Write $u_Q=|Q|^{-1}\int_Qu$. Since $u-u_Q$ has <arithmetic mean> zero,
$$
\|u\|_{L^2(Q)}^2=|Q|u_Q^2+\|u-u_Q\|_{L^2(Q)}^2,
$$
and the pairwise-difference identity gives
$$
\|u-u_Q\|_{L^2(Q)}^2
=\frac1{2|Q|}\int_Q\int_Q|u(x)-u(y)|^2\,dx\,dy.
$$
First take $u$ <smooth>. Join $x$ to $y$ by changing one coordinate at a time and apply the <Cauchy-Schwarz inequality>:
$$
|u(x)-u(y)|^2
\leq n\sum_{i=1}^n
|u(x_1,\ldots,x_i,y_{i+1},\ldots,y_n)-u(x_1,\ldots,x_{i-1},y_i,\ldots,y_n)|^2.
$$
For a one-dimensional slice $v$, the <fundamental theorem of calculus> yields
$$
|v(s)-v(t)|^2\leq |s-t|\int_{\min(s,t)}^{\max(s,t)}|v'(r)|^2\,dr
\leq L\int_0^L|v'(r)|^2\,dr.
$$
Integrating the $i$th summand over $x,y\in Q$ therefore gives at most $L^{n+2}\|D_i u\|_{L^2(Q)}^2$. Hence
$$
\|u-u_Q\|_{L^2(Q)}^2\leq\frac n2L^2\|Du\|_{L^2(Q)}^2.
$$
The <density of smooth functions in a Sobolev space> extends the estimate to every $u\in H^1(\mathbb R^n)$. Thus
$$
\boxed{\|u\|_{L^2(Q)}^2\leq |Q|u_Q^2+\frac n2L^2\|Du\|_{L^2(Q)}^2.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-105/1/a/solution