"The Laplacian is an unbounded operator": I read this in a book. But on Wikipedia it says:

The Laplace operator $$\Delta:H^2({\mathbb R}^n)\to L^2({\mathbb R}^n) \,$$ (its domain is a Sobolev space and it takes values in a space of square integrable functions) is bounded.

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    $\begingroup$ Whether it's bounded or unbounded depends on what topologies you choose. When you endow the domain $H^2(\mathbb{R}^n)$ with the subspace topology induced by $L^2(\mathbb{R}^n)$, the operator is unbounded. $\endgroup$ – Daniel Fischer Jul 12 '15 at 9:42
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    $\begingroup$ What is $H^2(\mathbb R^n)$? $\endgroup$ – Ooker Nov 9 '17 at 0:53

For the sake of this question having an answer ...

Functions come with domains. Often we forget this, as the domain is fixed. In functional analysis, however, half the battle and much of the art is finding the correct domain.

The Laplacian on a domain $\Omega$ is ordinarily defined as mapping some linear set of smooth functions into $L^2(\Omega)$. Boundedness/unboundedness is a statement about how the operator relates to a norm. When we regard the domain as a subspace of $L^2(\Omega)$, thereby endowing it with the $L^2$ norm, then the Laplacian is an unbounded operator.

When we regard the domain as a subspace of $H^2(\Omega)$, however, endowing it with the $H^2$ norm, the Laplacian is a bounded operator.

Some intuition. The $L^2$ norm doesn't know anything about how smooth a function is; when the Laplacian is applied to a function, it can "uncover" hidden "roughness." However the $H^2$ norm does take into account how smooth the function is --- a function with very large second derivatives may have a small $L^2$ norm, but will have a very large $H^2$ norm.

So while the Laplacian may uncover hidden roughness in the $L^2$ norm, and so take small vectors and arbitrarily lengthen them, those vectors are already very long in the $H^2$ norm, and so the Laplacian is bounded.


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