If $u \in L^2(0,T;H^1)$ with $u_t \in L^2(0,T;H^1)$ and $u$ and $u_t$ are bounded, is $u \in C([0,T];L^\infty$)?

Question

Let $u \in L^2(0,T;H^1(\Omega))$ with $u_t \in L^2(0,T;H^{-1}(\Omega))$ so $$\int_0^T u(t)\varphi'(t) = -\int_0^T u'(t)\varphi(t)$$ holds for all $\varphi \in C_c^\infty(0,T)$.

Suppose we know additionally that $u, u_t \in L^\infty(0,T;L^\infty(\Omega))$ .

Does it follow that $u \in C^0([0,T];L^\infty(\Omega))$?

Answer 1

Yes, the boundedness of weak derivative implies Lipschitz continuity. Let $X$ be any Banach space and suppose $u\in L^\infty(0,T;X)$ is a function such that $u_t\in L^\infty(0,T;X^*)$. Let $M$ be the $L^\infty$ norm of $u_t$. Take any two Lebesgue points of $u$, say $t<s$. Choose as $\varphi$ a test function that quickly grows from $0$ to $1$ near $t$, then stays at $1$, then drops down from $1$ to $0$ near $s$. The integral $\int_0^T u'\varphi $ is bounded by $M(s-t)$, no matter how quickly $\varphi$ rises/drops. So, $\int_0^T u\varphi'$ is bounded by $M(s-t)$. But this integral converges to $u(s)-u(t)$.