# 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.

• I have explained more, hoping it will be helpful. – Ma Ming Mar 20 '15 at 2:16

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.

• Thank you for your comment. Using the above argument we can prove that $\lim_{t_1\rightarrow\infty,t_2\rightarrow\infty}\int_{t_1}^{t_1+t_2}{e^{- \lambda (t_1+t_2-s)}x(s)ds}=0$. The original question however appears unanswered. – RTJ Mar 15 '15 at 16:14
• @CTNT After correcting a typo, I suppose my argument stands. – Ma Ming Mar 15 '15 at 16:23

I will elaborate on this. Following Ma's suggestion, from the Cauchy-Schwarz inequality we have

\begin{align} \bigg|\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}\bigg| &\leq \int_{t}^{t+\delta}{\big|e^{- \lambda(t+\delta-s)}x(s)\big|ds}\\ &\leq \sqrt{\int_{t}^{t+\delta}{e^{-2\lambda (t+\delta-s)}ds}\int_{t}^{t+\delta}{x^2(s)ds}}\\ &= \sqrt{\frac{1-e^{-2\lambda \delta}}{2\lambda}}\sqrt{\int_{t}^{t+\delta}{x^2(s)ds}}. \end{align}

For a square integrable $x$ it holds true that $\lim_{t\rightarrow\infty}\int_t^{t+\delta}{x^2(s)ds}=0$.

Using this fact and taking the limit $t\rightarrow\infty$ in the inequality above we obtain $\lim_{t\rightarrow\infty}\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}=0$.

We can write $\int_0^{t+\delta}{e^{-\lambda(t+\delta-s)}x(s)ds}=e^{-\lambda\delta}\int_0^t{e^{-\lambda (t-s)}x(s)ds}+\int_{t}^{t+\delta}{e^{- \lambda(t+\delta-s)}x(s)ds}$.

If we now define $S:=\lim_{t\rightarrow\infty}\int_0^t{e^{-\lambda (t-s)}x(s)ds}$ and take the limit in the identity above we have $S=e^{-\lambda\delta}S$ that yields the desired $S=0$. However, one has to prove that the limit $S$ actually exists.