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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?

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    $\begingroup$ 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
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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

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

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    $\begingroup$ 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

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