Need help to prove $\frac{dg}{dx}$ is unbounded and $\frac{d^2g}{dx^2}$ is bounded for $g\in L^2[-1,1]$

Question

Hi I am trying to prove $\frac{dg}{dx}$ is unbounded and $\frac{d^2g}{dx^2}$ is bounded for $g\in L^2[-1,1]$. So far I am thinking of taking $g$ as a quadratic polynomial and therefore $\frac{dg}{dx}$ is a linear function which is unbounded whereas $\frac{d^2g}{dx^2}$ is a constant which is bounded. I feel like I am missing something here. Any help would greatly appreciated. Thanks in advance.

Answer 1

Assume $g\in L^2([-1,1])$ is two times differentiable on $[-1,1]$ with bounded second derivative. We will show that there is no other way but for $\frac{dg}{dx}$ to be bounded.

As $\frac{dg}{dx}$ is differentiable its derivative $\frac{d^2g}{dx^2}$ is measurable and because $$ \int_{-1}^1\Big|\frac{d^2g}{dx^2}(t)\Big|\,dt\leq \Big\|\frac{d^2g}{dx^2}\Big\|_\infty \int_{-1}^1 1\,dt=2\Big\|\frac{d^2g}{dx^2}\Big\|_\infty\overset{\frac{d^2g}{dx^2}\text{ bounded}}<\infty $$ we find that $\frac{d^2g}{dx^2}$ is Lebesgue integrable (and obviously has an antiderivative). Thus the fundamental theorem of calculus for all $t\in[-1,1]$ tells us that \begin{align} \Big|\frac{dg}{dx}(t)\Big|&=\Big|\frac{dg}{dx}(-1)+\int_{-1}^t \frac{d^2g}{dx^2}(s)\,ds \Big|\\ &\leq\Big|\frac{dg}{dx}(-1)\Big|+\int_{-1}^t\Big| \frac{d^2g}{dx^2}(s)\Big|\,ds \\ &\leq \Big|\frac{dg}{dx}(-1)\Big|+\Big\|\frac{d^2g}{dx^2}\Big\|_\infty \int_{-1}^t 1\,dt\leq \Big|\frac{dg}{dx}(-1)\Big|+2\Big\|\frac{d^2g}{dx^2}\Big\|_\infty=:c\,. \end{align} But $\frac{d^2g}{dx^2}$ is bounded so $\|\frac{dg}{dx}\|_\infty=\sup_{t\in[-1,1]}|\frac{dg}{dx}(t)|\leq c<\infty$ showing that $\frac{dg}{dx}$ has to be bounded.