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 $\dot z_\ast$?

  • $\begingroup$ Weak convergence in $H^1$ implies weak convergence of derivatives in $L^2$. Or, to argue another way, if $z_j$ are bounded in $H^1$, then $\dot z_j$ are bounded in $L^2$, so there is a subsequence of $\dot z_j$ converging weakly in $L^2$. This means their integrals against any fixed $L^2$ functions converge... But it's possible I am missing something: you appear to be talking about $\mathbb{R}^{2n}$ denoting it $S^1$ and integrating from $0$ to $2\pi$, which is quite confusing. $\endgroup$ – user147263 Oct 5 '15 at 11:13
  • $\begingroup$ @MiceElf whoops, I swapped codomain and domain :). I had figured the convergence out already but the book seems to use the weak limit of the der.s is the der. of the weak limit, which is why I'm asking this. $\endgroup$ – MickG Oct 5 '15 at 11:26
  • $\begingroup$ If the weak convergence is in $H^1$, then yes, the derivative of weak limit is the weak limit of derivative. $\endgroup$ – user147263 Oct 5 '15 at 11:28
  • $\begingroup$ How do I prove that? I mean, w.c. means any linear functional applied to them gives a succession converging to the functional applied to the limit. How does that imply the same thing for the derivatives? Write it out as integral against a function (Riesz representation) and integrate by parts... oh right, if the function we integrate against is $H^1$ the derivative is $L^2$, and any $L^2$ funct is the weak derivative of its integral funct which is $H^1$, right? $\endgroup$ – MickG Oct 5 '15 at 11:36
  • 1
    $\begingroup$ I posted a simpler argument as an answer. Anyway, the $H^1$ inner product is $(f,g) = \int \nabla f\cdot \nabla g + \int fg$. $\endgroup$ – user147263 Oct 5 '15 at 11:45

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.

| cite | improve this answer | |
  • $\begingroup$ Indeed its operator norm is at most 1 since the squared norm in $H^1$ is the some of the s.n.s in $L^2$ so the gradient's $L^2$ norm is less than the $H^1$ norm of the func. $\endgroup$ – MickG Oct 5 '15 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.