Suppose $f(x)\geq 0$, and $\int_0^{+\infty} f^2(x)dx$ is convergent. Prove that $\lim\limits_{x \to \infty}\dfrac{\int_0^x e^t f(t) dt}{e^x}=0.$
Notice that we are not given the continuity of $f(x)$. Hence L' Hospital's rule can not work here. If we consider apply AM-GM inequality, we obtain $$e^{x}f(x)\leq \frac{f^2(x)+e^{2x}}{2},$$ where $\int_0^{+\infty} f^2(x)dx<+\infty$ but $\int_0^{\infty} e^{2t}=+\infty$, which gives nothing helpful.
How to solve it? Thanks.
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Since $\displaystyle\int_0^{+\infty}f^2(x)dx$ is convergent，by Cauchy's convergence test, we have $$\forall \varepsilon>0,\exists \xi>0,\forall x>\xi ~~~s.t.~~~ \int_{\xi}^{x} f^2(t)dt< 2\varepsilon^2.$$ Thus，as per Cauchy-Schwarz's inequality，we obtain $$\int_{\xi}^x e^t f(t)dt \leq \left(\int_{\xi}^x f^2(t)dt \cdot \int_{\xi}^x e^{2t}dt\right)^{\frac{1}{2}}< \left(2\varepsilon^2 \int_{\xi}^x e^{2t}dt\right)^{\frac{1}{2}}=\varepsilon \left(e^{2x}-e^{2\xi}\right)^{\frac{1}{2}},$$ which implies $$\frac{\int_{\xi}^x e^t f(t)dt}{e^x}\leq \varepsilon \left(1-e^{\frac{\xi}{x}}\right)^{\frac{1}{2}}\leq \varepsilon$$ holds for all $x>\xi$. Therefore， taking the limits of both sides as $x \to +\infty$, we have $$\lim_{x \to +\infty}\frac{\int_{\xi}^x e^t f(t)dt}{e^x}\leq \varepsilon.\tag{1}$$ Meanwihle, notice that，for the fixed $\xi$, $$\lim_{x \to +\infty}\frac{\int_0^\xi e^t f(t)dt}{e^x}=0.\tag{2}$$ $(1)$ plus $(2)$, we obtain $$\lim_{x \to +\infty}\frac{\int_0^x e^t f(t)dt}{e^x}\leq\varepsilon,$$ by the arbitariness of $\varepsilon>0$，which implies $$\lim_{x \to +\infty}\frac{\int_0^x e^t f(t)dt}{e^x}=0.$$
Let $\epsilon >0$. Choose $\Delta $ such that $\int_{\Delta} ^{\infty} f(x)^{2}dx <\epsilon^{2}$. By C-S inequality we have $|\int_{\Delta} ^{x} e^{t}f(t)dt| \leq \epsilon (\int_{\Delta} ^{x} e^{2t}dt)^{1/2}=\epsilon (\frac {e^{2x}-e^{2\Delta}} 2)^{1/2}$. Hence $|\frac {\int_{\Delta} ^{x} e^{t}f(t)dt} {e^{x}}| <\epsilon 2^{-1/2}$. Next note that $\frac {\int_0^{\Delta} e^{t}f(t)dt} {e^{x}} \to 0$ as $x \to \infty$. Combining these two we get the result.