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Consider $f : [0,\infty) \rightarrow \mathbb{R}$ be a function such that $\lim_{t\rightarrow \infty} f(t) = 0$. I was wondering if the following relation holds $$lim_{t\rightarrow\infty}\int_0^t |f(\tau)|d\tau = lim_{t\rightarrow\infty}\int_\lambda^t |f(\tau-\lambda)|d\tau$$ for some constant $\lambda>0$. Intuitively I'm thinking it is reasonable as guess, since if $f$ tends to zero and $\lambda$ if finite, as $t$ grows to infinity then, very informally, $f([t-\lambda,t])$ tends not to contribute to the left integral. However I'm looking for a quite rigorous proof for that (if it exists) and also to some conditions $f$ must satisfy. What I thought was to rewrite the first integral as $$ \int_0^t |f(\tau)|d\tau = \int_\lambda^{t+\lambda} |f(\xi - \lambda)|d\xi = \int_\lambda^t|f(\xi -\lambda)|d\xi + \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi$$ and then to show the the two latter integral does not contribute as $t\rightarrow\infty$, however I don't know how to deal with them. For example i can write, for $t\ge \lambda$ $$\int_t^{t+\lambda}|f(\xi-\lambda)|d\xi = \int_0^{\lambda}|f(\chi -\lambda+t)|d\chi$$ and the trying to show that $$lim_{t\rightarrow\infty}\int_0^{\lambda}|f(\chi -\lambda+t)|d\chi = \int_0^{\lambda}lim_{t\rightarrow\infty}|f(\chi -\lambda+t)|d\chi = 0$$ however I don't know what conditions I should pose for $f$ in such a way to be able to do that. I looked at the Lebesgue dominated convergence theorem, however I'm not dealing with successions of functions. Any hint?

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  • $\begingroup$ The problem is that you have defined $f$ only on $[0,+\infty[$ but you are taking $f(t-\lambda)$ with $\lambda>0$ and $0<t<\infty$ $\endgroup$
    – Martigan
    Commented May 29, 2015 at 12:00
  • $\begingroup$ ok I corrected the question, $\endgroup$
    – milo
    Commented May 29, 2015 at 13:12

2 Answers 2

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Ok, let's start with $$ \int_0^t |f(\tau)|d\tau = \int_\lambda^{t+\lambda} |f(\xi - \lambda)|d\xi = \int_\lambda^t|f(\xi -\lambda)|d\xi + \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi$$ Then $$\exists t_1: \forall t>t_1: f(t-\lambda)<\frac{1}{\lambda} \Rightarrow \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi < 1$$ and $$\exists t_2: \forall t>t_2: f(t-\lambda)<\frac{1}{\lambda^2} \Rightarrow \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi < \frac{1}{\lambda}$$ and so forth $$\exists t_n: \forall t>t_n: f(t-\lambda)<\frac{1}{\lambda^n} \Rightarrow \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi < \frac{1}{\lambda^{n-1}}$$ This sequence converges obviously for $\lambda > 1$ and your guess is correct.
The other cases shouldn't be too hard.

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  • $\begingroup$ smart. let me recap to see if I understood. From the definition of limit, it is $\forall \epsilon >0, \exists t_0(\epsilon) s.t. \forall t>t_0(\epsilon) |f(t-\lambda)| < \epsilon$ (and thus the integral is less than $\epsilon\lambda$). Thus in particular this is true for $\epsilon=\lambda^{-1}$, ..., $\epsilon = \lambda^{-n}$ And from the arbitrary of $n$ it follows that the integral is zero as t goes to inifinite, right? no more conditions are required for $f$ ? $\endgroup$
    – milo
    Commented May 29, 2015 at 14:18
  • $\begingroup$ You understood and I don't see any conditions required. I didn't work with convergence for quite some time (years) though, so anyone is free to correct me :p $\endgroup$
    – Inuyaki
    Commented May 29, 2015 at 15:09
  • $\begingroup$ great, what are the "other cases" you're referring to? $\endgroup$
    – milo
    Commented May 29, 2015 at 15:10
  • $\begingroup$ $\lambda = 1$ and $\lambda < 1$, but they should be similar $\endgroup$
    – Inuyaki
    Commented May 29, 2015 at 15:12
  • $\begingroup$ ah ok, thanks a lot! $\endgroup$
    – milo
    Commented May 29, 2015 at 15:31
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You can see that with the new equation you propose you have:

$\int_\lambda^t |f(\tau-\lambda)|d\tau=\int_0^{t-\lambda} |f(u)|du$

So $\int_0^t |f(\tau)|d\tau-\int_\lambda^t |f(\tau-\lambda)|d\tau=\int_{t-\lambda}^t |f(\tau)|d\tau$

You use the definition of $\lim_{t\rightarrow \infty} f(t) = 0$

That is $\forall \epsilon,\lambda >0$ , $\exists t_0 \forall t>t_0-\lambda$, $|f(t)|<\dfrac{\epsilon}{\lambda}$

etc.

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