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Dear all,

I would like to know whether the following claim is true. In particular, if it is true, then I would like to know if there is some textbook that contains the statement and maybe even the proof:

Claim
$\forall n \in \mathbb{N}, \ \exists k_n > 0$ such that

for any real numbers $A_1, \ldots, A_n, \varphi_1, \ldots, \varphi_n$ and any distinct positive numbers $\lambda_1, \ldots, \lambda_n$,

$\sup_{t \in \mathbb{R}} \left| \sum_{i=1}^n A_i \cos(\lambda_i t + \varphi_i) \right| \geq k_n \max(|A_1|, \ldots, |A_n|)$.

I think I have a (quite complicated) proof for $n \leq 3$, but I would like to know whether the statement holds in general.

Many thanks indeed,
Julian.

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Thank you for the replies that I have received so far. Just to clarify: k_n MUST ONLY DEPEND ON n. Anyone know? Many thanks, Julian. –  Julian Newman Jun 3 '11 at 15:02
    
You probably want the $\lambda_i$ distinct modulo $2\pi$. –  Juris Steprans Jun 3 '11 at 15:17
    
@Juris: What difference does it make? The inequality is invariant under scaling of all the $\lambda_i$’s by a nonzero constant. –  Emil Jeřábek Jun 3 '11 at 16:06

1 Answer 1

up vote 5 down vote accepted

Yes, this is true. It follows from the fact that the average $$\lim_{T\to\infty} {1\over T}\int_0^T\cos(\lambda_i t+\phi_i)\sum_j A_j\cos(\lambda_j t+\phi_j)\,dt$$ is $A_i/2$. Clearly the average is bounded by the supremum of the function.

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Many thanks for the answer. I presume the first two subscript i are meant to be a fixed number, and the second two are meant to be what you sum over. –  Julian Newman Jun 3 '11 at 16:11
    
I have edited it accordingly. –  Michael Renardy Jun 3 '11 at 17:11

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