$L^\infty(0,T;X) \cap C([0,T];Y) \subset C([0,T];X)$ for $X \subset Y$ dense?

Question

is the Inclusion stated in the title true? In my case the spaces (essentially) are $X = H^1(\Omega)$ and $Y = L^2(\Omega)$ for $\Omega \subset \mathbb{R}$ bounded. My first try was to show

$\lim_{t_1 \to t_2} \Vert v(t_2) - v(t_1) \Vert_X^2 = \lim_{t_1 \to t_2} \left( \Vert v(t_2) - v(t_1) \Vert_Y^2 + \Vert \nabla v(t_2) - \nabla v(t_1) \Vert_Y^2 \right) = 0 $

for $v \in L^\infty(0,T;X) \cap C([0,T];Y) $. But there I only get, that the first norm converge. Is there anyone who can give me a hint?

Answer 1

The inclusion is not true. For example, take the sequence spaces $\ell^2$ and $H^1$, where the $H^1$ norm is $$ ||(x_k)||_{H^1} = ||(kx_k)||_{\ell^2}. $$ If you want, this situation is achievable from $L^2(0,1)$, $H_0^1(0,1)$ by examining Fourier coefficients. Let $\{ e_k \} \subset \ell^2$ be the usual orthonormal basis and consider the sequence $$M_n = n^{-1}e_n \in \ell^2,$$ consisting of a single bump at mode $n$. Notice that $M_n \to 0$ in $\ell^2$ as $n \to +\infty$, while $M_n$ does not converge in $H^1$ as $n \to +\infty$.

Now define a function $f \in L^\infty(0,T;H^1) \cap C([0,T];\ell^2)$ as follows. Let $f(t) = 0$ except at times $t_n=T-1/n$, where we have $f(t_n) = M_n$ and $f$ decaying smoothly to zero in a small neighborhood about $t_n$ (this can be achieved by multiplying with an appropriate bump function in time). We have that $\lim_{t \to T_-} f(t_n) = 0$ in $\ell^2$, but the limit in $H^1$ does not exist. Thus, the function $f$ is not in $C([0,T];H^1)$.