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Define $W(X,Y) = \{ u \in L^2(0,T;X) \mid u' \in L^2(0,T;Y)\}$ where $u'$ is the usual weak derivative.

Let $V \subset H \subset V^*$ be a Hilbert triple. If $u \in W(V,V)$ (so $$u \in L^2(0,T;V) \quad\text{with}\quad u' \in L^2(0,T;V)$$), then is $$\lVert{u}\rVert_{C^0([0,T];V)} \leq C\left(\lVert{u}\rVert_{L^2(0,T;V)} + \lVert{u'}\rVert_{L^2(0,T;H)}\right)?$$ (Notice the norm on $u'$ is taken with respect the weaker space $H$).

We know that $$W(V,V) \hookrightarrow C([0,T];V).$$ So the quantities in the desired inequality make sense, but does the inequality hold?

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  • $\begingroup$ It sounds like you should probably look for references on vector-valued Sobolev embedding theorems. $\endgroup$
    – Christopher A. Wong
    Commented Feb 19, 2014 at 20:44
  • $\begingroup$ @ChristopherA.Wong Indeed. I searched with no luck. $\endgroup$
    – maximumtag
    Commented Feb 20, 2014 at 10:40
  • $\begingroup$ Try looking at Wloka's "Partial Differential Equations" somewhere around page 390 or so. What is the motivation of the question, i.e. what happens if the answer is yes/no? I think it should be relatively easy to design examples showing that both the embedding $W(V,H) \hookrightarrow C([0,T];V)$ and the corresponding norm inequality are not generally true. Without motivation, I am not sure if it is worthwhile bothering. $\endgroup$
    – TaQ
    Commented Feb 20, 2014 at 19:00
  • $\begingroup$ @TaQ The motivation is my first question (I will edit my post to remove the question about the embedding. I think you're right in that the embedding is not true). But maybe you're right also about my norm-inequality question.. Anyway the motivation for that is difficult to explain but I need it to apply a Minty-Browder monotone trick. $\endgroup$
    – maximumtag
    Commented Feb 20, 2014 at 20:20

1 Answer 1

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As mentioned by @TaQ, the embedding $W(V,H) \hookrightarrow C([0,T];V)$ is, in general, not true. However, if your estimate would be true, you can extend the embedding operator from the dense subspace $H^1(0,T;V)$ to $L^2(0,T;V) \cap H^1(0,T; H) = W(V,H)$. This yields a contradiction.

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    $\begingroup$ I think you mean furthermore rather than however $\endgroup$
    – username
    Commented Feb 21, 2014 at 12:16

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