| Posted on Sunday, 11 January, 2004 - 01:33 pm: |
Why doesn`t the sequence of functions {fk (x) = sin kx, x in [0, 2pi), k in
} have a convergent
subsequence?
| Posted on Sunday, 11 January, 2004 - 01:40 pm: |
Convergent in what sense? They trivially don't have an L2-convergent subsequence as they're orthogonal and have the same norm; hence you certainly can't get L¥ (uniform) convergence either. Do you mean, for example, 'everywhere pointwise convergent' or something along those lines? David
| Posted on Sunday, 11 January, 2004 - 03:18 pm: |
Just to add to what David has said - the sequence converges weakly to 0 in Lp , but that is probably not what you are after. Convergence of a sequence depends on which space it lives.
Kerwin
| Posted on Monday, 12 January, 2004 - 05:15 am: |
Sorry for the ambiguity - I meant pointwise convergence.
| Posted on Monday, 12 January, 2004 - 11:56 am: |
If a subsequence converges to 0 pointwise then it also converges to 0 to L2 (for example by the dominated convergence theorem, though that's almost certainly overkill). Another way, slightly more direct, would be to use a 'nested sequences' type argument. If fkr® 0 pointwise then you can inductively construct a subsequence fkrn and closed nested intervals In such that fkrn ³ 1/2 on In. The intervals have a point of intersection x, and fkr(x) clearly doesn't tend to 0 - a contradiction.
| Posted on Monday, 12 January, 2004 - 12:07 pm: |
If a subsequence of fk converges pointwise, it must converge in L2 [0,2pi) (dominated convergence). Now fk are orthogonal and have the same nonzero norm - contradiction.
Alternatively, if a subsequence converges pointwise, it must be the same as the weak limit in Lp . But this is obviously nonsense since the integral of |fk | is the same nonzero number.
Kerwin