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I'm trying to understand the proof that (c) implies (a) here in the following proposition (here, $\mathcal{V} = L^2(0,T;V)$). See the very last line in the image for that part:

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I give here Proposition 1.1 which the proof uses.

enter image description here

I do not understand how it's used in the proof. I'd appreciate an explanation. Thank you.


Here, $$ W_2(0,T) = \{u \in \mathcal V : u' \in \mathcal V'\} $$ and

enter image description here

(All images cut from Showalter's book Monotone Operators in Banach Space)

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It would be helpful to change the question title to make it more informative. –  Rasmus Bentmann Jul 17 '13 at 11:08
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1 Answer 1

In my opinion, OP has not given enough information in order for the question to be able to be answered properly. Maybe the author has made somewhere implicit assumptions that are not explicitly stated in Proposition 2.1. In (c) of that proposition, nothing implies that the weak derivative of $u$, interpreted as a function $[0,T]\to V'$ (or more properly, an a.e. equivalence class of such) would be in (possibly) the space $L^2([0,T],V')$. However, something like this would be needed to apply Proposition 1.1.

If we knew that $\mathcal A\,{.\,}u:t\mapsto\mathcal A(t)u(t)$ is in $L^2([0,T],V')$, and if we knew that $V$ is reflexive, then (c) would imply that $u:[0,T]\to V'$ is weakly differentiable with the weak derivative in $L^2([0,T],V')$. This also would give the superfluous assumption of weak absolute continuity, provided that continuity of $u:[0,T]\to V'$ is implicit somewhere. Thence $u$ being in $W^{1,2}([0,T],V')$ would follow. The definition of the latter space then directly would give (a) of Proposition 2.1.

Remark. OP has not given what the author means by $f:[0,T]\to B$ being weakly differentiable. Above, I have interpreted it so that for every $v\in B'$ the function $v\circ f:[0,T]\owns t\mapsto v(f(t))\in\mathbb R$ is differentiable in the distributional sense with derivative $v\circ g$ for some $g:[0,T]\to B$.

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By "Thence $f$ being in ..." I meant "Thence $u$ being in ..." –  TaQ Jul 21 '13 at 8:44
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Dear @TaQ: I incorporated the correction in your comment into your answer. Please remember that you can edit your own answers. –  Ricardo Andrade Aug 20 '13 at 8:45
    
@ Ricardo Andrade: I know that but I thought that if there were need for more edits, I would do them all at the same time so that it would count only one edit. Once I edited my answer so many times that it became CW unintentionally as the edits were ten or more in number. –  TaQ Aug 20 '13 at 23:57
    
Dear @TaQ: Good point. That has also happened to me a couple of times. On the other hand, I have found that waiting a couple of days before editing to incorporate any corrections and changes is usually sufficient. It gives me enough time to find most obvious problems and then one or two edits should suffice. Personally, I don't always abide by this principle, but it tends to work well when I do. Your mileage may vary... :) –  Ricardo Andrade Aug 21 '13 at 0:06
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