The example of weak convergence but not strong convergence

Asked Sep 14 '17 at 03:29

Active Sep 14 '17 at 03:29

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I am reading the following book:

Optimization by Vector Space Methods written by D. G. Luenberger

My problem comes from the Example 1. in p.127

In $X = l_2$, consider the elements $x_n=\{0,0,\ldots,0,1,0,\ldots\}$ with $1$ in the $n$-th place. For any $y=\{\eta_1,\eta_2,\ldots\}\in l_2=X^*$, we have $(x_n\mid y)=\eta_n \rightarrow 0$ as $n\rightarrow \infty$. Thus $x_n\rightarrow 0$ weakly. However $x_n$ does not converge to $0$ strongly since $\|x_n\|=1$.

I am confused about why $\eta_n \rightarrow 0$? If I pick $\eta_i = 0$ for all $i\neq n$ and $\eta_n=1$?

Kindly put some similar questions:

asked Sep 14 '17 at 03:29

sleeve chen

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2

y is fixed, it does not change with $x_n$. So your example would really be $y=(1,1,...)$ which is not in $l^2$. It might help to properly distinguish between indices, since $x_n$ is a sequence of sequences. – Ian Sep 14 '17 at 03:34
@Ian I am still confused why $y$ is fixed. To say ${x_n}$ converges weakly to 0 if for any $y$ such that $\langle x_n,y\rangle \rightarrow \langle 0,y\rangle$. So we have to test it for every $y$. This is the point I am still confused. – sleeve chen Sep 14 '17 at 03:40
2

Yes but you freeze y and then have a sequence of scalars. Thus $y_i \to 0$, and since $(x_n,y)=y_n$ you get the result. – Ian Sep 14 '17 at 03:42
@Ian I see thanks! – sleeve chen Sep 14 '17 at 03:43
What's $X^*$? Is it some dual? – mavavilj Feb 10 '19 at 18:19
@mavavilj yes! dual space of $X$ – sleeve chen Feb 10 '19 at 22:10
@sleevechen What is it? – mavavilj Feb 10 '19 at 22:13
@mavavilj you can read that book p.106. The short answer is the space of all bounded linear functionals on $X$. – sleeve chen Feb 10 '19 at 22:17

The example of weak convergence but not strong convergence

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