Edwin Koh

Posted on Sunday, 11 January, 2004 - 01:33 pm:

Why doesn`t the sequence of functions {f_k (x) = sin kx, x in [0, 2pi), k in

} have a convergent subsequence?

David Loeffler

Posted on Sunday, 11 January, 2004 - 01:40 pm:

Convergent in what sense? They trivially don't have an

L^{2}

-convergent subsequence as they're orthogonal and have the same norm; hence you certainly can't get

L^{\infty}

(uniform) convergence either. Do you mean, for example, 'everywhere pointwise convergent' or something along those lines?

David

Kerwin Hui

Posted on Sunday, 11 January, 2004 - 03:18 pm:

Just to add to what David has said - the sequence converges weakly to 0 in L^p , but that is probably not what you are after. Convergence of a sequence depends on which space it lives.

Kerwin

Edwin Koh

Posted on Monday, 12 January, 2004 - 05:15 am:

Sorry for the ambiguity - I meant pointwise convergence.

Michael Doré

Posted on Monday, 12 January, 2004 - 11:56 am:

If a subsequence converges to 0 pointwise then it also converges to 0 to

L^{2}

(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 $f_{k_{r}} \to 0$ pointwise then you can inductively construct a subsequence $f_{k_{r_{n}}}$ and closed nested intervals $I_{n}$ such that $f_{k_{r_{n}}} \geq 1 / 2$ on $I_{n}$ . The intervals have a point of intersection $x$ , and $f_{k_{r}} (x)$ clearly doesn't tend to 0 - a contradiction.

Kerwin Hui

Posted on Monday, 12 January, 2004 - 12:07 pm:

If a subsequence of f_k converges pointwise, it must converge in L² [0,2pi) (dominated convergence). Now f_k 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 L^p . But this is obviously nonsense since the integral of |f_k | is the same nonzero number.

Kerwin