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For parabolic sobolev spaces I follow the following definition:

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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.

Thanks in advance!

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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.

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    $\begingroup$ 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. $\endgroup$
    – Skeeve
    Mar 25 '19 at 19:18
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    $\begingroup$ @Skeeve Thank you for your comment. Yes, that could also be. kaithkolesidou can you please clarify whether my answer solved your question, or are you asking about a different aspect e.g. the measurability as pointed out by Skeeve. $\endgroup$
    – Fritz
    Mar 25 '19 at 19:24

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