weakly convergence imply strong convergence when $ \|f_n\| \rightarrow \|f\| $ in $l^2([0,1])$?

Question

I know in general weakly convergence do not imply strong convergence in $L^p$,but in $L^2[0,1]$ space which if we have additional condition do this condition plus the weak convergence will give us strong convergence?

The additional condition is $f \in L^2[0,1]$ and $ \|f_n\| \rightarrow \|f\| $

You are right.I will delete my post when I will be more sure about it.thanks — henry, Jan 05 '15 at 03:18

score 2 · Accepted Answer · answered Jan 05 '15 at 03:21

2

This is true in an arbitrary Hilbert space:

$$\|f_n - f\|^2 = \langle f_n-f,f_n-f\rangle = \langle f_n, f_n\rangle -\langle f_n,f\rangle - \langle f,f_n\rangle +\langle f,f\rangle\to 0.$$

answered Jan 05 '15 at 03:21

Math1000

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weakly convergence imply strong convergence when $ \|f_n\| \rightarrow \|f\| $ in $l^2([0,1])$?

1 Answers1