Would you please give me a hint on how to solve this problem:
Suppose $f(x)$ continuous in $[0,\infty)$ and for each $a,b>0$ and $c>b$, we have \begin{equation*} ab \left|\int_0^1 f\left(ax+c \right) dx \right|<1. \end{equation*} Prove that $\int_0^\infty f(x)dx$ converges. I do not understand how to start.
Thanks.
We have for all $a,b,c$ with $0 <a,b$ and $b < c$ that $$ \left|\int_{0}^1 f(ax + c) \,dx \right| < \frac{1}{ab} $$ Set $u = ax + c$, then $\frac{du}{dx} = a$ and $$ \left|\frac{1}{a} \int_c^{c+a} f(u) \,du\right| = \left|\int_{0}^1 f(ax + c) \,dx \right| < \frac{1}{ab} \text{.} $$
If you fix $c=1$, $b=\frac{1}{2}$ this yields $$ \left|\int_1^{1+a} f(u) \,du\right| < 2 \quad\Rightarrow\quad \lim_{a\to\infty} \left|\int_1^{a} f(u) \,du\right| < 2 \text{.} $$ Since $f$ is continuous on $[0,\infty)$, you can assume $\int_0^1 f(x) \,d$ converges, and then you're done.