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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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    Yes I changed $w$ into $m$ as it should be.2012-10-21

1 Answers 1

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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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    @alvoutila like that but not necessarily with identical coefficients of course. So linear combinations over $\mathbb{C}$ of the basis functions $e^{i k t}$.2012-10-22