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How to show that for arbitrary( complex) trigonometric polynomials $P$ and $Q$ holds

$$\frac{1}{2\pi} \int_{-\pi}^{\pi} P(t)Q(mt)dt = \frac{1}{2\pi} \int_{-\pi}^{\pi} P(t)dt \frac{1}{2\pi} \int_{-\pi}^{\pi} Q(t)dt$$

always when $m \in \mathbb{Z}$ is sufficiently large.

Hint: Use information that

$$\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{int}dt = 0\,\,,\,\,when \,\,n\neq 0\,\,\,,\,\, and \,\,\,\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{int}dt = 1\,\,\, when\,\,n=0$$ Just some hint. I know that complex trigonometric polynomials is

$$\sum_{n=-m}^{m} c_n e^{int}$$

but I also know that if $\,f(t)=t\,$ and $\,g(t)=t+1\,$ , then

$$\int_{-\pi}^{\pi} f(t)g(t)dt \neq \int_{-\pi}^{\pi} f(t)dt \cdot \int_{-\pi}^{\pi} g(t)dt$$

So I doubt that it is true for complex trigonometric polynomial.

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$w$ occurs in the premise, but not in the claim. $m$ occurs in the claim, but not in the premise. Copied from illegible handwriting? –  Hagen von Eitzen Oct 21 '12 at 13:01
    
What is that $\,w\,$ there? Something related to both functions' periodicy? –  DonAntonio Oct 21 '12 at 13:03
    
Without taking that still weird $\,w\,$ into account, I think your example proves the claim's utterly false... –  DonAntonio Oct 21 '12 at 13:08
    
Yes I changed $w$ into $m$ as it should be. –  laovultai Oct 21 '12 at 16:58

1 Answer 1

up vote 1 down vote accepted

The functions $t \mapsto e^{i k t}$ for $k \in \mathbb{Z}$ form an orthonormal set on the interval $[-\pi, \pi]$ with inner product

$$ \langle Q, P \rangle = \frac{1}{2 \pi} \int_{-\pi}^{\pi}Q(t)\overline{P(t)}dt. $$

If $m$ is sufficiently large in your expression, the only term from $Q$ that contributes to the stated inner product is therefore its constant term. This constant term is expressed by the latter integral of $Q$ (the inner product of $Q$ with the constant function $\mathbf{1}$ if you wish). The complex conjugate that appears in the inner product is irrelevant: simply consider $\langle Q, \overline{P} \rangle$.

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How would you define both $P(t)$ and $Q(t)$? $$P(t)= \sum_{n=-m}^{m} a_n e^{int}$$ and $$Q(t)=\sum_{n=-m}^{m} a_n e^{int}$$? –  laovultai Oct 21 '12 at 19:54
1  
@alvoutila like that but not necessarily with identical coefficients of course. So linear combinations over $\mathbb{C}$ of the basis functions $e^{i k t}$. –  WimC Oct 22 '12 at 5:18

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