Showing that the $L^2$ inequality does not hold

Question

I'm having difficulty with the following problem. Let $U$ be the unit ball in $\mathbb{R}^2$. Suppose $f,g \in C^\infty(\bar{U})$ satisfy,

$$ \begin{cases} \Delta f = \Delta g & x \in U \\ f = 0 & x \in \partial U\end{cases} $$

By Poincare's inequality, there exists a constant $C$ such that $||f||_{H^1(U)} \leq C||g||_{H^1(U)}$. Prove, however, that there cannot exist a constant $M$ (independent of $f,g$) such that,

$$||f||_{L^2(U)} \leq M||g||_{L^2(U)} $$

A hint given in the problem suggests to take sequences $\{f_n\}$ and $\{g_n\}$ satisfying $f_n - g_n = 1$.

I'm not sure what the hint is saying. Do we need to find sequences $\{f_n\}$ and $\{g_n\}$ explicitly? Or do we take any arbitrary sequences $\{f_n\}$ and $\{g_n\}$ satisfying the BVP and $f_n-g_n =1$?

Answer 1

Let's consider the functions $f_n(x)=1-|x|^n$, $g_n=-|x|^n$, and record the integrals

1. $\|f_n\|_2^2=2\pi (\frac{1}{2}+\frac{1}{2n+2}-\frac{2}{n+2})$.
2. $\|g_n\|_2^2=2\pi(\frac{1}{2n+2})$

Clearly $\Delta f_n=\Delta g_n$ and $f|_{\partial U}=0$. Then by our hypothesis,

$$ \|f_n\|_2\le M\|g_n\|_2. $$

This is clearly absurd, so the inequality cannot be true.

In the $H_1$ case, the derivatives of $g_n$ blow up, so the inequality is restored.