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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