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Let $y: \mathbb{R}_+ \to \mathbb{R}$

Suppose that $\bar y(t) = \dfrac{1}{t}\int\limits_0^t y(s)ds$, where $t$ Is time.

How does $$\lim\limits_{t \to \infty} \bar y(t) = \lim\limits_{t \to \infty}\dfrac{1}{t}\int\limits_0^t y(s)ds$$

compare with $$\lim\limits_{t \to \infty} y(t)?$$

Under what condition are they equal?

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    That the second limit may fail to exist when the first one does is analogous to the existence of the Cesaro Mean $C=\lim_{n\to \infty}\frac {1}{n}\sum_{j=1}^nA_n$ not implying that $\lim_{n\to \infty}A_n$ exists. E.g. if $A_{n!}=n$, and $A_m=0$ when $m$ is not a factorial, then $ C=0$ but $(A_n)_n$ is not even a bounded sequence.... Suppose $\int_{n!}^{n!+1}f(x)dx=n$ but $f(x)=0$ when $x\in [m,m+1]$ when $m$ is not a factorial. – 2017-01-31

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Always, provided the latter limit exists. If $\lim_{t\to\infty}y(t)$ exists, then $\int_0^ty(s)ds$ goes to either $0$, a nonzero constant $c$, or $\pm\infty$. If it goes to $0$ or a nonzero constant $c$, then $y(t)$ must go to zero; $\lim_{t\to\infty}\frac{1}{t}\int_0^ty(s)ds = 0 = \lim_{t\to\infty}y(t)$. If it goes to $\pm\infty$, then we can use L'Hopital's rule on the first limit; recall that, by the Fundamental Theorem of Calculus, $\frac{d}{dt}\int_0^ty(s)ds = y(t)$. So, by L'Hopital, $\lim_{t\to\infty}\frac{1}{t}\int_0^ty(s)ds = \lim_{t\to\infty}\frac{y(t)}{1} = \lim_{t\to\infty}y(t)$.

Note that it is possible for $\lim_{t\to\infty}\frac{1}{t}\int_0^ty(s)ds$ to converge without $\lim_{t\to\infty}y(t)$ converging; $y(t) = \sin{t}$ is a good example. I don't know whether that's relevant to your question.

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    Hi is there a book that states this result? Thanks! – 2017-01-31
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    @DonaldJ.Panda This particular result? No, this is far too specific. But L'Hopital's rule and the Fundamental Theorem of Calculus can be found in any decent calculus textbook. Personally, I recommend Stewart or Larsen. – 2017-01-31
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    If y is monotonic and the first limit exists then the the second one does too, and they are equal. – 2017-01-31