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Math Help - Prove Opial's Property (Theorem) in Hilbert space

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    Prove Opial's Property (Theorem) in Hilbert space

    Let H be a Hilbert space and suppose $x_{n} \rightharpoonup x$ . Then $\liminf_{n\rightarrow\infty}||x_n-x|| < \liminf_{n\rightarrow\infty}||x_n-y||,$ for all y \in H with x \neq y.

    please! ... I don't understand its.
    Last edited by mr fantastic; October 1st 2011 at 02:02 PM. Reason: Re-titled.
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  2. #2
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    Re: Prove Opial's Property (Theorem) in Hilbert space

    Quote Originally Posted by konna View Post
    Let H be a Hilbert space and suppose x_{n} \rightharpoonup x. Then \liminf_{n\rightarrow\infty}||x_n-x|| < \liminf_{n\rightarrow\infty}||x_n-y||, for all y \in H with x \neq y.

    please! ... I don't understand its.
    Presumably the notation x_{n} \rightharpoonup x means that x_n converges weakly to x. Then

    \|x_n-y\|^2 = \|(x_n-x) + (x-y)\|^2 = \|x_n-x\|^2 + 2\text{Re}\langle x_n-x, x-y \rangle + \|x-y\|^2.

    Now take the liminf, noting that the inner product term goes to 0 because of the weak convergence.
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    Re: Prove Opial's Property (Theorem) in Hilbert space

    Thank you so much. I think it is easily proof, when you show for me.

    By the way, I have other proof of this theorem . Can you check it for me?

    From Parallelogram Law, we have

    2||x_n-x||^2+2||x-y||^2=||x_n-y||^2+||x_n-x-(x-y)||^2.

    That is 2||x_n-x||^2+2||x-y||^2=||x_n-y||^2+||x_n-x||^2-2\langle x_n-x,x-y\rangle+||x-y||^2,

    or ||x_n-y||^2-||x_n-x||^2=2 \langle x_n-x,x-y\rangle+||x-y||^2.--------(*)

    Now take inferior limit as (*), we have

    \liminf_{n \rightarrow \infty}(||x_n-y||^2-||x_n-x||^2)=2\liminf_{n \longrightarrow \infty}\langle x_n-x,x-y\rangle+\liminf_{n \longrightarrow \infty}||x-y||^2.

    Since x_n \rightharpoonup x , we obtains that \liminf_{n \longrightarrow \infty}\langle x_n-x,x-y\rangle = 0.

    It follow from above and our assumption ; x \neq y, that \liminf_{n \longrightarrow \infty}(||x_n-y||^2-||x_n-x||^2)> 0.

    Thus, there is a positive integer N such that ||x_n-y||^2-||x_n-x||^2> 0, for all n \geq N.

    Consider, for all n \geq N, (||x_n-y||-||x_n-x||)(||x_n-y||+||x_n-x||)> 0,
    we can put M=\sup_{n \geq N}\{||x_n-y||+||x_n-x||\}.

    This implies that M(||x_n-y||+||x_n-x||) \geq 2 \langle x_n-x,x-y\rangle+||x-y||^2,

    or M||x_n-y|| \geq M||x_n-x||+ 2 \langle x_n-x,x-y\rangle+||x-y||^2

    Taking the inferior limit, we get

    M\liminf_{n \longrightarrow \infty}||x_n-y|| \geq M\liminf_{n \longrightarrow \infty}||x_n-x||+ 2 \liminf_{n \longrightarrow \infty}\langle x_n-x,x-y\rangle+||x-y||^2,

    and so M\liminf_{n \longrightarrow \infty}||x_n-y|| \geq M\liminf_{n \longrightarrow \infty}||x_n-x||+||x-y||^2.

    Thus M(\liminf_{n \longrightarrow \infty}||x_n-y|| - \liminf_{n \longrightarrow \infty}||x_n-x||) \geq ||x-y||^2 > 0.

    It follows that \liminf_{n \longrightarrow \infty}||x_n-y|| - \liminf_{n \longrightarrow \infty}||x_n-x||  > 0, and hence

    \liminf_{n \longrightarrow \infty}||x_n-y||   >  \liminf_{n \longrightarrow \infty}||x_n-x||. #

    I tried to do it. But not sure what to do.
    Last edited by konna; October 3rd 2011 at 05:18 AM.
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