mathrm{L}^{1}\left(\mathbb{R}_{+} \times \mathbb{R}^{n}\right)$. Hence applying Fubini-Lebesgue's Theorem, we have
\begin{align*}
\left\langle\mathcal{F} T_s, \phi\right\rangle=\frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{s}{2}\right)} \int_{\mathbb{R}^{n}}\left(\int_{0}^{+\infty} t^{\frac{s}{2}-\frac{n}{2}-1} e^{-\frac{|x|^{2}}{4 t}} \mathrm{~d} t\right) \phi(x) \mathrm{~d} x \text{ .}
\end{align*}
So it remains to compute for any $x\neq 0$, the integral
\begin{align*}
\int_{0}^{+\infty} t^{\frac{s}{2}-\frac{n}{2}-1} e^{-\frac{|x|^{2}}{4 t}} \mathrm{~d} t \text{ .}
\end{align*}
First since $t\longmapsto\frac{1}{t}$ is a $C^1$-diffeormorphism with jacobian $t\longmapsto \frac{1}{t^2}$, one has
\begin{align*}
\int_{0}^{+\infty} t^{\frac{s}{2}-\frac{n}{2}-1} e^{-\frac{|x|^{2}}{4 t}} \mathrm{~d} t = \int_{0}^{+\infty} t^{\frac{n}{2}-\frac{s}{2}-1} e^{-t\frac{|x|^{2}}{4}} \mathrm{~d} t
\end{align*}
so that applying the result from the beginning of the proof to $\xi=\frac{x}{2}$, and changing $s$ into $n-s$, we finally have
\begin{align*}
\int_{0}^{+\infty} t^{\frac{s}{2}-\frac{n}{2}-1} e^{-\frac{|x|^{2}}{4 t}} \mathrm{~d} t = \frac{2^{n-s}\Gamma\left(\frac{n-s}{2}\right)}{\left\lvert{x}\right\rvert^{n-s}}\text{ .}
\end{align*}
The result follows from the equality ${\pi^\frac{n}{2}2^{n-s}}=(2\pi)^{\frac{n}{2}}\frac{2^\frac{n-s}{2}}{2^{\frac{s}{2}}}$. $\blacksquare$
We get back on the well-define character and the boundedness of the fractional Laplacian $(-\Delta)^{-\frac{s}{2}}\,\text{, } s>0\text{.}$
- For what function spaces for u
is it valid? I assume that u
must belong to a space such that the Laplacian is invertible (i.e. a bijection), but for which spaces is this true?
The following result gives the $\mathrm{L}^p-\mathrm{L}^q$ boundedness. This is a sharp result and one cannot expect any other kind of boundedness on Lebesgues spaces.
- Theorem (Hardy-Littlewood-Sobolev Inequality): Let $s\in(0,n)$, and let $p,q\in(1,+\infty)$ such that
\begin{align*}
\frac{1}{q}=\frac{1}{p}-\frac{s}{n}\text{ .}
\end{align*}
Then the following linear convolution operator
\begin{align*}
(-\Delta)^{-\frac{s}{2}} : \mathcal{S}(\mathbb{R}^n) & \longrightarrow \mathrm{C}^\infty(\mathbb{R}^n)\cap\mathcal{S}'(\mathbb{R}^n)\\
f & \longmapsto \left[ x\mapsto \frac{2^\frac{n-s}{2}\Gamma\left(\frac{n-s}{2}\right)}{(2\pi)^\frac{n}{2}2^\frac{s}{2}\Gamma\left(\frac{s}{2}\right)}\int_{\mathbb{R}^n} \frac{1}{\left\lvert{x-y}\right\rvert^{n-s}}f(y)\mathrm{~d}y \right]
\end{align*}
is also a well defined bounded linear operator from $\mathrm{L}^p({\mathbb{R}^n})$ to $\mathrm{L}^q({\mathbb{R}^n})$, i.e. there exists a constant $C>0$ depending on $n, s, p$ and $q$ such that for any $f\in\mathrm{L}^p({\mathbb{R}^n})$, one has
\begin{align*}
\lVert (-\Delta)^{-\frac{s}{2}} f\rVert_{\mathrm{L}^q({\mathbb{R}^n})}\leqslant C \left\lVert f\right\rVert_{\mathrm{L}^p({\mathbb{R}^n})}.
\end{align*}
See [3, Section 1.2.1, Theorem 1.2.3] for a proof and the sharpness or the result, or [1, Theorem 1.38] for the simpler case $p=2$. I don't give any proof here because it requires too much additional technology.
Notice, since the result is sharp, one NEVER has $\mathrm{L}^p$-$\mathrm{L}^p$ boundedness of $(-\Delta)^{-1}$, even if $p=2$.
In particular, the operator $(-\Delta)^{-1}$ NEVER MAPS $\mathrm{L}^p$ to $\mathrm{W}^{2,p}(\mathbb{R}^n)$ ! (Here, $\mathrm{W}^{2,p}$, also denoted sometimes $\mathrm{H}^{2,p}$, is the standard Sobolev space made of $\mathrm{L}^p$ functions such that all weak derivatives up to the order $2$ also lies in $\mathrm{L}^p$).
- For what function spaces for u
is it valid? I assume that u
must belong to a space such that the Laplacian is invertible (i.e. a bijection), but for which spaces is this true?
- I assume that we gain two degrees of regularity through this operator, but how is this proven? What is the image of this operator?
Those two questions are deeply related and the possible answers are highly non-trivial.
The main idea comes from the next very important theorem, that says that essentially $\mathrm{L}^p$ norm of the Laplacian contains all the informations about the $\mathrm{L}^p$ of all second order derivatives, so that knowledge of the laplacian or all the second order derivatives makes no differences on $\mathbb{R}^n$.
- Theorem (Schauder estimates): Let $p\in(1,+\infty)$. There exists a constant $C>0$ depending on $n$ and $p$, such that for all $u\in \mathrm{W}^{2,p}(\mathbb{R}^n)(=\mathrm{H}^{2,p}(\mathbb{R}^n))$, one has
$$\lVert \nabla^2 u\rVert_{\mathrm{L}^p(\mathbb{R}^n)} \leqslant C\lVert \Delta u\rVert_{\mathrm{L}^p(\mathbb{R}^n)}\leqslant C \lVert \nabla^2 u\rVert_{\mathrm{L}^p(\mathbb{R}^n)}.$$
For a proof with all the steps, check for instance [2, Definition 5.1.13 & Proposition 5.1.14 p.325, Proposition 5.1.17 p.328, Corollary 5.2.8 p.340, Exercise 5.2.10 (a) p.354].
- Definition (of Homogeneous Sobolev spaces): Let $p\in(1,+\infty)$. We define the homogeneous Sobolev space of order $2$ over $\mathrm{L}^p(\mathbb{R}^n)$ to be the following completion in tempered distribution modulo polynomials
$$\dot{\mathrm{H}}^{2,p}(\mathbb{R}^n) := \widehat{\mathrm{H}^{2,p}(\mathbb{R}^n)}^{ \lVert \nabla^2 \cdot\rVert_{\mathrm{L}^p}}\subset \mathcal{S}'(\mathbb{R}^n)/\mathcal{P}(\mathbb{R}^n).$$
We set $\lVert u \rVert_{\dot{\mathrm{H}}^{2,p}(\mathbb{R}^n)} := \lVert \nabla^2 u \rVert_{\mathrm{L}^p(\mathbb{R}^n)}$.
The reason why the completion is taken in tempered distributions modulo polynomials, is that we want the semi-norm $\lVert \nabla^2 \cdot\rVert_{\mathrm{L}^p}$ to be an actual norm. That is, $\lVert \nabla^2 \cdot\rVert_{\mathrm{L}^p}$ does not distinguishe two elements of $\mathcal{S}'(\mathbb{R}^n)$ if they differ from a polynomial of degree $1$.
Notice also that we NEVER have $\dot{\mathrm{H}}^{2,p}(\mathbb{R}^n)\subset \mathrm{L}^p(\mathbb{R}^n)$.
In this case, reasoning by density, we obtain the following result from the Schauder estimates above.
- Theorem: Let $p\in(1,+\infty)$. Then $$ (-\Delta)\,:\, \dot{\mathrm{H}}^{2,p}(\mathbb{R}^n)\longrightarrow \mathrm{L}^p(\mathbb{R}^n) $$ is an isomorphism of Banach spaces.
However, due to the quotient structure involving polynomials, one can no longer compute $(-\Delta)^{-1}f$ pointwise using the convolution formula. The problem being that there is in general no way to choose canonically the polynomial part to obtain a representative being an element of $\mathcal{S}'(\mathbb{R}^n)$. This is a really technical and hard question.
However, in the specific case $(1,\frac{n}{2})$, thanks to the Hardy-Littlewood-Sobolev inequality when $s=2$, we can modify the construction to deal with actual elements of $\mathcal{S}'(\mathbb{R}^n)$.
- Definition (of Homogeneous Sobolev spaces): Let $p\in(1,\frac{n}{2})$. We (re)define the homogeneous Sobolev space of order $2$ over $\mathrm{L}^p(\mathbb{R}^n)$ to be
$$\dot{\mathrm{H}}^{2,p}(\mathbb{R}^n) := \{\, u\in\mathrm{L}^{\frac{pn}{n-2p}}(\mathbb{R}^n)\,|\,\lVert \nabla^2 u \rVert_{\mathrm{L}^p(\mathbb{R}^n)}<+\infty\,\}.$$
We still set $\lVert u \rVert_{\dot{\mathrm{H}}^{2,p}(\mathbb{R}^n)} := \lVert \nabla^2 u \rVert_{\mathrm{L}^p(\mathbb{R}^n)}$.
One can check that this space is a Banach space for the norm $\lVert \cdot \rVert_{\dot{\mathrm{H}}^{2,p}(\mathbb{R}^n)}$, and that Schwartz functions are dense for such a construction.
In this specific case, there is no more ambiguity, and one can compute for any $f\in\mathrm{L}^p(\mathbb{R}^n)$, almost every $x\in\mathbb{R}^n$
$$(-\Delta)^{-1}f(x)=\frac{\Gamma\left(\frac{n-2}{2}\right)}{4\pi^\frac{n}{2}}\int_{\mathbb{R}^n} \frac{1}{\left\lvert{x-y}\right\rvert^{n-2}}f(y)\mathrm{~d}y \in \dot{\mathrm{H}}^{2,p}(\mathbb{R}^n) \subset \mathrm{L}^{\frac{pn}{n-2p}}(\mathbb{R}^n) \text{.} $$
I just finish my post here, saying that similarly one can define similarly homogeneous Sobolev spaces $\dot{\mathrm{H}}^{s,p}(\mathbb{R}^n)$ but encounters the same kind of issues whenever $s\geqslant \frac{n}{p}$.
[REFRENCES]
- Bahouri, Hajer; Chemin, Jean-Yves; Danchin, Raphaël, Fourier analysis and nonlinear partial differential equations, Grundlehren der Mathematischen Wissenschaften 343. Berlin: Heidelberg (ISBN 978-3-642-16829-1/hbk; 978-3-642-16830-7/ebook). xvi, 523 p. (2011). ZBL1227.35004.
- Grafakos, Loukas, Classical Fourier analysis, Graduate Texts in Mathematics 249. New York, NY: Springer (ISBN 978-0-387-09431-1/hbk). xvi, 489 p. (2008). ZBL1220.42001.
- Grafakos, Loukas, Modern Fourier analysis, Graduate Texts in Mathematics 250. New York, NY: Springer (ISBN 978-1-4939-1229-2/hbk; 978-1-4939-1230-8/ebook). xvi, 624 p. (2014). ZBL1304.42002.
