By the Riemann-Lebesgue lemma, I have shown that for any finite interval measurable set $I$ of finite measure, any $h \in \mathbb{R}$, $$\lim_{n \to\infty}\int_I \cos (n(x+h)) \mathop{dx} = 0.$$

I also have an arbitrary real sequence $\{a_n\}$, and I would like to show $$\lim _{n \to \infty} \int_I \cos\left(n\left(x+\frac{a_n}{n}\right)\right)\mathop{dx} = 0$$

How can we see this? I tried by noting that for fixed $k$, we have $$\lim _{n \to \infty} \int_I \cos\left(n\left(x+\frac{a_k}{k}\right)\right)\mathop{dx} = 0$$ by the first equation above. I tried to do take this and do a "diagonalization argument," but I convinced myself that it would not work (if a countable number of sequences converge to a common limit, its "diagonal" sequence does not necessarily converge to that limit, if at all). I guess I haven't exploited the periodicity of cosine yet... any suggestions? Thanks!


1 Answer 1


Notice that your function can be easily integrated, so we don't need the Riemann-Lebesgue lemma.

Say, your interval is $I=[a,b]$. Then

$$\int_a^b \cos(nx+a_n) dx=\left.\frac{1}{n}\sin(nx+a_n)\right|_a^b$$

which converges to $0$ as $n\rightarrow\infty$.

More generally:

Claim: Let $f\in L^1(I)$ and $(a_n)_n$ any sequence of real numbers. Then $$\lim_{n\rightarrow\infty}\int_I \cos(nx+a_n) f(x) dx=0$$

The proof is the same as for the Riemann-Lebesgue lemma:

  • Assume without loss of generality that $f$ is smooth (smooth functions are dense in $L^1(I)$).
  • Do an integration by parts, integrating the $\cos$-term and giving a factor $1/n$.
  • Notice that this converges to $0$.
  • $\begingroup$ What happens if we generalize to measurable sets $I$ of finite measure? $\endgroup$
    – angryavian
    Commented Feb 28, 2014 at 16:53
  • $\begingroup$ The same, we also don't need finite measure. In fact this holds also on all of $\mathbb{R}$. $\endgroup$
    – J.R.
    Commented Feb 28, 2014 at 16:56
  • $\begingroup$ The point of Riemann-Lebesgue and the oscillatory integrals in general is to integrate by parts. That really is the whole story. $\endgroup$
    – J.R.
    Commented Feb 28, 2014 at 16:59

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