Example that the Poincare inequality fails for higher exponent

Question

I am trying to bound a function $u$ on a ball $B\subset \mathbb R^n$ using the norm of its gradient. I found in many literatures that $\lVert u-\bar u\rVert_{L^q(B)}$$\lesssim_ {p,q,n}\lVert\nabla u\rVert_{L^p(B)}$ for $q\in[1,np/(n-p)]$ and $p\in[1,n)$, where $\bar u$ is the average of $u$ on $B$. So I guess this inequality in general fails for larger $q$. However I cannot come up with an example, and counterexamples of this inequality I found on the net only involve altering the domain rather than the exponent. Can anyone help?

Answer 1

Let $u\in C_c^\infty(B)$ have zero average and for $r<1$ consider $u_r(x):=u(x/r)$. Then $u_r\in C_c^\infty(B)$ and so, if there was a Poincaré inequality, $$ \| u_r\|_{L^q}\lesssim \|\nabla u_r\|_{L^p}. $$ Now by chain rule and a change of variables we see $\| u_r\|_{L^q}= r^{n/q}\| u\|_{L^q}$ and $\|\nabla u_r\|_{L^p}= r^{n/p-1}\|\nabla u\|_{L^p}$. Therefore $$ r^{n/q}\| u\|_{L^q}\lesssim r^{n/p-1}\|\nabla u\|_{L^p}, $$ and this holds for all $r<1$. The only way this can happen is if $n/p-n/q-1\leq 0$, and this is exactly the restriction.

This'll give you counterexamples of the following type: Assume the Poincaré inequality holds with some constant $C>0$, then by scaling any non-zero function you'll break the inequality.

If you instead want a "explicit" example you can also do the following:

Set $u(x):=|x|^\alpha$. For $(\alpha-1)p>-n$ we have finite gradient norm, i.e. $\alpha>1-n/p$ (notice that $1-n/p>-n$ so $u\in L^1$). On the other hand if $\alpha q<-n$ then we'll have $u\notin in L^q$. In other words we want $\alpha$ satisfying $$ 1-\frac{n}{p}<\alpha<-\frac{n}{q}, \quad \text{or}\quad \frac{1}{q}<-\frac{\alpha}{n}<\frac{1}{p}-\frac{1}{n}. $$ To finish, simply notice that $q>np/(n-p)$ is equivalent to $1/q<1/p-1/n$, and so for example we can take $$ \alpha=-n\left( \frac{1}{2q}+\frac{1}{2p}-\frac{1}{2n}\right), $$ the average between the two.

Since $u\in L^1$ we have that the average, being a constant, is in $L^q$. By the above then $u-\bar{u}\notin L^q$, but $\nabla u\in L^p$.