Is the Laplacian operator $\Delta: C^2([a,b],\mathbb{R}) \to L^2(a,b)$ unbounded? Here $a,b \in\mathbb{R}$, $C^2([a,b],\mathbb{R})$ associated with $L^2$ norm.
If yes, how can we "modify" these spaces to obtain a bounded operator?
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$\begingroup$ An illustrative example of what, why its unbounded? $\endgroup$– Chris RamseyCommented Sep 21, 2015 at 19:30
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$\begingroup$ @Chris Ramsey: To make it clearer, I have edited my question. Thanks. $\endgroup$– Akai ShuichiCommented Sep 21, 2015 at 19:39
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$\begingroup$ If $C^2$ means the space of two times continuously differentiable functions, then this operator is bounded, since it acts from $C^2[a,b]$ to $C[a,b]$, and the last space is continuously embedded into $L^2[a,b]$. Or what do you mean? $\endgroup$– Sergei AkbarovCommented Sep 21, 2015 at 19:48
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1 Answer
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$C^2([a,b])$ is not closed with respect to the $L^2$-norm. Boundedness of $\Delta$ on $C^2([a,b])$ is equivalent to boundedness on its closure, and in fact its closure is $L^2(a,b)$ itself, on which $\Delta$ is obviously unbounded, as is very easy to see considering a sequence of steeper and steeper functions with smaller and smaller support that converge in $L^2(a,b)$ but whose Laplacians become larger and larger.
In order to make the Laplacian bounded there are two main ways: you can endow the domain $C^2([a,b])$ with the classical $C^2$-norm (or even just with the easier norm $\||f|\|_\infty:=\|f\|_\infty+\|f''\|_\infty$) and replace the target space $L^2(a,b)$ by $C([a,b])$; or else you can consistently turn to the Hilbert space theory by endowing $C^2([a,b])$ with the norm $\||f|\|_2:=\|f\|_2+\|f''\|_2$ and closing it up with respect to this norm -- thus obtaining as new domain the Sobolev space $H^2(a,b)$ (you can then keep $L^2(a,b)$ as target space).
