<s>[INST] Calculation related to second partial derivative Problem: let $u(x,t) = \frac{1}{2\sqrt{\pi t}} e^{-\frac{x^2}{4t}}$, prove when $t \gt 0$, we have $|\frac{\partial^2u}{\partial x^2}| \le Ct^{-3/2}e^{-\lambda\frac{x^2}{t}}$, where $C \gt 0$ and $\lambda \le \frac{1}{4}$.
My Attempt:
\begin{align}
\frac{\partial u}{\partial x} &= \frac{-x}{4\sqrt{\pi}t^{3/2}}e^{-\frac{x^2}{4t}} \\
\frac{\partial^2 u}{\partial x^2} &= (\frac{x^2}{2t}-1)\frac{1}{4\sqrt{\pi}t^{3/2}}e^{-\frac{x^2}{4t}}
\end{align}
How to get the inequality from here? Can someone give me a hint? Thanks!
 [/INST] We start from the question simplification by equivalent transformations, for all real $x$ and positive $t$ and $C$.
$$\left|\frac{\partial^2u}{\partial x^2}\right| \le Ct^{-3/2}e^{-\lambda\frac{x^2}{t}}$$
$$\left|\left(\frac{x^2}{2t}-1\right)\frac{1}{4\sqrt{\pi}t^{3/2}}e^{-\frac{x^2}{4t}}\right|\le Ct^{-3/2}e^{-\lambda\frac{x^2}{t}}$$
$$\left|\frac{x^2}{2t}-1\right|\le 4\sqrt{\pi}Ce^{(1-4\lambda)\frac{x^2}{4t}}$$
After a substitution $y=\frac{x^2}{4t}$ the last inequality becomes
$$\left|2y -1\right|\le 4\sqrt{\pi}Ce^{(1-4\lambda)y}.$$
Now we see that for each $\lambda<\frac 14$, there is $C=C(\lambda)$ such that the last inequality holds for all $y\le 0$. Since $e^{(1-4\lambda)y}\ge 1+(1-4\lambda)y $, it suffices to put $C(\lambda)=\max\left(\frac 1{4\sqrt{\pi}},\frac 2{1-4\lambda}\right)$. Then, for all real $x$ and positive $t$ 
$$\left|\frac{\partial^2u}{\partial x^2}\right| \le C(\lambda)t^{-3/2}e^{-\lambda\frac{x^2}{t}}.$$
</s>