Part (c)(i) already gives weak continuity of $t\mapsto\theta(t)$ in $L^2$. In the assumed energy equality, the dissipation integral
$$
t\longmapsto\int_0^t\|\nabla\theta(\tau)\|_2^2\,d\tau
$$
is absolutely continuous. The forcing integrand is in $L^1(0,T)$ because $u\in L^\infty(0,T;L^2)$ and $\theta\in L^\infty(0,T;L^2)$. The equality therefore makes $t\mapsto\|\theta(t)\|_2^2$ continuous.

Whenever $t_n\to t$, weak continuity gives $\theta(t_n)\rightharpoonup\theta(t)$ and the energy equality gives convergence of their norms. The <Radon-Riesz theorem>, or directly the , now gives $\theta(t_n)\to\theta(t)$ in $L^2$. Thus
$$
\boxed{\theta\in C([0,T];L^2_{\rm per}(\Omega))}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 359 / 1 / e / Solution

If $u:[0,T]\to H$ is weakly continuous and an energy equality makes $t\mapsto\|u(t)\|_H$ continuous, then $u$ is strongly continuous. This follows from the <Radon-Riesz theorem> applied whenever $t_n\to t$.

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 Incoming links: Strong continuity from weak continuity and an energy equality