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What's the point of Hölder spaces (and parabolic Hölder spaces)? I can understand when solving some PDE say $$u_t = au_x + bu_{xx}$$ you may want the solution to lie in $C^{2,1}$ (indexed in space then time) but why do people sometimes want something like $C^{2+\alpha, 1+\alpha}$?

Also being in a Hölder space is stronger than being in the same continuous (non-Hölder) space so this is not very useful for solving PDEs either when you want as big a space as possible to show existence of solutions (I think).

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People don't necessarily want to do that, that's what they can prove resp work with. If $Lu = f$ with continuous $f$ and second order elliptic $L$, then in general $u$ is not in $C^2$. If, however, $f\in C^{0, \alpha}$, then it is often possible to show $u\in C^{2, \alpha}$. Similarly for parabolic equations. It just turns out that these spaces are particularly suited for PD operators, while the $C^k$ spaces are not.

And many will admit that this is not always convenient, since, for example, the Hölder spaces refuse to be separable.

Edit/added: (on a more technical note, it turns out that the Newton potential of a function $f$, defined by $$N(f)(x):=C(n) \int \frac{f(y)}{|x-y|^{n-2}} dy$$ (with a constant $C(n)$ depending on the dimension) is not a $C^2$ function if $f$ is merely continuous. The Newton potential is the basic building block for the Greens function of the Laplacian. (This is only correct if $n>2$, for $n=2$ a $\log|x-y|$ term replaces the $1/|x-y|^{n-2}$ term). $N(f)$ is, however, in $C^{2,\alpha}$ if $f\in C^{0,\alpha}$. Again, similar observations hold true in the parabolic case. A counterexample for the case $f\in C^0$ is given in exercise 4.9 in Gilbarg and Trudingers 'Elliptic Partial Differential Equations of Second Order').

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I have a doubt , is there a relation between $C^2$ and $C^{2,\alpha}$ ie some kind of inclusion or something ? –  Theorem Jun 16 '12 at 16:36
    
Sure, $C^{2,\alpha}\subset C^2$. –  user20266 Jun 16 '12 at 16:42
    
Can you elaborate a bit about parabolic Holder spaces? Their definition is slightly different and I find it all a bit confusing. –  Court Jun 17 '12 at 18:48
    
@blackcat unfortunately, no. I do know that the results from elliptic theory influence the parabolic theory in the way indicated, but I would not venture to comment on parabolic equations in detail, that's not my area of expertise. Sorry. –  user20266 Jun 17 '12 at 19:15

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