20210515-Morrey's_inequality

Morrey's inequality

Statement of the result

Given f:\mathbb R^n\to\mathbb R $f:\mathbb R^n\to\mathbb R$ , Morrey's inequality asserts that

\int_{\mathbb R^n}|\nabla f|^p dx<\infty \text{ for some }p>n,

\int_{\mathbb R^n}|\nabla f|^p dx<\infty \text{ for some }p>n,

then

\sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|^{1-n/p}}\le C_{n,p} \left(\int_{\mathbb R^n} |\nabla f|^p dx\right)^{1/p}

\sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|^{1-n/p}}\le C_{n,p} \left(\int_{\mathbb R^n} |\nabla f|^p dx\right)^{1/p}

Suppose for simplicity that f $f$ is dimensionless, then the W^{1,p} $W^{1,p}$ norm of f $f$ has dimension

\left[\|f\|_{W^{1,p}}\right]={\rm L}^{n/p-1}.

\left[\|f\|_{W^{1,p}}\right]={\rm L}^{n/p-1}.

On the other hand, the Holder norm of f $f$ has dimension

\left[\|f\|_{C^{0,\alpha}}\right]={\rm L}^{-\alpha}.

\left[\|f\|_{C^{0,\alpha}}\right]={\rm L}^{-\alpha}.

In an infinite there are no dimensions involved, hence the constant C_{n,p} $C_{n,p}$ is dimensionless. Thus, dimensional consistency enforces \alpha=1-n/p $\alpha=1-n/p$ .

Another argument is based on rescaling: if we rescale lengths by a factor L $L$ , the right-hand side of the inequality scales as L^{n/p}-1 $L^{n/p}-1$ , while the left-hand side scales as 1/L^\alpha $1/L^\alpha$ .

The ingredients of the proof of Morrey's inequality are:

The Fundamental Theorem of Calculus, along with Fubini's theorem, which are used to derive the following inequality:^[1]

\int_{B(x,r)}|f(y)-f(x)| dy\le \frac{r^n}{n}\int_{B(x,r)}\frac{|\nabla f(y)|}{|x-y|^{n-1}}dy

\int_{B(x,r)}|f(y)-f(x)| dy\le \frac{r^n}{n}\int_{B(x,r)}\frac{|\nabla f(y)|}{|x-y|^{n-1}}dy

Holder's inequality, along with the observation that

\int_{B(0,1)}\frac 1 {|x|^\alpha} dx<\infty \text{ for }\alpha<n.

\int_{B(0,1)}\frac 1 {|x|^\alpha} dx<\infty \text{ for }\alpha<n.

The triangle inequality, which is used to deduce that if r>|x-y| $r>|x-y|$ and U=B(x,r)\cap B(y,r) $U=B(x,r)\cap B(y,r)$ then

|f(x)-f(y)|\le -\hskip{-1em}\int_{U}|f(x)-f(z)|dz+|f(y)-f(z)|dz

|f(x)-f(y)|\le -\hskip{-1em}\int_{U}|f(x)-f(z)|dz+|f(y)-f(z)|dz

Here we provide some details on the derivation of the inequailty \displaystyle \int_{B(x,r)}|f(y)-f(x)| dy\le \frac{r^n}{n}\int_{B(x,r)}\frac{|\nabla f(y)|}{|x-y|^{n-1}}dy $\displaystyle \int_{B(x,r)}|f(y)-f(x)| dy\le \frac{r^n}{n}\int_{B(x,r)}\frac{|\nabla f(y)|}{|x-y|^{n-1}}dy$ . This inequality is obtained by first showing that \forall 0<s<r: \int_{w\in\partial B(0,1)}|f(x+sw)-f(x)| dS\le \int_{t\in(0,s)}\int_{w\in\partial B(0,1)}|\nabla f(x+tw)|dS dt $\forall 0<s<r: \int_{w\in\partial B(0,1)}|f(x+sw)-f(x)| dS\le \int_{t\in(0,s)}\int_{w\in\partial B(0,1)}|\nabla f(x+tw)|dS dt$ . Then bound with \forall 0<s<r: \int_{w\in\partial B(0,1)}|f(x+sw)-f(x)| dS\le\int_{t\in(0,\color{red}r\color{black})}\int_{w\in\partial B(0,1)}|\nabla f(x+tw)|dS dt $\forall 0<s<r: \int_{w\in\partial B(0,1)}|f(x+sw)-f(x)| dS\le\int_{t\in(0,\color{red}r\color{black})}\int_{w\in\partial B(0,1)}|\nabla f(x+tw)|dS dt$ . Integrate again to get the boxed equation. ↩︎