# Question on relation between a parabolic sobolev space and a sobolev bochner space

For parabolic sobolev spaces I follow the following definition:

According to this definition, we have that $$W^{1,1,2}(I \times \Omega)=L^2(I; W^{1,2}(\Omega)) \cap W^{1,2}(I; W^{-1,2}(\Omega))$$

Now my question is: If we have a function such that $$f \in W^{1,2}(I; L^2(\Omega))$$ with, in addition, $$\nabla f(x,\cdot) \in L^2(I;L^2(\Omega))$$, can we claim that $$f \in W^{1,1,2}(I \times \Omega)$$?

Instinctively I would say yes, but I need a math confirmation.

Any hint or help is much appreciated.

Yes, it is true. You have $$f \in W^{1,2}(I;L^2(\Omega)) \cap L^2(I;W^{1,2}(\Omega))$$ and you are asking whether this function is in $$W^{1,1,2}(I \times \Omega)=L^2(I;W^{1,2}(\Omega))\cap W^{1,2}(I;W^{-1,2}(\Omega)).$$
This follows from the fact $$L^2 \hookrightarrow W^{-1,2}$$ where we identified $$L^2$$ with its dual as usual in the corresponding Gelfand triple. See also that question in MSE. Then it is easy to check that $$W^{1,2}(I;L^2(\Omega))$$ is continuously embedded in $$W^{1,2}(I;W^{-1,2}(\Omega))$$, just use the norm definition.
• Probably it remains to justify that $f\in L^2(I;W^{1,2}(\Omega))$. This is nearly obvious, since $f\in L^2(I;L^2(\Omega))$ and $\nabla f \in L^2(I;L^2(\Omega))$, but I think in the OP some technical details on this point are also needed (in particular, about the measurability as in the definition of Bochner integral). For instance one could mollify $f$ with respect to $\Omega$-valued variable and show that such approximations converge in $L^2(I;W^{1,2}(\Omega))$, but maybe there is a better way. Mar 25 '19 at 19:18