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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.

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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.

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  • $\begingroup$ But how to express this ? what is the expresion of $x^*$ ? $\endgroup$ – Vrouvrou Jul 30 '15 at 21:59
  • $\begingroup$ Not sure what you mean. It's a functional in the dual; what kind of expression do you expect? $\endgroup$ – Martin Argerami Jul 30 '15 at 22:00
  • $\begingroup$ can we express it with integral ? $\endgroup$ – Vrouvrou Jul 30 '15 at 22:01
  • $\begingroup$ when we work with a Hilbert space $\langle,\rangle$ is the inner pruduct and here what is it ? $\endgroup$ – Vrouvrou Jul 30 '15 at 22:02
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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 "."

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  • $\begingroup$ when we work with a Hilbert space ⟨,⟩ is the inner pruduct and here what is it ? $\endgroup$ – Vrouvrou Jul 30 '15 at 22:10
  • $\begingroup$ 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 $W_0^{1,p}$ $\endgroup$ – Svetoslav Jul 30 '15 at 22:11
  • $\begingroup$ yes but how to compute this ? $\endgroup$ – Vrouvrou Jul 30 '15 at 22:13
  • $\begingroup$ What is $||v_n||_{(W^{1,p}_0)^*}\rightarrow0$ equal to ? $\endgroup$ – Vrouvrou Jul 30 '15 at 22:14
  • $\begingroup$ 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 $V^*$. $\endgroup$ – Svetoslav Jul 30 '15 at 22:25
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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).

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    $\begingroup$ Also, the duality pairing is frequently written $\langle f,v \rangle_{X^*\times X}$ $\endgroup$ – Svetoslav Jul 30 '15 at 22:03
  • $\begingroup$ 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... =) $\endgroup$ – bartgol Jul 30 '15 at 22:06
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    $\begingroup$ And I prefer the other way. Maybe it is just how one first met it and got used to it. $\endgroup$ – Svetoslav Jul 30 '15 at 22:08
  • $\begingroup$ when we work with a Hilbert space ⟨,⟩ is the inner pruduct and here what is it ? $\endgroup$ – Vrouvrou Jul 30 '15 at 22:10
  • $\begingroup$ @bartgol what is $d$ in $\psi\in \left[L^q(\Omega)\right]^d$? $\endgroup$ – Vrouvrou Jul 31 '15 at 12:47

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