# A question in Sobolev spaces involving time

Let $$X$$ be a Banach space, we understand $$L^1(0, T, X)$$ is the space of strongly measurable functions from $$[0, T]$$ valued in $$X$$, that is integrable. Assume $${\bf u}\in L^1(0, T, X)$$, we say $${\bf v}\in L^1(0, T, X)$$ is the (weak time) derivative of $${\bf u}$$ if for any $$\varphi\in C_0^\infty(0, T)$$ we have $$\int_0^T {\bf u}(t) \varphi'(t)dt=-\int_0^T {\bf v}(t)\varphi(t)dt,$$ and we will write $${\bf v}={\bf u}'$$.

Now in Evans' PDE book, he discussed a situation where $${\bf u}\in L^2(0, T, H_0^1(\Omega))$$ while $${\bf u}'\in L^2(0, T, H^{-1}(\Omega))$$. It seems there are two possible interpretations:

(A). The situation implies $${\bf u}\in L^1(0, T, H_0^1(\Omega))$$, so we follow the DEFINITON of derivative I mentioned in the beginning and understand that $${\bf u}'\in L^1(0, T, H_0^1(\Omega))$$. However in this situation is possible $${\bf u}'\notin L^2(0, T, H_0^1(\Omega))$$, while from the continuous injection $$H_0^1(\Omega)\to H^{-1}(\Omega)$$ it happens that $${\bf u}'\in L^2(0, T, H^{-1}(\Omega))$$. In conclusion, $${\bf u}'\in L^1(0, T, H_0^1(\Omega))\cap L^2(0, T, H^{-1}(\Omega)).$$

(B). It just means $$\int_0^T {\bf u}(t) \varphi'(t)dt=-\int_0^T {\bf u}'(t)\varphi(t)dt$$ literally. This looks like a little miracle - the right hand side is supposed to be in $$H^{-1}(\Omega)$$, but the equation tells us it is indeed in $$H_0^1(\Omega)$$.

My question is, Assume one uses (B) as definition, is there a theorem which implies that indeed $${\bf u}'$$ is also in $$L^{1}(0, T, H_0^1(\Omega))$$ ? Or counterexamples?

• (B) is the correct definition, in particular, there is no requirement (or theorem) that $u'\in L^1(0,T,H^1_0)$. The "little miracle" comes from the smoothness of $\varphi$ which makes the integration by parts possible. – Michael Renardy Oct 3 at 2:54