My instructor briefly discussed a result in lecture that I need help with. Here was the set up. We

cdsommegolfzp 2022-07-01 Answered
My instructor briefly discussed a result in lecture that I need help with. Here was the set up.
We assumed that { f n } is an orthonormal system in L 2 ( 0 , 1 ) and we supposed that there exists an M > 0 such that | f n ( x ) | M a.e for all n N . Furthermore, we let { c n } be a sequence of real numbers such that n = 1 c n f n cnfn converges a.e.
They said that it was "clear" that
lim n c n = 0.
This is not clear or intuitive for me.
After reading through the page, I am still a bit confused. Can someone put together a working proof to clear things up?
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Answers (1)

Shawn Castaneda
Answered 2022-07-02 Author has 17 answers
Suppose the sum of an infinite series is S ,and we denote by S n the n −th element of the sequence of partial sums of the series. By definition, the series converges iff the limit of S n exists finitely ( and in this case S n S ) , so:
S = n = 1 a n , S n := k = 1 n a k a n = S n S n 1 n S (arith. of limits) S = 0
and we have the basic necessary condition for convergence of infinite series.
In your case you have | c n f n | n 0 (no matter whether real of complex numbers), but { f n } is a bounded sequence...! Finish the argument.
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