Integral convergence question

Question

Let $ f: [a,\infty ) \to \mathbb R $ be $ f\in C([a,\infty)) $, with a cycle $ T > 0 $,

and let $ g: [a,\infty ) \to \mathbb R $ be a monotonic function, and $$ \lim_{x\to \infty} g(x) = 0$$

Assume that $$ \int_a^{a+T} f(x)\,dx = 0 $$

Prove that: $$ \int_a^\infty f(x)g(x)\,dx $$ converges.

Help?

Answer 1

Hint: if $a_{n}$ and $b_{n}$ are sequences such that

$$ \lim_{n\to\infty}a_{n} = 0 \\ $$ and there exists a $M$ such that for all $N$ $$ | \sum_{n=1}^{N} b_{n} | \leq M, $$ i.e. the partial sum is bounded, then $$ \sum a_{n} b_{n} $$ converges.

Answer 2

Assume without loss of generality that $g(x)\ge 0$.

First, let $a=\int_0^T \lvert f(x)\rvert\,dx$, then \begin{align} \int_s^{s+T}f(x)\,g(x)\,dx&=\int_s^{s+T}f(x)\,\big(g(x)-g(s+T)\big)\,dx+\int_s^{s+T}f(x)\,g(s+T)\,dx \\ &=\int_s^{s+T}f(x)\,\big(g(x)-g(s+T)\big)\,dx+g(s+T)\int_s^{s+T}f(x)\,dx \\ &=\int_s^{s+T}f(x)\,\big(g(x)-g(s+T)\big)\,dx, \end{align} and hence $$ \Big|\int_s^{s+T}f(x)\,g(x)\,dx\,\Big|\le \big(g(s)-g(s+T)\big)\int_s^{s+T}\lvert f(x)\rvert\,\,dx=a\big(g(s)-g(s+T)\big). $$

We need to show that for any $\varepsilon>0$, there exists an $M>0$, such that, for every $N>M$, $$ \Big|\int_M^N f(x)\,g(x)\,dx\,\Big|<\varepsilon. $$ As $\lim_{x\to\infty}g(x)=0$, let $M>0$, such that $$ag(M)<\varepsilon.$$ Let now $N>M$, and assume that $$ M<M+s<\cdots<M+ks\le N<M+(k+1)s. $$ Then \begin{align} \int_M^N f(x)\,g(x)\,dx&=\sum_{j=1}^k \int_{M+(j-1)s}^{M+js}f(x)\,g(x)\,dx+\int_{M+ks}^{N}f(x)\,g(x)\,dx \\ &=\sum_{j=1}^k \int_{M+(j-1)s}^{M+js}f(x)\,\big(g(x)-g(M+js)\big)\,dx+\int_{M+ks}^{N}f(x)\,g(x)\,dx, \end{align} and hence \begin{align} \Big|\int_M^N f(x)\,g(x)\,dx\,\Big|&\le \sum_{j=1}^k \int_{M+(j-1)s}^{M+js} \lvert f(x)\rvert\,\big(g(x)-g(M+js)\big)\,dx+\int_{M+ks}^{N}\lvert f(x)\rvert\,g(x)\,dx \\ &\le \sum_{j=1}^k a\big(g(M+(j-1)s)-g(M+js)\big)+ag(M+ks)=ag(M)<\varepsilon. \end{align}