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I have a simple question : What it means $$||v_n||_{(W^{1,p}_0)^*}\rightarrow 0$$

Where $(W^{1,p}_0)^*$ is the dual space of $W^{1,p}_0$

  • I know that $v_n\rightarrow 0$ in $(W^{1,p}_0)^*$ mease that $\langle x^*,v_n\rangle\rightarrow 0, \forall x^*\in (W^{1,p}_0)^*$, but in this case what is $\langle.,.\rangle$? $W^{1,p}_0$ is a Banach space not a Hilbert space.

Thank you.

Vrouvrou
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3 Answers3

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The symbols $\langle x^*,v_n\rangle$ just express $x^*(v_n)$, the functional $x^*$ evaluated at $v_n$. It is a common notation, inspired in the Hilbert space case, where the dual is the same original space.

Martin Argerami
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As Martin pointed out, $\langle f,v,\rangle$ is just notation for the action of the functional $f\in X^*$ on the element $v\in X$; it does not mean (in general) an inner product. To be less ambiguous, one can write the pairing as

$$ _{X^*}\langle f,v\rangle_X $$ to emphasize the fact that $f$ and $v$ live in two different spaces, so the action is not an inner product.

Edit: here's an example of a functional $f$ for you case:

$$ _{X^*}\langle f,v\rangle_X = f(v) = \int_\Omega (v\phi+\nabla v\cdot \psi)dx $$ with $\phi\in L^q(\Omega)$ and $\psi\in \left[L^q(\Omega)\right]^d$, and $1/p+1/q=1$. You can see that it is linear in $v$, and bounded (use Holder inequality).

bartgol
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    Also, the duality pairing is frequently written $\langle f,v \rangle_{X^*\times X}$ – Svetoslav Jul 30 '15 at 22:03
  • Yes, I've seen that too, though I prefer the one where the dual is on the left, cause to me it's more self-explanatory. De gustibus... =) – bartgol Jul 30 '15 at 22:06
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    And I prefer the other way. Maybe it is just how one first met it and got used to it. – Svetoslav Jul 30 '15 at 22:08
  • when we work with a Hilbert space ⟨,⟩ is the inner pruduct and here what is it ? – Vrouvrou Jul 30 '15 at 22:10
  • @bartgol what is $d$ in $\psi\in \left[L^q(\Omega)\right]^d$? – Vrouvrou Jul 31 '15 at 12:47
  • Can you say that $f(v)=\int_{\Omega} \nabla v\nabla\psi dx, \forall \psi\in W^{1,p}_0$ – Vrouvrou Jul 31 '15 at 14:13
  • Well, if $\nabla \psi \in L^{q}$ then yes. In the special case of zero trace, that would be a bounded operator since the $L^p$ norm of the gradient is equivalent to the $W^{1,p}$ norm of the function. – bartgol Aug 04 '15 at 14:38
  • @Vrouvrou, it is the (geometric) dimension of $\Omega$ (the length of the gradient vector). – bartgol Aug 04 '15 at 14:39
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In what you wrote first, $v_n$ should be an element of the dual space and then it means convergence in the dual norm (norm of the functionals over $W_0^{1,p}$). The second thing, that you wrote, i.e $\langle x^*,v_n\rangle\rightarrow 0\quad \forall x^*\in (W_0^{1,p})^*$ means weak convergence of the sequence $\{v_n\}_{n=1}^\infty\subset W_0^{1,p}$ to $0\in W_0^{1,p}$. The notation $\langle x^*,.\rangle$ means evaluation of the functional $x^*\in (W_0^{1,p})^*$ at the element "."

Svetoslav
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  • when we work with a Hilbert space ⟨,⟩ is the inner pruduct and here what is it ? – Vrouvrou Jul 30 '15 at 22:10
  • Here is a notation for $x^(v_n)$, i.e for the action of the functional $x^$ from the dual space $(W_0^{1,p})^*$ on the element $v_n$ from the primal space $W_0^{1,p}$ – Svetoslav Jul 30 '15 at 22:11
  • yes but how to compute this ? – Vrouvrou Jul 30 '15 at 22:13
  • What is $||v_n||_{(W^{1,p}_0)^*}\rightarrow0$ equal to ? – Vrouvrou Jul 30 '15 at 22:14
  • In order to avoid a possible confusion, I would write a $^$ to $v_n$, i.e $|v_n^|{(W_0^{1,p})^}$. The computation depends on the particular case that you have. But you should now that $|v^|{(W_0^{1,p})^}=\sup\limits_{v\in W_0^{1,p},v\neq 0}{\frac{|\langle v^,v \rangle|}{|v|V}}=\sup\limits{|v|\leq 1}{|\langle v^,v\rangle |}$ for any space $V$ and any functional $v^$ in $V^*$. – Svetoslav Jul 30 '15 at 22:25
  • and what is exactly $\langle v^*,v\rangle $ – Vrouvrou Jul 31 '15 at 12:49
  • This is the notation for the action of the functional $v^*$ on the element $v$. This notation is also called duality pairing! – Svetoslav Jul 31 '15 at 13:40
  • Ok but can we write it with integrals ? – Vrouvrou Jul 31 '15 at 13:53
  • See representation of linear functionals. In Hilbert spaces the dual space can be identified with the original space by the Riesz representation theorem. See http://math.stackexchange.com/questions/246735/dual-space-of-the-sobolev-spaces – Svetoslav Jul 31 '15 at 14:24
  • I know that in Hilbert space we use the inner prudact but what about banach space ? – Vrouvrou Jul 31 '15 at 14:31