8
$\begingroup$

The proof of Morrey's Inequality in page 266 of Evans's PDE book really puzzles me a lot. I cannot get the general idea of the proof.

I know a simple proof just in the case of $n=1$:

For any $u(x)\in C_0^{\infty}(a,b)$, we have $u(y)-u(x)=\int_x^y u'(t)dt$.

$|u(y)-u(x)|\le\int_x^y |u'(t)|dt\le ||u'||_{L^p}|y-x|^{\frac{p-1}{p}}$.

Then we can get the estimate of $[u]_{C^{0,\gamma}}$.

Q1: Can this proof be extended to the case of $n>1$?

Q2: How to comprehend the general idea of the subtle proof in Evans's book?

$\endgroup$

1 Answer 1

15
$\begingroup$

The proof does extend to higher dimensions, but the application of the fundamental theorem is done with directional derivatives of $u$ at $x$. The idea behind the proof is to examine how much the value of $u(x)$ varies from the average value of $u$ on a ball $B$ that contains $x$. To start, let
$B = B(z,r)$ be a ball of radius $r$ and let $x \in B$. Then $$|u(x) - \bar u_B| = \left|u(x) - \frac{1}{|B|} \int_{B} u(y) \, dy \right| \le \frac{1}{|B|} \int_{B} |u(x) - u(y)| \, dy.$$ It is helpful to re-center the integral at $x$. Since $B \subset B(x,2r)$ and $|B| = 2^{-n}|B(x,2r)|$ we have $$\frac{1}{|B|} \int_{B} |u(x) - u(y)| \, dy \le \frac{2^n}{|B(x,2r)|} \int_{B(x,2r)} |u(x) - u(y)| \, dy.$$

The problem at this point is to estimate the last integral using the gradient of $u$. It is convenient to switch to spherical coordinates. Write $S^{n-1}$ for the unit sphere in $\mathbb R^n$. Points $y \in B(x,2r)$ have the form $y = x + t \omega$, where $0 \le t < 2r$ and $\omega \in S^{n-1}$. Moreover, the change of variables formula yields $$\int_{B(x,2r)} |u(x) - u(y)| \, dy = \int_0^{2r} \int_{S^{n-1}} |u(x) - u(x+t\omega)| \, dt d\omega.$$ Fix $\omega \in S^{n-1}$. The fundamental theorem of calculus applied to the function $\phi(s) = u(x + s\omega)$ states that $$u(x+t\omega) - u(x) = \int_0^t Du(x + s\omega) \cdot \omega \, ds, \quad 0 < t < 2r.$$ Since $|\omega| = 1$ this leads to $$|u(x) - u(x+t\omega)| \le \int_0^t |Du(x + s\omega)| \, ds \le \int_0^{2r} |Du(x + s\omega)| \, ds.$$

At this point we can go back to Cartesian coordinates to get \begin{align*}|u(x) - \bar u_B| &\le \frac{2^n}{|B(x,2r)|} \int_0^{2r} \int_{S^{n-1}} \int_0^{2r} |Du(x + s\omega)| \, ds d\omega dt \\ &= \frac{2^n}{|B(x,2r)|} 2r \int_{S^{n-1}} \int_0^{2r} |Du(x + s\omega)| \, ds d\omega \\ &= \frac{2^{n+1}r}{|B(x,2r)|} \int_{B(x,2r)} |Du(y)| \, dy.\end{align*}

Holder's inequality implies $$\int_{B(x,2r)} |Du(y)| \, dy \le \left( \int_{B(x,2r)} |Du(y)|^p \, dy \right)^{1/p} |B(x,2r)|^{1/p'}$$ and, since $|B(x,2r)| = c_n 2^nr^n$, $$|u(x) - \bar u_B| \le \frac{2}{c_n}r^{1-n} \left( \int_{B(x,2r)} |Du(y)|^p \, dy \right)^{1/p} c_n^{1/p'} r^{n/p'} \le 2 c_n^{-1/p} r^{1-n/p} \|Du\|_{p}.$$

Finally, if $x,y \in \mathbb R^n$ let $B$ be a ball with diameter (barely larger than) $|x-y|$ containing both $x$ and $y$ to obtain $$|u(x) - u(y)| \le |u(x) - \bar u_B| + |u(y) - \bar u_B| \le 2^{1+n/p} c_n^{-1/p} |x-y|^{1-\frac np} \|Du\|_p.$$

$\endgroup$
1
  • $\begingroup$ Hello, would you have a reference where this calculation is done?^^ (with the explicit constant) $\endgroup$
    – Huggy
    Oct 4, 2023 at 15:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .