What is $\lim_{n \to +\infty} \frac{1}{\log(n)}\sum_{k=1}^n \frac{1}{k}f(\frac{k}{n} ) = f(0)$

Question

I am working on the following question : let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function. Prove that: $$\lim_{n \rightarrow +\infty} \frac{1}{\log(n)}\sum_{k=1}^n \dfrac{1}{k}f\left(\frac{k}{n} \right) = f(0).$$

My thoughts: First one can assume that $f(0)=0$, since $ \log(n) \sim \sum_{k=1}^n \frac{1}{k}$. Then I don't really know how to approach the question: I am pretty confident that one should use the uniform continuity of $f$. I try to do a summation by parts to make appear some differences like $f(k/n)-f((k+1)/n)$, that I can control with uniform continuity, but I cannot conclude.

It also recalls me some Stolz-Cesàro-like theorems, but I am not able to use them directly in this situation.

Answer 1

Here I show that

For any bounded function function $f$ on $[0,1]$ that is continuous at $0$ $$\frac{1}{\log n}\sum^n_{k=1}\frac1k f(\frac{k}{n})\xrightarrow{n\rightarrow\infty}f(0)$$

Recall that there is a constant $\gamma$ such that \begin{align} 0\leq \sum^n_{k=1}\frac1k-\int^n_1 \frac1x\,dx - \gamma \leq \frac1n\tag{1}\label{one} \end{align}

Denote $s_n=\sum^n_{k=1}\frac1k$. From \eqref{one} we have that for $1\leq m<n$ \begin{align} -\frac1m+\log(n/m)\leq s_n-s_m\leq \log(n/m)+\frac1n\tag{2}\label{two} \end{align} Since $$\sum^n_{k=1}\frac1kf(\frac{k}{n})= \sum^n_{k=1}\frac1k\big(f(\frac{k}{n})-f(0)\big) + f(0)\sum^n_{k=1}\frac1k, $$ it is enough to consider the case $f(0)=0$. Suppose $|f(x)|\leq B$ for all $x\in[0,1]$. Given $\varepsilon>0$, choose $0<\delta<1$ such that $|f(x)|<\varepsilon$ whenever $|x|<\delta$. and for $n>\delta^{-1}$, define $K_n=\lfloor n\delta\rfloor$. Then \begin{align} \left|\frac{1}{\log n}\sum^n_{k=1}\frac1kf(\frac{k}{n})\right|&\leq\frac{1}{\log n}\sum^{K_n}_{k=1}\frac1k\big|f(\frac{k}{n})\big|+\frac{1}{\log n}\sum^n_{k=K_n+1}\frac1k\big|f(\frac{k}{n})\big|\\ &<\varepsilon\frac{s_{K_n}}{\log n}+B\frac{s_n-s_{K_n}}{\log n} \end{align} Since $\delta^{-1}\leq \frac{n}{K_n}<\frac{n\delta^{-1}}{n-\delta^{-1}}$, and $K_n\xrightarrow{n\rightarrow\infty}\infty$, it follows from \eqref{one} and \eqref{two} that \begin{align} \frac{s_{K_n}}{\log n}&\xrightarrow{n\rightarrow\infty}1\\ \frac{s_n-s_{K_n}}{\log n}&\xrightarrow{n\rightarrow\infty}0 \end{align} Therefore $$\limsup_n\Big|\sum^n_{k=1}\frac1kf(k/n)\Big|\leq\varepsilon$$ and the desired conclusion follows.

Answer 2

Assume $f(0)=0$. Let $\epsilon >0$. Choose $j$ such that $|f(x)|<\epsilon$ for $ x <\frac 1j$. Now split $\dfrac{1}{\log(n)}\sum_{k=1}^n \dfrac{1}{k}f\left(\dfrac{k}{n} \right)$ into sum over $k <\frac n j$ and $k \ge \frac n j$. It is easy to see that the first term divided by $\ln n$ is less than $\epsilon$ in absolute value since $\frac k n <\frac 1 j$ here. For the second term use thee fact that $f$ is bounded. Note that $\frac {\ln n-\ln (n/j)} {\ln n}\equiv \frac {\ln j} {\ln n} \to 0$ as $ n\to \infty$.