Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 23H (Analysis of Functions)

$H^s({\mathbb{R}}^d)=\{u\in {\mathcal{S}}^{\prime }:\int (1+|\xi {|}^2{)}^s|\hat{u}(\xi ){|}^{2}d\xi <\infty \}$. For integer $s\ge 0$, an equivalent norm is ${\sum }_{|\alpha |\le s}\Vert D^{\alpha }u{\Vert }_2^2$.

If $s>d/2$, then $H^s({\mathbb{R}}^d)\hookrightarrow L^{\infty }$. Indeed Fourier inversion and Cauchy-Schwarz give $|u(x)|\le C\Vert (1+|\xi {|}^2{)}^{s/2}\hat{u}{\Vert }_2\Vert (1+|\xi {|}^2{)}^{-s/2}{\Vert }_2$, and the second factor is finite exactly when $s>d/2$.

$H_0^1(\Omega )$ is the closure of $C_c^{\infty }(\Omega )$ in $H^1(\Omega )$.

For smooth compactly supported $u$, multiply by $1-\phi (nx)$. The removed term has $L^2$ norm $O(n^{-3/2})$, while its gradient terms are $O(n^{-1/2})$ in dimension three. Thus it tends to zero in $H^1$. Density then proves $H_0^1({\mathbb{R}}^3\setminus \{0\})=H^1({\mathbb{R}}^3)$.

No. Point evaluation is continuous on $H^1(\mathbb{R})$, and every limit of functions supported away from zero has trace $u(0)=0$. A smooth function with nonzero value at zero belongs to $H^1$ but not to this $H_0^1$.