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This exercise comes from a past PDE qual problem. Assume $u(x,t)$ solves $$ \left\{\begin{array}{rl} u_{t}-\Delta u=0&\text{in}\mathbb{R}^{n}\times(0,\infty)\\ u(x,0)=g(x)&\text{on}\mathbb{R}^{n}\times\{t=0\}\end{array}\right. $$ and $g$ is Holder continuous with continuity mode $0<\delta\leq1,$ that is $$|g(x)-g(y)|\leq|x-y|^{\delta}$$ for every $(x,y)\in\mathbb{R}^{n}$. Prove the estimate $$|u_{t}|+|u_{x_{i}x_{j}}|\leq C_{n}t^{\frac{\delta}{2}-1}.$$

I have quite a few pages of scratch work in trying to prove this estimate, but I have not been able to arrive at a situation where it is even obvious how to exploit the Holder continuity of $g$. Because of translation invariance in space, we can just prove it for the case $x=0$, so that at least simplifies some things. But again, there is a key observation that has apparently eluded me, and a hint would be appreciated!

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Sorry, yes! Corrected. – Taylor Martin Feb 24 '13 at 5:31
up vote 4 down vote accepted

This calls for a scaling argument.

As you noticed, it suffices to consider $x=0$. Replace $g$ with $g-g(0)$; this does not change the derivatives. Now we know that $$|g(x)|\le |x|^\delta\tag1$$

Prove an estimate of the form $$|u_{t}(0,1)| + |u_{x_ix_j}(0,1)| \le C_n\tag2$$ This requires writing the derivatives as convolutions of $g$ with the derivatives of $\Phi$, and a rough estimate such as $|g(x)|\le 1+|x|$.

For every scaling factor $\lambda$ the function $u_\lambda=\lambda^{-\delta} u(\lambda x,\lambda^2 t)$ solves the heat equation with the initial data $g_\lambda(x)=\lambda^{-\delta} g(\lambda x)$. Notice that $g_\lambda$ also satisfies (1). Therefore, $u_\lambda$ satisfies (2). All of a sudden, we're done.

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Thanks 5pm; didn't even think about argument scaling! – Taylor Martin Feb 24 '13 at 7:11
I will post a fully fleshed proof in a little later in case anyone else encounters a similar question. – Taylor Martin Feb 24 '13 at 7:30

I don't think this is true, as it stands. The Cauchy problem has nontrivial solutions $u$ with $u(x,0) = 0$. If this statement is true, it would imply that $u_t$ is bounded and in particular that $u(\cdot, t)$ is bounded for each $t$. But this is not true for such $u$; indeed, Tychonoff's uniqueness theorem says they grow faster than $e^{c|x|^2}$.

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I guess it is implicitly assumed that $u$ is a "physically relevant" solution, i.e., convolution with the heat kernel. – user53153 Feb 24 '13 at 5:21
Right, $u(x,t)=\frac{1}{(4\pi t){\frac{n}{2}}}\int_{\mathbb{R}^{n}}\Phi(x-y,t)g(y)\;dy$ where $\Phi$ is the fundamental solution as usual. – Taylor Martin Feb 24 '13 at 5:35

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