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

1. 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.

2. Re: Prove Opial's Property (Theorem) in Hilbert space

Originally Posted by konna
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.

3. 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.