$f$ is a real and continuous function defined by $f(x+1)=f(x) \forall x\in\mathbb{R}$. $g(t)=\int_0^tf(x)\,dx$, $t \in \mathbb{R}$. Then $h(t)=$$\lim\limits_{n\to\infty}\frac{g(t+n)}{n}$ is defined for :

(a) $t=0$ only

(b) integer $t$ only

(c) $\forall t \in \mathbb{R}$ and $h(t)$ is independent of $t$

(d) none of the above is true

My attempt:

$g(t)= \int_0^tf(x)dx=f(c)(t-0) $ for some $c\in(0, t)$ by mean value theorem. $\implies g\prime(t)=f(c)$.Now, $h(t) = \lim\limits_{n\to\infty}\frac{g(t+n)}{n} = \lim\limits_{n\to\infty}\frac{g\prime(t+n)}{1}$. Now, as $g\prime=f $ is periodic, we can say, $h(t)=\lim\limits_{n\to\infty} g\prime(t)=f(c)$. Therefore option (c) is the correct answer. But the problem with the proof is that nowhere it is said that $n$ is a natural number. But I am using the periodic property taking $n$ as a natural number. So, how can I correct this proof? Should I ignore the non-integer part as $\lim\limits_{n\to\infty}\frac{\{n\}}{n}$ goes $0$? Thanks.


At first, you make $2$ mistakes:

First one: The number $f(c)$, in fact, is dependent to variable $t$, that may changes by different $t$, so the true derivation: $$g'(t) = t\frac{d}{dt}f(c) + f(c)$$ Second one: You use L'Hôpital's Rule when your denominator goes to $\infty$, so it was allowed iff the nominator also goes to $\infty$, that was not true, for example let $f(x)=0$.

But you can do as follow: $$h(t) = \lim_{n \longmapsto \infty} \frac{g(t+n)}{n} = \lim_{n \longmapsto \infty} \frac{\int_0^{t+n}f(x)dx}{n} = \lim_{n \longmapsto \infty} \frac{\int_0^nf(x)dx + \int_n^{t+n}f(x)dx}{n} =$$ $$\lim_{n \longmapsto \infty} \frac{n \int_0^1f(x)dx + \int_0^tf(x)dx}{n} = \lim_{n \longmapsto \infty} (\int_0^1f(x)dx + \frac{g(t)}{n}) = \int_0^1f(x)dx + \lim_{n \longmapsto \infty} \frac{g(t)}{n}$$ Now as $f$ is a continuous periodic function, it's bounded, and so for every $t$, $g(t)$ is finite, so we have: $$h(t) = g(1)$$ So the option c is correct, but in this way you can prove it.

  • $\begingroup$ L'Hospital's Rule works when denominator tends to $\infty$. There is no need to worry about numerator in this case. The problem here is that the limit after L'Hospital's Rule may not exist. $\endgroup$ – Paramanand Singh Sep 18 at 13:54
  • $\begingroup$ No, it's a critical hypothesis. Once can a function doesn't goes to $\infty$, and real limit was $0$, but its drivation goes to $\infty$ and make wrong results. For example suppose $f(x)=\sqrt{1-x^2}$ on $[-1,1]$ that repeated in every $[2n-1,2n+1]$. $\endgroup$ – Ali Ashja' Sep 18 at 14:15
  • $\begingroup$ You should check the proof for L'Hospital's Rule in this case rather than trying to find examples. A good reference is Wikipedia : en.wikipedia.org/wiki/L'H%C3%B4pital's_rule $\endgroup$ – Paramanand Singh Sep 18 at 14:17
  • $\begingroup$ In your reference also it's mentioned that BOTH of them must goes to $\infty$. $\endgroup$ – Ali Ashja' Sep 18 at 14:23
  • 1
    $\begingroup$ Wikipedia mentions this explicitly. "In case 2 the assumption that $f(x) $ diverges to infinity was not used in the proof..." $\endgroup$ – Paramanand Singh Sep 18 at 14:27

From $g(t) = tf(c)$ for some $c \in (0,t)$, it does $\color{red}{\text{not}}$ follow that $g'(t) = f(c)$ for that same $c$. In fact, by the fundamental theorem of calculus, we have that $g'(t) = f(t)$ for all $t$. So that statement is wrong.

Everything that follows from here is then flawed by the above mistake.

What you need to do is notice that for any positive integer $n$ we have $g(n) = \underbrace{g(1)+g(1)+... + g(1)}_{n \text{ times}} = ng(1)$, by periodicity of $f$ (draw a diagram of any $f$ and see this for yourself).

In a similar fashion, for any $t \in \mathbb R$ and positive integer $n$, we have $g(t+n) - g(t) = ng(1)$, by noting that $g(t+n) - g(t) = \int_t^{t+n} f(x)dx$ and then using periodicity.

Therefore, for any $t \in \mathbb R$ and $n$ a positive integer, we have : $$ \frac{g(t+n)}{n} = \frac{g(t+n) - g(t)}{n} + \frac{g(t)}{n} = g(1) + \frac{g(t)}{n} $$

Now, what happens as $n \to \infty$? Can you conclude?

  • $\begingroup$ I didn't say that $g(t)=f(c)$. From the mean value theorem, I concluded that $g(t)=tf(c)$. So, then $g\prime(t)=f(c)$ $\endgroup$ – tomriddle99 Sep 18 at 5:26
  • 1
    $\begingroup$ Even that is not true! For example, let $f(x) = x$ on $[0,1]$ and extended to a periodic continuous function on $\mathbb R$. We have $g(1) = \int_0^1 x dx = \frac 12 =f(\frac 12)$, but then it is not true that $g'(1) = f(\frac 12)$, in fact it equals $1 = f(1)$ by the fundamental theorem of calculus. $\endgroup$ – астон вілла олоф мэллбэрг Sep 18 at 5:32
  • $\begingroup$ Okay. I got it. Thanks. $\endgroup$ – tomriddle99 Sep 18 at 5:36
  • 1
    $\begingroup$ You are welcome! $\endgroup$ – астон вілла олоф мэллбэрг Sep 18 at 5:49

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