if $u \in C([0,T]; H^{1}(\Omega)$ then $\partial_{x} u \in$?

Question

If $u \in C([0,T]; H^{1}(\Omega))$ what can we say about $\partial_{x} u$?

My issue is figuring out what happens to the time regularity. I would guess it to be something like $\partial_{x} u \in X(0,T; L^{2}(\Omega))$ however I don't know what $X$ should intuitively be.

Answer 1

We know that $\partial_x$ maps $H^1$ into $L^2$. This mapping is clearly (linear and) bounded.

Now, you can lift $\partial_x$ to operate from $C([0,T]; H^{1})$ to $C([0,T]; L^2)$, by defining $(\partial_x u)(t):=\partial_x (u(t))$. You still get a linear, bounded operator.

Some technical remarks:

• the definition is well posed (we are implicitly picking the unique representative of $u$ that is continuous, if you think in a measure theoretic framework)
• $\partial_xu \in C([0,T]; L^{2})$ because $\partial_x : H^1\rightarrow L^2$ is linear and bounded. In detail, if $t\rightarrow s$, then $u(t) \rightarrow u(s)$ in $H^1$, so that $\partial_x u(t)\rightarrow \partial_xu(s) $ in $L^2$