Let $a>0$ and $f,g:[0,+\infty) \to \Bbb{R}$ where $f$ is a Lebesgue integrable function and $g$ has the property: $$|\frac{g(t)}{t}| \leq a, \forall t \geq 1$$. Prove that $\lim_{t \to +\infty}\frac{1}{t}\int_1^tf(x)g(x)dx = 0$
Here is my proof:
$$|\frac{1}{t}\int_1^tf(x)g(x)dx| \leq \frac{1}{t}\int_1^t|f(x)||g(x)|dx$$ $$=\frac{1}{t}\int_1^{\sqrt{t}}|f(x)||g(x)|dx+\frac{1}{t}\int_{\sqrt{t}}^t|f(x)||g(x)|dx$$ $$\leq \frac{a||f||_1}{\sqrt{t}}+a\int_{\sqrt{t}}^t|f(x)|dx \to 0$$ as $t \to +\infty$ because $\int_{\sqrt{t}}^t|f(x)|dx \to 0$ from integrability of $f$.
Is this proof correct,or i am missing something?
Thank you in advance.
Your proof is correct. But you can also get this as an immediate consequence of DCT: $\frac 1 t I_{(1,t)} f(t)g(t) \to 0$ for each $x$ and this function is dominated in absolute value by $a|f|$ which is integrable.