Fourier analysis — Proving an equality given $f, g \in L^1[0, 2\pi]$ and $g$ bounded

Question

We were given a challenge by our Real Analysis professor and I've been stuck on it for a while now. Here's the problem:

Consider the $2\pi$-periodic functions $f, g \in L^1[0, 2\pi]$. If $g$ is bounded show that $$ \lim_{n \to +\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(t)g(nt)\,dt = \frac{1}{2\pi}\int_{0}^{2\pi}f(t)\,dt \cdot \frac{1}{2\pi}\int_{0}^{2\pi}g(t)\,dt. $$

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I first thought of using the Riemann-Lebesgue theorem, i.e. if $f \in L^1[a, b]$ then $$\lim_{R \to +\infty}\int_{a}^{b}f(t)\cos{Rt} \, dt = 0 \;\; \text{and} \;\; \lim_{R \to +\infty}\int_{a}^{b}f(t)\sin{Rt} \, dt = 0$$

but it didn't get me far.

Answer 1

Let $\epsilon > 0$ be given.

Let $M$ be a bound for $|g|$, which is assumed to exist. Define $f_{N}(t)=f(t)\chi_{\{ x : |f(x)| \le N\}}(t)$. Let $g_{\delta}$ be a standard mollification of $g$. Then $g_{\delta}\in\mathcal{C}^{\infty}(\mathbb{R})$ is $2\pi$-periodic, $|g_{\delta}| \le M$, and $$ \lim_{\delta\rightarrow 0}\int_{0}^{2\pi}|g(t)-g_{\delta}(t)|dt =0. $$ Let $S_{g_{\delta}}^{K}$ be the truncated Fourier series for $g_{\delta}$; the Fourier series converges uniformly to $g_{\delta}$ as $K\rightarrow\infty$. Then \begin{align} \int_{0}^{2\pi}f(t)g(nt)dt &= \int_{0}^{2\pi}(f(t)-f_{N}(t))g(nt)dt \tag{1}\\ & +\int_{0}^{2\pi}f_{N}(t)(g(nt)-g_{\delta}(nt))dt \tag{2}\\ & +\int_{0}^{2\pi}f_{N}(t)(g_{\delta}(nt)-S_{g_{\delta}}^{K}(nt))dt \tag{3}\\ & +\int_{0}^{2\pi}f_{N}(t)S_{g_{\delta}}^{K}(nt)dt.\tag{4} \end{align} The first term on the right is bounded by $$ \int_{0}^{2\pi}|f(t)-f_{N}(t)||g(nt)|dt \le M\int_{0}^{2\pi}|f(t)\chi_{\{x : |f(x)| > N\}}|dt, \tag{1} $$ which tends to $0$ as $N\rightarrow\infty$. Choose $N$ large enough that the above is strictly bounded by $\frac{\epsilon}{4}$. Then, for this fixed $N$, \begin{align} \left|\int_{0}^{2\pi}f_N(t)(g(nt)-g_{\delta}(nt))dt\right| & \le N\int_{0}^{2\pi}|g(nt)-g_{\delta}(nt)|dt \\ & = N\int_{0}^{2\pi/n}|g(nt)-g_{\delta}(nt)|d(nt) \\ & = N\int_{0}^{2\pi}|g(t)-g_{\delta}(t)|dt \rightarrow 0 \mbox{ as } \delta\rightarrow 0. \tag{2} \end{align} Choose $\delta > 0$ small enough that the above is strictly bounded by $\frac{\epsilon}{4}$. And $(3)$ is bounded by $$ 2\pi N\sup_{0\le t\le 2\pi}|g_{\delta}(t)-S_{g_{\delta}}^{K}(t)|, $$ which is strictly bounded by $\frac{\epsilon}{4}$ for $K$ large enough because $g_{\delta}\in\mathcal{C}^{\infty}(\mathbb{R})$ is periodic. Finally, $(4)$ can be bounded by $\frac{\epsilon}{4}$ by taking $n$ large enough, which follows from the Riemann-Lebesgue lemma. Hence, it follows that the following holds for all large enough $n$: $$ \left|\int_{0}^{2\pi}f(t)g(nt)dt\right| < \epsilon $$

Answer 2

Here is an argument using formal manipulations without justifying the interchange of various limits. It allows us to see why the result should be true, provided that the interchanges can be justified.

Let us suppose that the Fourier series of $g$ converges to $g$, so we may write $$g(t) = \sum_{k=-\infty}^{\infty}a_k e^{ikt}$$ where $$a_k = \frac{1}{2\pi}\int_0^{2\pi}g(t) e^{-ikt}dt$$ Then $$g(nt) = \sum_{k=-\infty}^{\infty}a_k e^{iknt}$$ Now substitute this into the desired integral and rearrange naively: $$\begin{aligned} \frac{1}{2\pi}\int_{0}^{2\pi}f(t) g(nt) dt &= \frac{1}{2\pi}\int_0^{2\pi}f(t) \sum_{k=-\infty}^{\infty} a_k e^{iknt} dt \\ &= \frac{1}{2\pi}\sum_{k=-\infty}^{\infty} a_k \int_0^{2\pi} f(t) e^{iknt} dt \\ \end{aligned}$$ Now take the limit as $n \to \infty$: $$\begin{aligned} \lim_{n \to \infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(t) g(nt) dt &= \lim_{n \to \infty}\frac{1}{2\pi}\sum_{k=-\infty}^{\infty} a_k \int_0^{2\pi} f(t) e^{iknt} dt \\ &= \frac{1}{2\pi}\sum_{k=-\infty}^{\infty} a_k \left(\lim_{n \to \infty} \int_0^{2\pi} f(t) e^{iknt} dt \right)\\ \end{aligned}$$ The term inside the parentheses is zero if $k \neq 0$, by the Riemann-Lebesgue lemma. For $k=0$ it is simply $\int_0^{2\pi} f(t) dt$, so right hand side reduces to the $k=0$ term, which is $$a_0 \frac{1}{2\pi}\int_0^{2\pi} f(t) dt = \frac{1}{2\pi}\int_0^{2\pi} g(t) dt \cdot \frac{1}{2\pi}\int_0^{2\pi}f(t) dt$$ as desired.

Of course the above is not a formal proof, and may not be correct with sufficiently badly behaved functions. For example, the Fourier series of an arbitrary $L^1$ function may not converge everywhere or even almost everywhere, so we're already in trouble at the first line of the "proof."

But with some additional limiting arguments you may be able to make it formally correct, and it's already correct for any "reasonable" function :-)

Answer 3

I do not have enough reputation to comment, so I write here a minor mistake of TrialAndError. The integral $$\int_0^{2\pi}f_N(t)S_{g_\delta}^K(nt)dt$$ may cannot be bounded by $\varepsilon/4$, as there may exists a constant term in $S_{g_\delta}^K(t)$. That is the reason why the limit is not $0$. We can assume though that the limit is equal to $0$, since $$\lim_{n \rightarrow \infty}\dfrac{1}{2\pi}\int_0^{2\pi}f(t)g(nt)dt= \dfrac{1}{2\pi}\int_0^{2\pi}f(t)dt\dfrac{1}{2\pi}\int_0^{2\pi}g(t)dt\iff$$

$$\lim_{n \rightarrow \infty}\dfrac{1}{2\pi}\int_0^{2\pi}f(t)\left(g(nt)-\dfrac{1}{2\pi}\int_0^{2\pi}g(x)dx\right)dt=0$$

Now, the molifications $g_\delta$ can be selected such that their constant terms are equal to $0$, and thus the constant terms of $S_{g_\delta}^K$ are equal to $0$.