Cauchy Problem for Heat Equation with Holder Continuous Data 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!
 A: 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. 
A: 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 $e^{c|x|^2}$.
