# How to define Laplacian on $L_2$

This might be a dumb question, but I thought the Laplacian (classical) is defined for $C^2$ functions. How do we extend that to be a self-adjoint operator on all of $L_2$? Is it the so called Friedrich's extension theorem? Is the extension explicit? How would I, for example, compute the Laplacian of |x|

Thanks

-

(1) Let me answer first to the last question: $\Delta \vert x\vert$ is homogeneous of degree $-1$ and radial. On $\mathbb R^d$ ($d\ge 2$) it is $$(\partial_r^2+\frac{d-1}{r}\partial_r)(r)=\frac{d-1}{\vert x\vert}$$ which is an $L^1_{loc}$ function.

(2) Now you can define $\Delta$ on Distributions $T$ on some open set $\Omega$ of $\mathbb R^d$ with $$\langle\Delta T,\phi\rangle_{\mathscr D'(\Omega), \mathscr D(\Omega)}= \langle T,\Delta \phi\rangle_{\mathscr D'(\Omega), \mathscr D(\Omega)}.$$ A particular case with $T$ in $L^2$ is given in the previous answer.

(3) You can also consider $\Delta$ as an unbounded operator on $L^2(\mathbb R^d)$ with domain

$D(\Delta)$={$u\in L^2(\mathbb R^d), \Delta u\in L^2(\mathbb R^d)$}

where the term $\Delta u$ is taken in the distribution sense as in (2). The point is to prove that the operator $(-\Delta)$ is non-negative selfadjoint, which means that it is symmetric nonnegative and that the domain of the adjoint is the same as the domain of $-\Delta$. Thanks to Friedrichs extension theorem, since $-\Delta$ is nonnegative, there is no other selfadjoint extension.

-

I can think of two differing viewpoints, and though many would prefer the latter due to its simplicity inherited from the Hilbert structure, I prefer the former for its connections to Sobolev spaces and PDE.

I. The Laplacian can be defined in the sense of distributions, since the Laplacian of an $L^2$ function is a distribution. In particular, for $u \in L^2(\Omega)$ define

$<\Delta u, \phi> := \int_\Omega u \Delta \phi\;dx$

for $\phi \in C^\infty_c(\Omega)$.

This would then say that in one dimension, for example, $\Delta |x| = 2\delta_0$, where I have used $\delta_0$ to denote the Dirac mass at zero, a distribution/measure.

II. For $u \in L^2(0,1)$, $sin(n\pi x)$ and $cos(n\pi x)$ form a basis and we can write

$u(x)= b_0 + \sum_n a_n sin(n\pi x) + b_n cos(n\pi x)$,

and we have $\sum_n a_n^2+b_n^2 <\infty$

Then we can define, formally, $\Delta u := \sum_n (n\pi)^2(-a_n sin(n\pi x) - b_n cos(n\pi x))$. In general, $\Delta u$ will not make sense pointwise, unless we know that $\sum_n n^4(a_n^2+b_n^2) <\infty$ (but as above, we can write $<\Delta u, \phi> = \int u \Delta \phi\;dx = \sum_n (n\pi)^2 (-a_n a_n^\prime-b_nb_n^\prime)$

where $a_n^\prime$ and $b_n^\prime$ are the Fourier coefficients of $\phi$. Now this makes sense for any $u \in L^2(0,1)$ if $\phi$ satisfies $\sum_n n^4((a_n^\prime)^2+(b^\prime_n)^2)<\infty$.

Of course, II can be done in higher dimensions - I have only chosen one dimension to illustrate with a simple example the basis functions. I do comment that I is more general, since it does not require the Hilbert structure.

-

Let me comment a bit on Daniel Spector's answer. Yet two further possible alternative approaches are based on the notion of weak derivative and derivative almost everywhere. For the sake of simplicity, I only discuss the case of functions defined over $(0,1)$.

1) An $L^2(0,1)$ function $f:(0,1)\to \mathbb R$ is said to be weakly differentiable if there exists a $g\in L^2(0,1)$ such that $$\int_0^1 f(x)h'(x)dx=-\int_0^1 g(x)h(x)dx\qquad \hbox{for all }h\in C^1[0,1],$$ and in this case $g$ is uniquely defined and called the weak derivative of $f$. So far so good. Now, you can define in the very same way the second (weak) derivative $f''$ of $f$ and define the (weak) Laplacian as the mapping $f\mapsto f''$.

2) If a function is Lipschitz continuous, then by Rademacher's theorem it is differentiable almost everywhere. Hence, a Lipschitz continuous functions that is differentiable with a Lipschitz continuous derivative is twice differentiable a.e. It is then natural to associate with each function its second derivative. This is possible at least a.e., giving rise to another notion of Laplacian that is well-behaved if one wants to solve integrated versions of elliptic problems.

-
Part 1) is equivalent to the assumption $u \in H^2$ for which we know we can give meaning to the Laplacian in precisely the sense you describe. Part 2) is the assumption $u \in W^{2,\infty}$, which is a stronger hypothesis even than $H^2$. $H^1$ is enough to consider weak solutions to Laplace's equation, though defining the Laplacian on $H^1$ is another matter. –  Daniel Spector Jan 29 '13 at 8:19
of course. well, clearly i was trying to avoid to use explicitly sobolev space theory, which i was assuming not to be known by the OP. in fact, another possibility would be to study everything in $W^{2,1}$, if one already knows what an absolutely continuous function is, and in which sense one can think of it as differentiable. –  Delio Mugnolo Jan 29 '13 at 14:18