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.

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


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.

  • $\begingroup$ 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. $\endgroup$ – RTJ Mar 15 '15 at 16:14
  • $\begingroup$ @CTNT After correcting a typo, I suppose my argument stands. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.