Let $H$ be Hilbert space, $x,y \in H$ and $x_n, y_n \in H$. Suppose $x_n$ weakly converge to $x$ and $y_n$ weakly converges to $y$. Is it true that $\langle x_n,y_n\rangle$ converges to $\langle x,y\rangle$?

I know that scalar product is a continous function, but does it help? Could if follow from Riesz representation theorem?


No. Let $H= l^2$ and let $(u_n)$ be the usual orthonormal basis of $H$.

Put $x_n=y_n=u_n$. Then $(u_n)$ converges weakly to $x=y=0$.

But $\langle u_n,u_n\rangle=||u_n||_2^2=1 $ and $\langle x,y\rangle=0$.

  • 1
    $\begingroup$ I think $y_n$ is assumed to converge strongly. $\endgroup$ – Kenny Wong Mar 31 '17 at 11:18
  • $\begingroup$ Hmmm... My interpretation was: $y_n$ weakly converges to $y$ $\endgroup$ – Fred Mar 31 '17 at 11:23
  • $\begingroup$ We should ask the OP! $\endgroup$ – Kenny Wong Mar 31 '17 at 11:24
  • $\begingroup$ I meant weakly. Edidted question just now. Sorry for not being precise enough. $\endgroup$ – maq Mar 31 '17 at 11:24
  • $\begingroup$ That is a good idea ! $\endgroup$ – Fred Mar 31 '17 at 11:25

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