I am working in $H^1(S^1)$, the space of absolutely continuous $2\pi$-periodic functions $\mathbb R\to\mathbb R^{2n}$ wih square integrable derivwtives. I have a sequence $z_j$ (for the record, it comes from minimizing a functional, in the middle of Hofer-Zehnder's proof that a Hamiltonian field has a periodic orbit on a strictly convex compact regular energy surface) which I know to be bounded and to have bounded derivatives, hence a subsequence will have to converge weakly to $z_\ast$ and the derivatives to $z'$. But HZ seems to use that $\dot z_j\to\dot z_\ast$ weakly, and I don't know how to prove it. How can I? More precisely, the use that is made is that the integral of $\int_0^{2\pi}\langle\nabla G(\dot z_\ast),\dot z_j-\dot z_ast\rangle\to0$, which is deduced only because that gradient is in $L^2$ (for an estimate on its norm I still have to convince myself about). So the actual question is: does that integral go to zero for any $L^2$ function? And how do I prove it? And is that not equivalent to weak convergence of the derivatives to $\dot z_\ast$?
[Math] Weak convergence and weak convergence of time derivatives
functional-analysisweak-convergenceweak-derivatives
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Best Answer
Claim: if $f_j\to f$ weakly in $H^1$, then $\nabla f_j\to\nabla f$ weakly in $L^2$.
Proof: the gradient operator $f\mapsto \nabla f$ is bounded from $H^1$ to $L^2$. Bounded linear operators preserve weak convergence, i.e., are continuous with respect to weak topology on both spaces. As is proved here.