Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 23F (Analysis of Functions)

Choose a nonzero $\varphi \in C_c^{\infty }({\mathbb{R}}^n)$ supported in a ball of radius $1$, and letTranslation preserves both the $L^2$ norm and the weak-derivative norms, sofor every $k$. Thus $({\varphi }_k)$ is bounded in the Sobolev space $H^1({\mathbb{R}}^n)$.

Distinct translates have disjoint supports. Hence, for $j\ne k$,No subsequence is Cauchy in $L^2$, so no subsequence converges there. The embedding $H^1({\mathbb{R}}^n)\hookrightarrow L^2({\mathbb{R}}^n)$ is therefore not a compact operator. This is the noncompactness of a Sobolev embedding by translation caused by loss of compactness at infinity.

Regard the space of continuous functions vanishing at infinity $C_0({\mathbb{R}}^n)$ as a subspace of $L^{\infty }({\mathbb{R}}^n)$. Point evaluation at the origin,is a bounded linear functional of norm one on this subspace because a continuous function's supremum and essential supremum agree. By the Hahn-Banach theorem, ${\ell }_0$ extends to a bounded linear functional

Suppose that some $g\in L^1({\mathbb{R}}^n)$ represented this extension:Choose $\phi \in C_c({\mathbb{R}}^n)$ with $0\le \phi \le 1$, $\phi (0)=1$, and support in the unit ball, and set ${\phi }_k(x)=\phi (kx)$. Then $\ell ({\phi }_k)=1$ for every $k$. On the other hand, ${\phi }_k(x)\to 0$ for almost every $x$ and $|{\phi }_{k}g|\le |g|$. The dominated convergence theorem givesa contradiction. Thus $\ell$ is a singular functional on L infinity and has no $L^1$ density.

We prove the estimate first for $u$ in the Schwartz space. By the Fourier inversion theorem and the Cauchy-Schwarz inequality,The integral is finite because $n>n/2$.

For $0<\alpha \le 1$, the elementary boundgivesThe last integral is finite near zero for every $\alpha >0$ and at infinity exactly when $2\alpha <n$. Choose, for example, $\alpha =1/4$, which works for every $n\ge 1$. We obtainDensity of the Schwartz space in $H^n$ extends the estimate and supplies a unique Hölder-continuous representative. Thereforecontinuously. This is the Fourier proof of Hölder regularity from a Sobolev norm.

Use the centered Hardy-Littlewood maximal functionIf $x>0$, choosing $0<r<x$ makes the whole interval lie in the positive half-line, so $M\theta (x)=1$. If $x=0$, every centered interval has exactly half its length in the positive half-line, so $M\theta (0)=1/2$.

If $x<0$, the average is zero for $r\le -x$. For $r>-x$, it isand these values tend to $1/2$ as $r\to \infty$. HenceFor the uncentered maximal-function convention, intervals extending arbitrarily far to the right make the supremum equal to $1$ at every $x$.