Find the upper bound of $|\frac{d^2}{dx^2}(e^{-x^2})|\leq6$ in $x\in[0,1]$

Question

Show by finding the second derivative of $e^{-x^2}$ that for all $x\in[0,1]$

$$|\frac{d^2}{dx^2}(e^{-x^2})|\leq6$$

(if you obtain a better bound, that is fine )

My Try

let, $f(x)=e^{-x^2}$

$f''(x)=2e^{-x^2}(2x^2-1)$

From the plot in wolfram alpha I could see that the bound is 2/e or 0.73. But how to solve this analytically?

Answer 1

We want to find $\max\{(\exp(-x^2))''\}$ so we evaluate its stationary points. This occurs when $$(\exp(-x^2))'''=(2(2x^2-1)\exp(-x^2))'=4x(3-2x^2)\exp(-x^2)=0$$ so $x=0,\sqrt{3/2}$. At the former, $(\exp(-x^2))''=-2$ and at the latter, $(\exp(-x^2))''=4\exp(-3/2)<1$ which is a much tighter bound than $6$.

Answer 2

Just use the triangle inequality. The given bound holds for all $x\in\mathbb{R}$, not just $x\in[0,1]$.

You know $$ f''(x)=2\cdot[2x^2\exp(-x^2)-\exp(-x^2)] $$ So \begin{align*} \lvert f''(x)\rvert &\leq 2\cdot\bigg[\sup_{x\in\mathbb{R}}2x^2\exp(-x^2)+\sup_{x\in\mathbb{R}}\exp(-x^2)\bigg]\\ &= 2\cdot\left[2\sup_{x\in\mathbb{R}}\frac{x^2}{\exp(x^2)}+1\right]\\ &\leq 2\cdot\left(2\sup_{x\in\mathbb{R}}\frac{x^2}{1+x^2}+1\right)\\ &\leq 2\cdot(2\cdot 1+1)=6. \end{align*}

For $x\in[0,1]$, you can obtain the bound a little more quickly using $x^2\leq 1$ and $\exp(-x^2)\leq 1$.