Problem proposed by prof. Gadjev

$\textbf{Problem.}$

 Let $f:[0,1]\to\mathbb{R}$ be a continuous function.

Find the limit 

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right).$$

 

$\textbf{Solution.}$ 

Fix $\varepsilon>0$. We aim to show that for large enough $n$  

$$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|<3\varepsilon$$

 which would mean that the limit is $f(1).$

Let $M$ be an upper bound for $|f|$ on $[0,1]$, i.e. $|f(x)|\le M$ for all $x\in [0,1].$

 For all $n\ge 1$ define $$k(n)=\left\lfloor\frac{n}{n^{1/\sqrt{\ln n}}}\right\rfloor.$$

The number is chosen in such a way that $$\lim_{n\to\infty}\frac{k(n)}{n}=0 \ \ \& \ \  \lim_{n\to\infty}\frac{\ln(k(n))}{\ln n}=1.$$

 Split the sum in two:

$$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|\le \left|\frac{1}{n}\sum_{k=1}^{k(n)} f\left(\frac{\ln k}{\ln n }\right)\right|+\left|\frac{1}{n}\sum_{k=k(n)+1}^{n} f\left(\frac{\ln k}{\ln n }\right)-f(1)\right| \ \ \ (*)$$

 Take $n_1$ such that for $n>n_1$ holds $\displaystyle \frac{k(n)}{n}<\frac{\varepsilon}{M}$. For the first summand on the right, using the triangle inequality, we obtain

$$\left|\frac{1}{n}\sum_{k=1}^{k(n)} f\left(\frac{\ln k}{\ln n }\right)\right|\le \frac{1}{n}\sum_{k=1}^{k(n)} \left|f\left(\frac{\ln k}{\ln n }\right)\right|\le \frac{Mk(n)}{n}<\varepsilon$$

when $n>n_1$. 

Now take $\delta>0$ such that when $|x-1|<\delta$ holds $|f(x)-f(1)|<\varepsilon$. Take $n_2$ such that for $n>n_2$ holds $\displaystyle \left|\frac{\ln(k(n))}{\ln n}-1\right|<\delta.$ Now for $n>\max\{n_1,n_2\}$ we obtain 

 $$\left|\frac{1}{n}\sum_{k=k(n)+1}^{n} f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|\le \sum_{k=k(n)+1}^{n} \frac{1}{n}\left|f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|+\left|\frac{k(n)}{n}f(1)\right|<\frac{n-k(n)}{n}\varepsilon+\varepsilon\le 2\varepsilon$$

 Thus for $n>\max\{n_1,n_2\}$ from $(*)$ we obtain 

$$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|\le \left|\frac{1}{n}\sum_{k=1}^{k(n)} f\left(\frac{\ln k}{\ln n }\right)\right|+\left|\frac{1}{n}\sum_{k=k(n)+1}^{n} f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|<\varepsilon+2\varepsilon=3\varepsilon.$$

$\textbf{Remark.}$ We only used that the function $f$ is bounded and continuous at $1$. The proof heavily relied on the existence of the function (sequence) $k$. Functions, for which such function $k$ exists (like $\ln$) are called super slowly varying. They arise in Probability theory. The same proof could be carried for any such function. 

The proof proposed from Prof. Gadjev relies on the following equivalences

$$\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)\equiv\int_1^n  f\left(\frac{\ln x}{\ln n }\right)dx\underbrace{=}_{x=n^y}\int_0^1  f\left(y\right)\ln(n)n^ydy$$

 and consequently $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)=\lim_{n\to\infty}\int_0^1  \frac{\ln n}{n}n^y f\left(y\right)dy.$$

 This reasoning could be made rigorous. After that, what remains is to observe that the sequence 

$ \displaystyle\frac{\ln n}{n-1}n^y$ tends to the Dirac Delta function at $1$.

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