In the answer to this question Is the Laplacian an unbounded operator?, there is some intuition, why the Laplacian is a bounded operator on $H^2$:
"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."
But I have read on page 64 of the book "A concise course on stochastic differential equations" by Prevot/Röckner, that the laplacian can also be extended to a bounded operator from $H^1$ to $(H^1)^{\prime}$ (the dual space).
My question: On $H^1$, the second derivative is not considered in the norm, so how can it still be a bounded operator? Is it not possible to find a function with a "small" $H^1$-norm, but a "large" second derivative?
Thank you in advance!
Best,
Luke
Best Answer
To answer this question in a satisfying manner would require a little bit of background. There are two key points to make:
Extensions of such operators are much more intangible than the original operators. In its classical form, the Laplacian of a smooth $f$ is directly calculated by computing the partial 2nd derivatives of $f$. An extension of the Laplacian no longer has anything to do with derivatives (in the classical sense), but rather just happens to coincide with the usual Laplacian when restricted to differentiable functions.
The actual construction relies on integration by parts (or more generally, a weak form). Let $\Omega$ be a bounded domain. Take $u$ smooth and $v \in H_0^1(\Omega)$. Then $$ \int_{\Omega} (-\Delta u) v \, dx = \int_{\Omega} Du \cdot Dv \, dx \quad (1) $$ so taking absolute values and applying a Poincare Inequality shows that the above bilinear form is coercive, and we also have the bound $$ |\langle -\Delta u, v \rangle| \le \| Du \|_{L^2} \|v\|_{H^1} $$ This realizes $-\Delta u$ as a bounded linear functional on $H_0^1(\Omega)$ whenever $u \in C^{\infty}$. Uniqueness comes from Lax-Milgram. Next, since the norm is controlled by $\|Du\|_{L^2}$ and $C^{\infty}$ is dense in $H_0^1$, then $-\Delta$ itself can be extended to $H_0^1(\Omega)$ such that $-\Delta u$ is simply defined to be the element of $H^{-1}$ with the action given by (1).