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Thread: Prove Opial's Property (Theorem) in Hilbert space

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    Prove Opial's Property (Theorem) in Hilbert space

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

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

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

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

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

    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

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

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

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

    Now take inferior limit as (*), we have

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

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

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

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

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

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

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

    Taking the inferior limit, we get

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

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

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

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

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

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