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I have read this relation: if $y\in C(0,T;L^2(0,L))\subset L^2((0,T)\times (0,L))$ then $y_t\in L^2(0,L;H^{-1}(0,T))$

can anyone explain how to get this?

I know $y(t)(x)\in L^2(0,L)$, but why $y_t(x)(t)\in H^{-1}(0,T)$?

I found it is straightforward:

$y\in C(0,T;L^2(0,L))\subset L^2((0,T)\times (0,L))=L^2(0,L;L^2(0,T))$, hence $y(x)(t)\in L^2(0,T)$, then $y_t(x)(t)\in H^{-1}(0,T)$

Lookout
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  • $H^1\subset L^2\Rightarrow L^2=(L^2)^\subset (H^1)^=H^{-1}$ EDIT: but i think your problem is something else. If $x\in [0,L]$ then $y_t(x)\in H^{-1}(0,T)$ for almost every $x$ by definition of $y_t\in L^{2}(0,L;H^{-1}(0,T))$. – Max Jan 01 '17 at 15:34
  • @Max As a remark: Generally $H^{-1}$ denotes the dual space of $H_0^1$ and not $H^1$. But of course it still holds $H_0^1 \subset L^2 \subset H^{-1}$. – Cahn Jan 01 '17 at 15:40
  • @Max of course I know this relationship, but one is about time, the other is about space. Is there nothing to do with the derivative wrt t? – Lookout Jan 01 '17 at 15:40
  • might well be i did not understand your question as intended. i'll have a look later. – Max Jan 01 '17 at 15:42
  • I am not sure if this is your problem exactly but in one case it is $y(t):[0,L] \to \mathbb{R}, x \mapsto y(t)(x):=y(x,t)$ and in the other case it is $y(x):[0,T] \to \mathbb{R}, t \mapsto y(x)(t):=y(x,t)$. But this is no problem to be exact since $y(x,t)=y(x)(t)=y(t)(x)$. – Cahn Jan 01 '17 at 15:43
  • @MarvinF. exactly – Lookout Jan 01 '17 at 15:44
  • @user166445 Do you still know where you read this? Normally I'd say since $y \in C(0,T;L^2(0,L))\subset L^2(0,T;L^2(0,L))$ then by definition $y_t \in L^2(0,T;L^2(0,L))$ is its weak time derivative provided $\int_0^T \phi'(t) y(t) \ dt = - \int_0^T \phi(t) y_t(t) \ dt$. But we have not used continuity of $y:[0,T] \to L^2(0,L)$ up to now. – Cahn Jan 01 '17 at 16:02
  • I would not expect this to be true in general. This is discussed in the context of a PDE, certainly -- would you tell us which one? – anonymous Jan 01 '17 at 16:03
  • @anonymous in transport equation's hidden regularity. also in kdv – Lookout Jan 01 '17 at 16:06
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    Right, so that's $u_t + cu_x = 0$? This means $u_t = - cu_x$, so that the regularity of $u_t$ is the same as of $u_x$. If $u$ is in $L^2(0,T,L^2(\Omega))$, then $u_x$ is in $L^2(0,T,H^{-1}(\Omega))$ and so is $u_t$ by these considerations. – anonymous Jan 01 '17 at 16:11
  • @MarvinF. unfortunately, I read this in my lecture notes, perhaps my professor is wrong :) – Lookout Jan 01 '17 at 16:12
  • You edited the question to something incorrect. You can't say $y(t)(x) \in L^2(0,L)$ (and so on) since the $x$-values are in $(0,L)$. In the same way you say $f \in L^2(\Omega)$ and not $f(x) \in L^2(\Omega)$. These are function spaces! – Cahn Jan 02 '17 at 08:49

1 Answers1

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We have $y \in C(0,T;L^2(0,L))\subset L^2(0,T;L^2(0,L))$ then by the definition of weak time derivatives we have $y_t \in L^2(0,T;L^2(0,L))$. So by the definition of the Bochner-norm and using Fubini-Tonelli (which works since $y$ is time-continuous) we get

$$ \begin{align} ||y_t||^2_{L^2(0,T;L^2(0,L))}&=\int_0^T ||y_t(t)||_{L^2(0,L)}^2 dt \\ &= \int_0^T \int_0^L y_t^2(t)(x) \ dx \ dt \\ &= \int_0^L \int_0^T y_t^2(x)(t) \ dt \ dx \\ &= \int_0^L ||y_t(x)||^2_{L^2(0,T)} \ dx=||y_t||^2_{L^2(0,L;L^2(0,T))} \end{align}$$

that means $y_t \in L^2(0,L;L^2(0;T))$. And since $H_0^1(0,T) \subset L^2(0,T) \subset H^{-1}(0,T)$ we get $y_t \in L^2(0,L;H^{-1}(0,T))$ as wanted.

Cahn
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  • can you explain why $y_t\in L^2(0,T;L^2(0,L))$ from $y\in L^2(0,T;L^2(0,L))$? – Lookout Jan 02 '17 at 02:27
  • We define weak time derivatives to have this property, for example have a look at http://math.stackexchange.com/questions/1717465/definition-of-weak-time-derivative . Generally we say $u' \in L^2(0,T;V')$ is the weak time derivative of $u \in L^2(0,T;V)$ if $\int_0^T u'(t) \phi(t)\ dt=-\int_0^T J(u(t)) \phi'(t)\ dt$ for all $\phi \in C_0^\infty (0,T)$ where $J$ is the bounded, linear, injective, dense mapping from $V$ to $V'$. Here it is quite nice since $V'=L^2(0,T)'$ is isometrically isomorphic to $L^2(0,T)$. – Cahn Jan 02 '17 at 08:46