# Prove $\sup_x|u_x(x,t)|\le Ct^{-\frac{3}{4}}\|f\|_2$ for all $t>0$.

Let $u$ be a bounded solution to the heat equation $u_t-u_{xx}=0$ in $-\infty<x<\infty,t>0$ with $u(x,0)=f(x)$ with $f\in L^2(\Bbb R)$. Prove that there is a constant $C>0$, independent of $u$, such that $$\sup_x|u_x(x,t)|\le Ct^{-\frac{3}{4}}\|f\|_2$$ for all $t>0$.

My attempt: The solution to the heat equation is $u(x,t)=\frac{1}{\sqrt{4\pi t}}\int e^{-\frac{|x-y|^2}{4t}}f(y)dy$. Then use the Holder's inequality, $u(x,t)=\frac{1}{\sqrt{4\pi t}}\int e^{-\frac{|x-y|^2}{4t}}f(y)dy\le \frac{1}{\sqrt{4\pi t}}(\int |e^{-\frac{|x-y|^2}{4t}}|^2dy)^{\frac{1}{2}}\|f\|_2$. Then can I take the differentiation inside the integral?

I don't know how to proceed, could someone kindly help? Thanks!

• You'll want to differentiate first and then use Holder's inequality. – Jose27 Aug 5 '15 at 1:35

## 1 Answer

Given: $$u(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{\mathbb{R}}e^{-\frac{|x-y|^2}{4t}}f(y)\,dy = \frac{1}{\sqrt{4\pi t}}\int_{\mathbb{R}}e^{-\frac{y^2}{4t}}f(x+y)\,dy$$ we also have (since $u=f*g$ implies $u'=f'*g$): $$u_x(x,t) = -\frac{1}{2t\sqrt{4\pi t}}\int_{\mathbb{R}}y\,e^{-\frac{y^2}{4t}}f(x+y)\,dy$$ hence by the Cauchy-Schwarz inequality: $$|u_x(x,t)|\leq \frac{\|f\|_2}{4 t^{3/2}\sqrt{\pi}}\sqrt{\int_{\mathbb{R}}y^2 e^{-\frac{y^2}{2t}}\,dy}=\frac{\|f\|_2}{\left(2^7\pi\, t^3\right)^{\frac{1}{4}}}\leq \frac{t^{-3/4}}{2\sqrt{5}}\,\|f\|_2.$$