# Lp weak convergence

1. Feb 10, 2012

### muzialis

Hello,

I am struggling to understand why L1 is not weakly compact, while Lp, p>1, is.

The example I have seen put forward is the function u_n (x) = n if x belongs to (0, 1/n), 0 otherwise, the function being defined on (0,1).

It is shown this u_n converging to the Dirac measure, and this shows L1 not being weakly compact (as integrating u_n times the charactersitic function of the interval yields 1 as a result, while the result is zero for a function which is zero at the boundary.

I can not see why this should not happen for Lp, p > 1 too. In the attached short notes there is an example after which (pag. 6) it is stated that this function converges weakly to zero in Lp.

I can not understand this.

Many thanks

Regards

#### Attached Files:

• ###### Weakconvergence L1.pdf
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2. Feb 10, 2012

### kai_sikorski

Sorry, not sure my response makes sense, edited it out. Will think about it some more.

Last edited: Feb 10, 2012
3. Feb 10, 2012

### kai_sikorski

Actually maybe what I said did make sense. So to be weakly compact any bounded sequence has to have a convergent sub-sequence from what I remember right? The sequence you defined is not bounded in Lp for p>1.

4. Feb 10, 2012

### muzialis

Thanks for replying, I am not sure I understand your reply. The sequence i mention is bounded in the L1 norm, and the example I mentioned, u_n = x ^(1/2) if x belongs to (0,1/n), 0 otherwise, is bounded in L2. But L2, on the contrary to L1, is weakly compact.

5. Feb 10, 2012

### kai_sikorski

"u_n (x) = n if x belongs to (0, 1/n), 0 otherwise"

So my point is that this is bounded in L1, and therefore it can be used as a counterexample to show L1 is not weakly compact.

However since it is not bounded in Lp for p>1, it can not be used as a counter example in those cases because the definition of weak compactness only puts restrictions on the behavior of bounded sequences.

6. Feb 10, 2012

### muzialis

Sorry, I was very unclear. I showed the example in L1, giving for granted that when considering the L2 case its analogue, mentioned in the attachment, would have considered.
The function u_n = x^(1/2) if x belongs to (0,1/n), 0 otherwise.
This is bounded in L2 and must converges to an element of L2, given that L2 is weakly compact.

But thank you anyhow, I think I discovered my silly mistake!

Bye