Show $|\partial_{x_i}\Phi(x,t)| \leq \frac{M_2}{\sqrt{t}}\Phi(x,2t)$, and $|\partial_{x_ix_j}\Phi(x,t)|\leq \frac{M_3}{t}\Phi(x,2t)$

Question

Recall that If $0<\alpha<1$, $0\leq \beta$, then there exists $M(\alpha,\beta)>0$ so that $$z^\beta e^{-z}\leq Me^{-\alpha z}$$ for all $z\geq 0$. Show $|\partial_t\Phi(x,t)| \leq \dfrac{M_1}{t}\Phi(x,2t)$, $|\partial_{x_i}\Phi(x,t)| \leq \dfrac{M_2}{\sqrt{t}}\Phi(x,2t)$, and $|\partial_{x_ix_j}\Phi(x,t)|\leq \dfrac{M_3}{t}\Phi(x,2t)$

Proof: Recall that $\Phi(x,t)=\dfrac{1}{(4\pi t)^{n/2}}\exp\left[-\dfrac{|x|^2}{4t}\right]$ for $x\in \mathbb{R}^n$ and $t>0$.

First, we will show that $|\partial_t\Phi(x,t)|\leq \dfrac{M_1}{t}\cdot \Phi(x,2t)$. By some calculus and part (a), \begin{equation*} \begin{aligned} |\partial_t\Phi(x,t)| & = \left| \dfrac{1}{(4\pi)^{n/2}}\cdot \dfrac{-n}{2}t^{-(n+2)/2} \exp\left[-\dfrac{|x|^2}{4t}\right] + \dfrac{1}{(4\pi t)^{n/2}} \exp\left[-\dfrac{|x|^2}{4t}\right] \cdot \dfrac{|x|^2}{4t^2}\right| \\ & = \dfrac{1}{2t(4\pi t)^{n/2}} \exp\left[-\dfrac{|x|^2}{4t}\right] \cdot \left| -n + \dfrac{|x|^2}{2t}\right| \\ & = \dfrac{1}{t}\cdot \dfrac{1}{(8\pi t)^{n/2}} \exp\left[-\dfrac{|x|^2}{4t}\right] \cdot \left| -n + \dfrac{|x|^2}{2t}\right| \\ & \leq \dfrac{1}{t}\cdot \dfrac{1}{(8\pi t)^{n/2}} \exp\left[-\dfrac{|x|^2}{8t}\right] \cdot M_1 \\ & = \dfrac{M_1}{t}\cdot \Phi(x,2t) \end{aligned} \end{equation*} with $\alpha=1/2$ and $M_1=\left| -n + \dfrac{|x|^2}{2t}\right|>0$.

Next, we will show that $|\partial_{x_i}\Phi(x,t)| \leq \dfrac{M_2}{\sqrt{t}}\Phi(x,2t)$. By some calculus and part (a), \begin{equation*} \begin{aligned} |\partial_{x_i}\Phi(x,t)| & = \left| \dfrac{1}{(4\pi t)^{n/2}} \exp\left[-\dfrac{|x|^2}{4t}\right] \cdot -\dfrac{2|x|}{4t} \right| \\ & = \dfrac{1}{(4\pi t)^{n/2}} \exp\left[-\dfrac{|x|^2}{4t}\right] \cdot \dfrac{|x|}{2t} \\ & = \dfrac{1}{\sqrt{t}} \cdot \dfrac{1}{(8\pi t)^{n/2}} \exp\left[-\dfrac{|x|^2}{4t}\right] \cdot \dfrac{|x|}{\sqrt{t}} \\ & \leq \dfrac{1}{\sqrt{t}} \cdot \dfrac{1}{(8\pi t)^{n/2}} \exp\left[-\dfrac{|x|^2}{8t}\right] \cdot M_2 \\ & = \dfrac{M_2}{\sqrt{t}}\Phi(x,2t) \end{aligned} \end{equation*} with $\alpha=1/2$ and $M_2=\dfrac{|x|}{\sqrt{t}}>0$.

I know that my derivative of $|x|$ is wrong what is it suppose to be?

Also for the third inequality, do I just take the derivative twice?

Answer 1

The partial derivative you want is $$\frac{\partial |x|^2}{\partial x_i}=2x_i.$$ You just need the inequality $|x_i|\le|x|$.

For the second order partial derivative, $$\frac{\partial^2}{\partial x_i\partial x_j} \exp\left(-\dfrac{|x|^2}{4t}\right)=\frac{e^{-\frac{|x|^2}{4t}}}{2t}\Big(\frac{x_ix_j}{2t}-\delta_{i,j}\Big),$$ where $\delta_{i,j}$ is the Kronecker delta function. Here you need the inequality $$2x_ix_j\le x_i^2+x_j^2\le |x|^2$$ to furnish your final inequality.

Your characterization of the $M_i$ is wrong. They are constants, independent of $(t,x)$.