calculus - Square integrable and related limit

Let a continuous function $x:[0,\infty)\rightarrow\mathbb{R}$. Does $x\in\mathcal{L}_2[0,\infty)$ (square integrable i.e. $\lim_{t\rightarrow\infty}\int_0^t{x^2(s)ds=c<\infty}$) implies $\lim_{t\rightarrow\infty}\int_0^t{e^{-\lambda(t-s)}x(s)ds}=0$ for every $\lambda>0$?

I can prove this if $x$ is bounded but does it also hold true for unbounded x?

Note that $\int_0^t{e^{-\lambda(t-s)}x(s)ds}$ is a bounded also square integrable function if $x$ is square integrable and no boundedness assumption on $x$ is needed for this.

Answer

Hint:

For any $\delta>0$, there exists $b>0$ such that $\int_{b}^\infty x^2(s)dx<\delta^2$. Denote $M=\max_{[0, b]} |x(s)|$.

As you noted, we have
$$\int_0^b e^{-\lambda(t-s)} x(s) ds\to 0 \text{ when $t\to \infty$.}$$

On the other hand, suppose $t\gg b$, we have

$$\int_b^t e^{-\lambda(t-s)} x(s) ds\le \sqrt{\int_b^t e^{-2\lambda(t-s)}ds\delta^2} \\ \le \frac{1}{2\lambda} \delta$$

It follows that the desired limit must be 0.