My overall aim is to show that, for a bounded solution $u(x,t)$ to the heat equation in $\mathbb{R}^n \times [0,T]$ with boundary condition $u(x,0) = f(x)$, $$\max |\nabla u(x,T) | \leq \frac C{\sqrt{T}} \max \left|f\right|.$$
I've been able to bound the term on the left by $$\left(\frac 1{(4 \pi t)^n 4t^2} \int_{\mathbb{R}^n} f(y)^2 e^{-\frac{|x-y|^2}{2t}} |x-y| dy\right)^{\frac12}.$$ How can I show that this term is less than or equal to the right side of my original inequality?
EDIT: I can actually get a slightly stronger bound on the left by using Cauchy-Schwarz in earlier calculations: $$\left(\frac 1{(4 \pi t)^n 4t^2} \int_{\mathbb{R}^n} f(y)^2 dy \int_{\mathbb{R}^n} e^{-\frac{|x-y|^2}{2t}} |x-y| dy.\right)^{\frac12}$$ If $y \in \mathbb{R}^1$ then the second integral becomes simply $2t$, though I'm not sure how the evaluation changes in higher dimensions. But the first integral is, of course, the $L^2$-norm for the function f (once you take into account the power of $1/2$ around the whole expression) so that's something at least.
From $$ u(x,t)=(4\,\pi\,t)^{-n/2}\int_{\mathbb{R}^n}e^{-\tfrac{|x-y|^2}{4\,t}}f(y)\,dy $$ get $$ \frac{\partial u}{\partial x_i}(x,t)=(4\,\pi\,t)^{-n/2}\int_{\mathbb{R}^n}\frac{x_i-y_i}{2\,t}e^{-\tfrac{|x-y|^2}{4\,t}}f(y)\,dy $$ and $$ \Bigl|\frac{\partial u}{\partial x_i}(x,t)\Bigr|\le\|f\|_\infty(4\,\pi\,t)^{-n/2}\int_{\mathbb{R}^n}\frac{|x_i-y_i|}{2\,t}e^{-\tfrac{|x-y|^2}{4\,t}}\,dy. $$ The change of variable $x-y=2\,\sqrt t\,z$ gives $$ \Bigl|\frac{\partial u}{\partial x_i}(x,t)\Bigr|\le\frac{\|f\|_\infty}{\sqrt t}\,(4\,\pi)^{-n/2}\int_{\mathbb{R}^n}|z_i|\,e^{-|z|^2}\,dz=C\,\frac{\|f\|_\infty}{\sqrt t},\quad t>0. $$ But from here you cannot get the inequality you want.