Function $f$ which tends to $0$ but the mean $\frac 1 x \int_1^x f$ doesn't tend to $0$

Question

Is there a continuous function $f : ]1, \infty[ \to \Bbb R_{>0}$ such that $f(x) \to 0$ when $x \to +\infty$ but $$\lim_{x \to +\infty} \dfrac 1 x \int_1^x f \;\neq\; 0$$ (either the limit doesn't exist but a better example for me would be when it exists and is $>0$)?

I tried something like $f(t) = \dfrac 1 {\log\big(\log(t)\big)}$ but I'm not sure how to handle the integral. The problem seems to be when $f$ doesn't go to $0$ sufficient quickly.

Answer 1

No.

If $f(x)\to 0$, then you can find an $x_0$ such that $f(x)<\varepsilon/2$ everywhere to the right of $x_0$, and then no matter how large $\int_1^{x_0} f(t)\,dt$ is, you can always push $\frac1x \int_1^x f(t)\,dt$ below $\varepsilon$ by choosing $x$ far enough to the right of $x_0$.

Of course you do need to assume that, say, $\int_1^2 f(t)\,dt$ exists (and is finite) at all. For example, if $f(t)=\frac{1}{t-1}$, then $\frac1x\int_1^xf(t)\,dt$ never exists (or is always infinite, depending on your conventions).

Answer 2

No, there is not.

Lets take $\varepsilon$ and prove that there exists $N$ s.t. for $x>N$ the mean $M(n) =\dfrac 1 x \int_1^x f(x)\,dx \in [-\varepsilon,\varepsilon]$

Suppose that for $x> Q~|f(x)| < \delta $. Then

$$M(2Q) = \frac12M(Q)+\frac1{2Q} \int_1^{2M} f(x)dx\le \frac12M(Q)+\frac1{2Q} \int_1^{2Q} \delta \,dx = \frac12M(Q)+\delta$$ Similarly $$M(4Q) \le \frac12M(2Q)+\delta \le \frac14M(Q)+(1+\frac12)\delta$$ and so on. So we will finally find the power $q$ such that $M(2^qQ)<\varepsilon$.

Answer 3

Writing the expression as

$$\tag 1 \frac{\int_1^x f(t)\,dt}{x}$$

we can apply L'Hopital (since the denominator $\to \infty$). The ratio of derivatives is simply $f(x)/1=f(x),$ which is given to $\to 0.$ Therefore the limit of $(1)$ is $0.$