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I was working on a question very similar to this post: Show that $\int_{-\pi}^\pi ~f(x) \cos (nx) \mathrm{d}\mu(x)$ converges to $0$ .

I want to show that $\lim_{n \rightarrow \infty}\int_{-M}^M f(x) \cos (nx) dx=0$ for a closed and bounded interval $[-M,M]$ provided $f\in L^1([-M,M])$. The answer to the link above says that for any step function $g$ in $[-M,M]$, $$\lim_{n \rightarrow \infty}\int_{-M}^M g(x) \cos (nx) dx=0.$$

I want to supply the details that will prove that the limit of the original integral will vanish. Since, step functions in $[-M,M]$ are dense in $L^1([-M,M])$, then there exists a sequence $(g_m)$ if step functions in $[-M,M]$ that converges to $f$ in the $L^1([-M,M])$ norm. We also know that up to a subsequence $(g_{m_{k}})$, it also converges pointwise a.e. to $f$. Therefore, since $$|g_{m_k}cos(nx)|\leq|g_{m_k}|,$$ and $$\lim_{k\rightarrow \infty}\int_{-M}^M|g_{m_k}|=\int_{-M}^M|f|,$$ then by the Generalized Lebesgue Dominated Convergence Theorem (we can 'pass the limit'): $$\lim_{n \rightarrow \infty}\int_{-M}^M f(x) \cos (nx) dx = \lim_{n \rightarrow \infty}\int_{-M}^M \lim_{k \rightarrow \infty} g_{m_k}(x) \cos (nx) dx $$ $$\lim_{n \rightarrow \infty}\int_{-M}^M f(x) \cos (nx) dx = \lim_{k \rightarrow \infty} \lim_{n \rightarrow \infty}\int_{-M}^M g_{m_k}(x) \cos (nx) dx=0.$$ This proves the theorem. Is my proof correct?

Community
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Kurome
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2 Answers2

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By the Riemann-Lebesgue Lemma we know that if $f\in L^1$ then $$\lim_{|n|\to\infty} \hat{f}(n)=0$$ this is to say that $$0=\lim_{n\to\infty} \int_\mathbb{R} f(x) e^{in x} dx=\lim_{n\to\infty}\int_\mathbb{R} f(x)\cos nx dx+i\int_\mathbb{R}f(x)\sin nx dx$$ therefore$$ \lim_{n\to\infty} \int_\mathbb{R} f(x)\cos nx dx=0\;\;\mbox{and}\;\; \lim_{n\to\infty} \int_\mathbb{R} f(x)\sin nx dx=0 $$

Sebastián Naranjo Álvarez
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This may be seen as a particular case of Féjer's Lemma:

Fejér's formula: Suppose $g$ is a bounded measurable $T$-periodic function on $\mathbb{R}$ ($T>0$). For any integrable function $f$ (denoted by $f\in\mathcal{L}_1(\mathbb{R})$) and numeric sequence $a_n\in\mathbb{R}$, $$ \lim_n\int_\mathbb{R} f(x)g(nx+a_n)\,dx=\Big(\frac{1}{T}\int^T_0 g\Big)\int_\mathbb{R} f \tag{1}\label{one} $$

A proof based on the density of step functions in $L_1(\mathbb{R})$ can be found in this posting

Mittens
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