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Let H be a Hilbert space, $(x_n)_{n \in \mathbb N} \subset H$ and $z \in H$. Prove that $x_n \rightarrow z$ if and only if $$x \stackrel{w}{\longrightarrow} z$$ and $$\Vert x_n \Vert \rightarrow \Vert{z}\Vert$$

I found this exercise in a master's admission exam, but I don't understand what it means $x \stackrel{w}{\longrightarrow} z$

I have searched for the definition of $x \stackrel{w}{\longrightarrow} z$ in some functional analysis books but I cannot find its meaning. Any information you can give me would be greatly appreciated.

Rafael Bello
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  • in this context it means $<x_n,y> \to <z,y>$ for every $y \in H$ – Conrad Dec 05 '22 at 23:55
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    It probably is weak convergence, i.e. $l\left(x_{n}\right) \to l\left(x\right)$ for every bounded linear functional, in the case of Hilbert spaces, due to Riesz Representation Theorem it is $\langle x_{n},v \rangle \to \langle x,v \rangle$ for every $v \in H$ – Raul Fernandes Horta Dec 05 '22 at 23:56

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If $x_n$ converges in norm then it converges weakly and norms of $x_n$ converge to the norm of the limit - this part is obvious. To see the other way consider the scalar product $<x_n-x,x_n-x>$ which is the same as $||x_n-x||^2$. One has $<x_n-x,x_n-x> = <x_n,x_n> - <x_n,x> -<x,x_n> + <x,x>$. Now, $<x_n,x_n>$ converges to $||x||^2$, $<x_n,x>$ converges to $<x,x> = ||x||^2$, and $<x,x_n>$ is the conjugate of $<x_n,x>$ so it goes to $||x||^2$ as well. In words, $||x_n-x||$ converges to $0$.

Salcio
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