I'm reading Reed & Simon's book and, at some point, the authors define the following object. If $\Delta$ denotes the Laplacian operator on $\mathbb{R}^{d}$, they set the domain $D_{\text{\max}}$ of $-\Delta$ as: $$D_{\text{max}} := \{\varphi \in L^{2}(\mathbb{R}^{d}): \hspace{0.1cm} \mbox{$\Delta\varphi \in L^{2}(\mathbb{R}^{d})$ in the sense of distributions}\}$$

Question: What does "$\Delta\varphi \in L^{2}(\mathbb{R}^{d})$ in the sense of distributions" means?

  • 1
    $\begingroup$ Do you know about distributions in the sense of Schwartz? There is an extension of $\Delta$ to distributions. So that $\Delta \varphi$ may be defined even for non-differentiable functions. In general, the result would be a distribution on $\mathbb R^d$. The condition says: that distribution is identified with an $L^2$ function. $\endgroup$
    – GEdgar
    Jul 15 at 16:55
  • $\begingroup$ @GEdgar I know about distributions, and I'm okay with $\Delta$ being defined in the sense of distributions. However, what does it mean $\Delta \varphi \in L^{2}(\mathbb{R}^{d})$ in the sense of distributions? I mean, to be an element of $L^{2}$ in the sense of distributions is a little bit odd to me. $\endgroup$
    – MathMath
    Jul 15 at 16:58
  • $\begingroup$ You can embed $L^2(\mathbb{R}^d)$ (or more generally $L^1_{\text{loc}}(\mathbb{R}^d)$ to the space of distributions via $$T(f) \quad := \quad \left[ C_c^{\infty}(\mathbb{R}^d) \ni \varphi \quad \mapsto \quad \int_{\mathbb{R}^d} f(x)\varphi(x) \, \mathrm{d}x \right].$$ That way you may identify $L^2(\mathbb{R})^d$ as a subspace of distributions. $\endgroup$ Jul 15 at 17:04

For any $L^2$ function $\varphi$, one can define the distribution derivatives as a functional $C^\infty_c(\mathbb R^d)\to \mathbb R$ by

$$ \partial_{x_i} \varphi (f) : = -\int_{\mathbb R^d} \varphi f_{x_i}. $$

Then $\Delta \varphi$ is just the functional

\begin{align} (\Delta \varphi) (f) &= \left( \varphi_{x_1x_1} + \cdots + \varphi_{x_dx_d}\right) (f) \\ &= \int_{\mathbb R^d} \varphi \left( f_{x_1x_1} + \cdots + f_{x_dx_d}\right) \\ &= \int_{\mathbb R^d} \varphi \Delta f. \end{align}

So $\Delta\varphi$ as defined is just a functional. We say that $\Delta \varphi \in L^2(\mathbb R^d)$ in the sense of distribution, if this functional is given by integration of a $L^2$ function: there is $g\in L^2(\mathbb R^d)$ so that

$$ (\Delta \varphi) (f) = \int _{\mathbb R^d} \varphi \Delta f = \int_{\mathbb R^d } g f, \ \ \ \forall f\in C^\infty_c(\mathbb R^d).$$

  • 1
    $\begingroup$ Thanks! So, basically $\varphi \in L^{2}$ need not to be differentiable, but one uses the analogy of integrating by parts; the $\Delta \varphi \in L^{2}$ in the sense of distributions if there exists some $g \in L^{2}$ such that $\int \varphi \Delta f = \int gf$,in which $g$ is interpreted as "$\Delta \varphi$ simbolically. $\endgroup$
    – MathMath
    Jul 15 at 17:13

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.