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Let $f,g\in PC(2\pi)$ continuous. Show that $$\lim_{n\to \infty}\int_{-\pi}^{\pi}f(x)g(nx)\,dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx\int_{-\pi}^{\pi}g(x)\,dx$$

Where if $f\in PC(2\pi)$ then $\int_{-\pi}^{\pi}f(x)\,dx=\int_c^{c+2\pi}f(x)\,dx \quad\forall c\in\mathbb{R}.$

I've been struggling with this proof. Any suggestions would be great!

DMcMor
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Sofía Contreras
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1 Answers1

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Here's a sketch:

Let $\mu_g = \int_{-\pi}^{\pi} g(x) \,dx$. Fix $\epsilon > 0$. Since $f$ is continuous, it's also uniformly continuous, so let $n$ be big enough so that $|f(x) - f(y)| < \epsilon$ whenever $|x-y| < 2\pi/n$. Now we can write \begin{align*} \int_{-\pi}^{\pi} f(x) g(nx) \,dx \ &= \ \frac1n \int_{-n\pi}^{n\pi} f(y/n) g(y) \,dy \ = \ \sum_{j=-n}^{n-2} \frac1n \int_{j\pi}^{(j+2)\pi} f(y/n) g(y) \,dy \\ &= \ \sum_{j=-n}^{n-2} \frac1n \int_{j\pi}^{(j+2)\pi} \Big( f(j\pi/n) \pm \epsilon \Big) g(y) \,dy \ = \ \mu_g \cdot \sum_{j=-n}^{n-2} \frac1n \Big( f(j\pi/n) \pm \epsilon \Big). \end{align*}

Now the summation can just be recognized as a Riemann sum which converges to $\frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \,dx$ as $n \to \infty$.

Adam
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  • how can I go from the second equality to the third one? I know that it's related to the continuity of $f$ but I can't fully understand how. Thanks! – Jaime_Borel Nov 17 '20 at 05:08
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    Since $f$ is continuous, $f(t)$ can change by at most $\epsilon$ as $t$ varies over an interval of length $2\pi/n$. So in the third equality i am saying that $f(y/n)$ changes by at most $\epsilon$ as $y$ ranges over the interval $[j\pi, (j+2)\pi]$ (there was a small typo in the original answer which is now fixed) – Adam Nov 17 '20 at 05:22
  • Why is the 2nd equality true? Let's take $n = 2$. Then you would divide $[-2\pi, 2\pi]$ into $[-2\pi, 0]$ and $[0, 2\pi]$. But if we look at your answer in the 2nd equality, we have $$\int_{-2\pi}^0 + \int_{-\pi}^{\pi} + \int_{0}^{2\pi}.$$ So the interval is divided into 3 intervals with length $2\pi$. – Vicky Jan 07 '21 at 09:57
  • Ah, good point. So have $j$ count in steps of 2 instead – Adam Jan 07 '21 at 16:58