Convergence on the dual of a Banach space

Question

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.

Answer 1

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.

Answer 2

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 "."

Answer 3

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).