# The proof of Morrey's Inequality in Evans Book

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?

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 $$|u(x) - u(x+t\omega)| \le \int_0^t |Du(x + s\omega)| \, ds \le \int_0^{2r} |Du(x + s\omega)| \, ds.$$
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 $$|u(x) - u(y)| \le |u(x) - \bar u_B| + |u(y) - \bar u_B| \le 2^{1+n/p} c_n^{-1/p} |x-y|^{1-\frac np} \|Du\|_p.$$