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There a lot of function spaces and would be nice if somebody can correct me if I am wrong in comparing a few. I want to compare $C^2,L^2,W^{2,2}$ (continuous up to third derivative, Hilbert space of square integrable functions and Sobolev with second weak derivative in $L^2$) of functions defined on a compact set. I think $C^2\subset W^{2,2}\subset L^2$. Is that the right inclusion? thanks!

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In the case of open bounded sets of $\mathbb{R}^n$, yes, this is true. – matgaio Apr 25 '12 at 20:09
I guess you cold take a look at the Evans's "Partial Differential Equations", chapter 5. There you will find lots of inclusions of space of functions as linear bounded (and sometimes compact) operators between these spaces. – matgaio Apr 25 '12 at 20:11
@matgaio Is it true for open bounded $\Omega$ (not closed bounded)? $C^2\subset W^{2,2}$ without assuming any integrability for the functions in $C^2$? – Jun 13 at 3:46
up vote 1 down vote accepted

Yes, this is correct. Note that a continuous function on a compact set is square integrable (because it is bounded and a compact set has finite measure), and if the classical derivative exists then the weak derivative exists and they are equal. Thus $C^2$-functions are twice weakly differentiable and the derivatives are again square integrable because they are continuous. In other words, $C^2 \subset W^{2,2}$. Functions in $W^{2,2}$ are square integrable by definition, so we have $W^{2,2} \subset L^2$.

An interesting question for you to ponder is: are the inclusion maps continuous? That is: is the map $f \mapsto f$ bounded from $C^2$ to $W^{2,2}$ and from $W^{2,2}$ to $L^2$?

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