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I need to find a continuous and bounded function $\mathrm{f}(x)$ such that the limit $$ \lim_{T\to\infty} \frac{1}{T}\, \int_0^T \mathrm{f}(x)~\mathrm{d}x$$ doesn't exist.

I thought about $\mathrm{f}(x) = \sin x$ but I am not sure if the fact that we divide by $T$ may some how make it converge to zero.

What do you think ?

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"such that the limit" what? exists? does not exist? – user2566092 Apr 10 '14 at 15:42
$sin(x)$ does not work since the antiderivative $-cos(x)$ is also bounded for all $T$. Hence the limit is zero... – Thorben Apr 10 '14 at 16:00
Such a function exists, but is a pain (for me) to write down. Basically the function can be +1 so that the integral hits ${1 \over 2}$ then transition to -1 so that the integral hits $-{1 \over 2} $, etc, etc. – copper.hat Apr 10 '14 at 16:02

$ \sin(x) $ won't do, but it's only a tad trickier. $$ f(x) = \sin\left(\ln(x)\right) $$ should do the job (you can integrate it exactly by elementary techniques and then show it is $ \mathcal{O}(T) $).

Note that the function is bounded and continuous in $ (0,+\infty) $.

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$\int_0^T f(x)dx=-1/2T(cos(ln(T))-sin(ln(T)))$... divide that by $T$ and you get $-1/2(cos(ln(T))-sin(ln(T)))\rightarrow -1$ hence the limit exists for this $f(x)$. Maybe I missed something... – Thorben Apr 10 '14 at 16:53
@Thorben: As $T \to \infty$, that oscillates between $\pm {1 \over 2 \sqrt{2}}$. I think this is a nice solution. – copper.hat Apr 10 '14 at 19:12
This only needs a minor tweak to make $f$ be continuous at $x=0$. – copper.hat Apr 10 '14 at 19:24
@copper.hat you are absolutly right! As I look again at the function I have no idea how I obtained this limit... sorry for that! – Thorben Apr 10 '14 at 20:05
@Thorben: No problem, I wouldn't have caught it if not for symbolic integration... – copper.hat Apr 10 '14 at 20:07

The idea here is simple, but the details laborious. Keep the function at $+1$ long enough so the average hits some positive value, then keep the function at $-1$ long enough so the average hits some negative value and repeat.

Let $t_n = 2^{n+2}-4$. Let $f_n(t) = \begin{cases} t-t_n, & t \in [t_n,t_n+1] \\ 1 , & t \in (t_n+1,t_{n+1}-1) \\ 1-(t-(t_{n+1}-1)) , & t \in [t_{n+1}+1, t_n] \\ 0, \text{otherwise}\end{cases}$.

Note that $f_n$ has support $(t_n,t_{n+1})$, and $\int_0^{t_{n+1}} f_n = t_n+3$.

Let $\phi(t) = \sum_{k=0}^\infty (-1)^kf_k(t)$. Then \begin{eqnarray} \int_0^{t_{n+1}} \phi_n &=& \sum_{k=0}^n (-1)^k (t_k+3) \\ &=& \sum_{k=0}^n (-1)^k (2^{k+2}-1) \\ &=& \sum_{k=0}^n (-2)^{k+2}- \sum_{k=0}^n (-1)^k \\ &=& {4 \over 3} (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \end{eqnarray} Then \begin{eqnarray} \sigma_n &=& {1 \over t_{n+1} }\int_0^{t_{n+1}} \phi_n \\ &=& {4 \over 3} { (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \over 2^{n+3}-4} \\ &=& {1 \over 3}{ (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \over 2^{n+1}-1} \\ &=& {1 \over 3} { {1 \over 2^{n+1} } - (-1)^{n+1} - {1 \over 2^{n+2} (1 - (-1)^{n+1} }\over 1 - {1 \over 2^{n+1} } } \end{eqnarray} We see that $\liminf_n \sigma_n = - {1 \over 3}$, $\limsup_n \sigma_n = {1 \over 3}$, hence the limit does not exist.

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The integral has to diverge to infinity if this limit is to not exist, (of course any other limit will give zero overall). Given the type $``\dfrac{\infty}{\infty}"$ limit, if we apply L'Hopitals; $$\displaystyle\lim_{T\to \infty}\dfrac{1}{T}\displaystyle\int_{0}^{T}f(x)\ dx = \lim_{T\to \infty} f(T)$$

This makes it clear that the limit doesn't exist iff $f$ is not bounded, so there is no suitable function.

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If the function has a limit as $t \to \infty$ this is true, but it doesn't have to have such a limit. It is straightforward, if tedious, to create a bounded $f$ that doesn't have an average limit. – copper.hat Apr 10 '14 at 18:34
"The integral has to diverge to infinity if this limit is to not exist". Nope. "does not converge " is not synonymous to "diverges to infinity". – leonbloy Apr 10 '14 at 21:12

What about $\mathrm{f}(x)=T$?

$$ \lim_{T\to\infty} \frac{1}{T}\, \int_0^T \mathrm{f}(x)~\mathrm{d}x=\lim_{T\to\infty} \frac{1}{T}\, \int_0^T T~\mathrm{d}x=\lim_{T\to\infty} \frac{1}{T}\, T^2=\infty$$

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What does this mean? $T$ is not a constant. – copper.hat Apr 10 '14 at 18:43
$T$ does not depend on $x$. Or am I misunderstanding something? – zighalo Apr 10 '14 at 18:53
This is genius!! So simple and brilliant. – rubik Apr 10 '14 at 18:54
@copper.hat: See*+%28integral+from+0+to+T+o‌​f+T+with+respect+to+x%29+as+T+-%3E+inf&a=*C.T-_*Variable-&a=UnitClash_*T.*Teslas.‌​dflt-- – rubik Apr 10 '14 at 18:54
@zighalo: $f(x)$ is taken to be defined prior to the creation of the expression containing the limit which binds the formal parameter $T$. At the earlier time, $T$ is a free variable. At the later time, the formal parameter of the limit must be unbound in its context, so cannot be any symbol already in the context. If you set $f(x)=T$, then $T$ cannot be the formal parameter in the limit; it must be some other free-at-that-time variable. – Eric Towers Apr 10 '14 at 19:42

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