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Consider the heat equation on $\mathbb{R}$ $$ u_t=u_{xx} $$ with boundary conditions $u(0,x)=g(x)$.

It is well-known that even if the function $g$ is "very bad'' (say, only bounded but not continuous), the function $u(t,\cdot)$ would be from class $C^{\infty}$even for very small $t$.

My question is whether it is possible to quantify this estimate. Namely, assuming that $g\in C^\gamma$ for $\gamma\in(0,1)$, I wonder whether one can prove something like this $$ \|u(t,\cdot)\|_{C^1}\le t^{-\lambda}\|g\|_{C^\gamma} $$ for some $\lambda>0$?

Thanks!

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    $\begingroup$ Yes. Just differentiate the heat kernel $\alpha$ times, and you can have it with $\alpha=0$. $\endgroup$ Nov 19, 2015 at 3:01
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    $\begingroup$ Try interpolate from $C^\alpha$ bound and $C^1$ bound, for instance. $\endgroup$
    – Fan Zheng
    Nov 19, 2015 at 3:04
  • $\begingroup$ @AlexandreEremenko , thanks for your comment. If $\alpha$ is integer, then it is indeed possible. But what to do if $\alpha$ is not integer? $\endgroup$
    – Oleg
    Nov 19, 2015 at 9:11
  • $\begingroup$ @Oleg: This is not important: the estimate is true for ALL alpha. So use it with some integer greater than your alpha. $\endgroup$ Nov 19, 2015 at 20:36
  • $\begingroup$ @AlexandreEremenko: this is not so clear to me. Say $\gamma=1/2$ and $\alpha=1$, how can I get the desired estimate (with $\lambda=(\alpha-\gamma)/2$ as suggested by Jean)? $\endgroup$
    – Oleg
    Nov 20, 2015 at 2:28

2 Answers 2

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Yes, you can. But things are not completely elementary. You need the following ingredients:

1) The second derivative generates a (not strongly continuous!) semigroup on $L^\infty(\mathbb R)$

2) This semigroups is analytic and hencce maps the ambient space into the domain of any power of its generator.

3) Such domains are known to be imbeddable in spaces of Hölder functions and hence the semigroup induces "restricted" semigroups on $C^\alpha(\mathbb R)$, for which your estimates can then be proved to hold.

You can find all this (and much more) in Chapter 3 of Lunardi's nice monograph Analytic Semigroups and Optimal Regularity in Parabolic Problems (Birkhäuser 1995).

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  • $\begingroup$ Delio, thanks for your answer. I have checked the Chapter 3 of this book but unfortunately I have not found the statement. The closest that I have found was Theorem 5.1.2 that proves the statement for the cases $\alpha=\gamma$ or $\gamma=0$. Could you please provide a more specific reference? Thanks! $\endgroup$
    – Oleg
    Nov 24, 2015 at 22:47
  • $\begingroup$ @Oleg Corollary 3.1.21, to begin with, which states that the operator is sectorial (and hence an analytic semigroup generator) on $L^\infty(\mathbb R^n)$ with domain embedded in suitable spaces $C^{1,\alpha}$. Your estimate then follows from Proposition 2.2.9. $\endgroup$
    – DeM
    Nov 25, 2015 at 7:45
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If you write the heat equation $u_t=\nu\ u_{xx}$ with $\nu$ a physical constant of dimension (length)$^2$ (time)$^{-1}$ -- considering $t$ as time and $x$ as length --, you can see that, for dimensional reasons, only $\lambda=\frac{\alpha-\gamma}2$ is possible for all $t$, and the inequality is $$\sup\frac{|u(t,x)-u(t,y)|}{|x-y|^\alpha}\le C(\gamma,\alpha)\ (\nu t)^{-(\gamma-\alpha)/2}\ \sup \frac{|g(x)-g(y)|}{|x-y|^\gamma}$$for $\alpha\le 1$, and mutatis mutandis for all $\alpha$.

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  • $\begingroup$ Thanks a lot! But do you know if it would be possible to prove (rigorously) such an inequality? $\endgroup$
    – Oleg
    Nov 20, 2015 at 2:29

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