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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?2012-10-21
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    What is that $\,w\,$ there? Something related to both functions' periodicy?2012-10-21
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    Without taking that *still weird* $\,w\,$ into account, I think your example proves the claim's utterly false...2012-10-21
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    Yes I changed $w$ into $m$ as it should be.2012-10-21

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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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    How would you define both $P(t)$ and $Q(t)$? $$P(t)= \sum_{n=-m}^{m} a_n e^{int}$$ and $$Q(t)=\sum_{n=-m}^{m} a_n e^{int}$$?2012-10-21
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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