under which conditions $\lim_{t\to \infty} \frac{1}{t}\int_0^t f(s)ds$ implies $\lim_{t\to \infty} f(t) $

Question

I am interested in knowing under which conditions $\lim_{t\to \infty} \frac{1}{t}\int_0^t f(s)ds$ implies that $\lim_{t\to \infty} f(t)$ exist.

It is known that $\lim_{t\to \infty} \frac{1}{t}\int_0^t f(s)ds$ can exist, but $\lim_{t\to \infty} f(t) $ not. For instace $f(t) = \cos t$.

First, I wonder if it is enough that $f(t)$ is continuous and $\lim_{t\to \infty} \frac{1}{t}\int_0^t f(s)ds$ converges to the maximum value of the image. However, I found this counter-example:

Let us define $f(t)$ by cases

$f(t):= 1 -2(t-(10^{n-1}-1))$ if $t \in [10^{n-1}-1,10^{n-1}-\frac{1}{2}]$,

$f(t):= 2(t-(10^{n-1}-\frac{1}{2}))$ if $t \in [10^{n-1}-\frac{1}{2},10^{n-1}]$,

$f(t) := 1$ if $t \in [10^{n-1},10^n-1] $

Where $n = 1,2,3,\dots$. We can provide a smooth example using the same argument and bump function. However $f(t)$ is not analytic. Since the proof relies that $f(t)$ is constant function in open interval. I wonder if $f(t)$ being analytic is enough.

Is true the following statement? Let $f(t):[0,\infty] \to [-1,1]$ be an analytic function. If $\lim_{t\to \infty} \frac{1}{t}\int_0^t f(s)ds = 1$, then $\lim_{t\to \infty} f(t) = 1$.

Answer 1

No, the statement is not true even for analytic functions: What we want to have is a function with mean converging to $1$ for $t \to \infty$ but whose value does not converge to $1$ for $t \to \infty$. Intuitively, this can be done easily by taking a periodic function which maps into $[0,1]$ and then push its values towards $1$ for large $t$ but still retaining some points which come close to $0$.

So let's take a periodic function, e.g. $\cos$ and scale it so that it maps into $[0,1]$:

$$ \frac{\cos(t)+1}{2} $$

Now we want that the mean value of this function converges to $1$ for $t \to \infty$. This can be done by applying an exponent which goes to $0$ for $t \to \infty$:

$$ \left(\frac{\cos(t)+1}{2}\right)^{\frac{1}{t+1}} $$

As the exponent pushes all nonzero values towards 1, one can easily see that the mean of this function is $1$ for $t \to \infty$. On the other hand, the function vanishes for all $t \in (\mathbb{Z} + \frac{1}{2}) \pi$. Consequently, it satisfies all requirements except that it fails to be analytic at its zeros. However, this can be fixed by lifting the function a little bit away from $0$ in such a way that it still comes arbitrarily close to $0$ for large $t$.

$$ f(t) = \left(\frac{\cos(t) + 1 + \left(\frac{1}{t+1}\right)^{t+1}}{2 + \left(\frac{1}{t+1}\right)^{t+1}}\right)^{\frac{1}{t+1}} $$

Note that the term added in the denominator servers just as a rescaling factor so that the function does not attain values larger than $1$. One can check directly that $\lim_{n \to \infty} f((n+1/2)\pi) = 0$, which finishes the proof.

Concerning your initial question about conditions: The only condition that I can think of is if $\lim_{t\to \infty} f(t)$ exists, then $$ \lim_{t\to \infty} \frac{1}{t}\int_0^t f(s)ds = \lim_{t\to \infty} f(t), $$ but this is rather elementary and I guess not really what you are looking for.

Edit: As alejandro pointed out, the derivative of the counterexample above is unbounded. However, one can also construct a counterexample with a bounded derivative. Instead of making the portions where $cos$ attains $0$ smaller (which increases the derivative), one can make the portions where $cos$ attains $1$ larger. Concretely, we want a function $f$ which looks like

This can be constructed analytically by adding shifted versions of the error function. Then precomposing this function with $cos$ looks like

and this is exactly what we want. I will leave out the details at this point because I think it's clear what's going on. All in all our counterexample looks approximately like

$$ \frac{1}{2} \left(\cos\left( \sum_{n \in \mathbb{N}} (\text{erf}(t - n^2) + 1)\pi \right) + 1\right), $$

where I guess one has to adjust the shifts and maybe also add a correction term for the error functions inside the sum so that the mean really converges to $1$ for $t \to \infty$.

Answer 2

The answer is no but I do not know an easy counterexample. I will give a sketch of a counterxample here. if the Riemann Hypothesis is true then one can use a translation of $1-c|\zeta(1+it)|/(\log \log t)$ and the respective theorems from Titchmarsh but otherwise one needs to work harder and use a function similar to $\zeta$ for which the analog of RH holds.

The idea is that if $\chi:\mathcal P \to S^1$ we obtain a character ($\chi(ab)=\chi(a)\chi(b), |\chi(a)|=1$) on $\mathbb Q^+$ defined by $\chi(\Pi p_k^{\alpha_k})=\Pi \chi(p_k)^{\alpha_k}$, where $\mathcal P$ is the set of primes. Then as usual, we can associate a Dirichlet series with an Euler product $$L_\chi(s)=\sum_{n \ge 1} \chi(n) n^{-s}=\Pi_{p \in \mathcal P}(1-\chi(p)p^{-s})^{-1}, \Re s >1$$

A famous theorem of Helson says that for almost all $\chi$ (in the sense of the Haar measure of $(S^1)^{\infty}$ which is a compact topological group with which the set of characters is naturally isomorphic) the Riemann Hypothesis for $L_\chi$ holds, meaning here of course that $L_{\chi}$ extends to a zero free analytic function in $\Re s >1/2$ and we pick such a $\chi$ and call $L$ the corresponding $L_{\chi}$ which is then analytic and non zero in $\Re s >1/2$

In particular, it follows (essentially by Kronecker approximation theorem and convexity arguments for $\log |L|$ - following the proof in Titchmarsh chapters 8, 14 for the usual Riemann Zeta ) that $\limsup_{t \to \infty} \frac{|L(1+it)|}{\log \log t}=\gamma >0$ and finite.

Also the fact that $\log L$ is analytic in $\Re s >1/2$ immediately implies by the usual arguments (three-circle theorem and convexity bounds) that $L$ satisfies the analogue of Lindelof hypothesis, so again by usual arguments (residue theorem, Perron etc - Titchmarsh chapters 7, 13) it follows that $L(1+it)$ is approximated by its Dirichlet polynomial up to length $t^{\delta}, \delta >0$ so the usual mean value theorem for Dirichlet polynomials shows that $$\frac{1}{T}\int_0^{T}|L(1+it)|^2dt \to \zeta(2)$$ (since $\sum |\chi(n)|^2 n^{-2}=\zeta(2)$)

But now we consider $f(t)=1-\frac{1}{\gamma}\frac{|L(1+it+T_0)|}{\log \log (t+T_0)}$ where $T_0>5$ is large enough so $|L(1+it)| <2\gamma \log \log t, t >T_0$

Then clearly $-1 < f(t) < 1$ and $f$ is real analytic since the absolute value of a nonzero analytic function is real analytic, while $\liminf_{t \to \infty} f(t) =0$

But now $$\int_1^Tf(t)dt=T-1-\frac{1}{\gamma}\int_1^T\frac{|L(1+it+T_0)|}{\log \log (t+T_0)}dt$$ and by Cauchy-Schwarz we have: $$\int_1^T\frac{|L(1+it+T_0)|}{\log \log (t+T_0)}dt \le (\int_1^T|L(1+it+T_0)|^2dt)^{1/2}(\int_1^T\frac{1}{(\log \log (t+T+0))^2}dt)^{1/2}$$ and the first integral is $cT$ while the second is easily seen to be $o(T)$ so $$\frac{1}{T\gamma}\int_1^T\frac{|L(1+it+T_0)|}{\log \log (t+T_0)}dt \to 0, T \to \infty$$ hence $\lim_{T\to \infty} \frac{1}{T}\int_0^T f(t)dt = 1$ as required