Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 23H (Analysis of Functions)

If $x=0$, the claim is immediate. Otherwise weak convergence givesBy Cauchy-Schwarz,Taking the lower limit and dividing by $\Vert x\Vert$ proves the weak lower semicontinuity of the Hilbert norm:

Strong convergence immediately implies convergence of norms by the reverse triangle inequality.

Conversely, suppose $x_n\rightharpoonup x$ and $\Vert x_n\Vert \to \Vert x\Vert$. ThenThus $x_n\to x$ strongly. This is the Radon-Riesz theorem for Hilbert spaces.

No. In $H={\ell }^2$, let $x_n=e_n$ be the standard orthonormal sequence. For any $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$. However,for $n\ne m$. No subsequence is Cauchy, hence none converges strongly. This is a weakly null orthonormal sequence.

Weak convergence $u_n\rightharpoonup u$ in $H_0^1(\Omega )$ impliesin $L^2$. The $L^2$ norm is weakly lower semicontinuous, so

The Rellich-Kondrashov compactness theorem for H01 upgrades the weak convergence tostrongly in $L^2(\Omega )$. Indeed, compactness gives this along every subsequence after passage to a further subsequence, and the weak limit uniquely identifies every such strong limit as $u$. Consequently,Combining the two terms proves the weak lower semicontinuity of a bounded-domain Schrodinger energy:

Choose a minimizing sequence $(u_n)$ withSincethe sequence is bounded in $H_0^1(\Omega )$. Passing to a subsequence, weak compactness givesin $H_0^1(\Omega )$. Rellich compactness gives $u_n\to w$ strongly in $L^2$, and hencePart (i) now yieldsThe definition of $\lambda$ gives the reverse inequality because $w$ is admissible. ThereforeThis is the constrained ground-state minimizer on a bounded domain obtained by the direct method.

Choose $\varphi \in C_c^{\infty }(\mathbb{R})$ with $\Vert \varphi {\Vert }_2=1$, and for $R>0$ setA change of variables givesThe latter tends to zero as $R\to \infty$. Since the energy is nonnegative,This is vanishing Dirichlet energy by dilation on the line.

The infimum is not attained. If an admissible $u$ had zero energy, then its weak derivative would vanish almost everywhere, so $u$ would be constant almost everywhere. The only constant in $L^2(\mathbb{R})$ is zero, contradicting $\Vert u{\Vert }_2=1$.