Uniform convergence and triangle inequality

Question

Following @mathcounterexamples.net's answer on my previous question, I'm trying to rigorously prove $$\left\vert \frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})-\frac{1}{n}\sum_{k=1}^{n}f_m(\frac{k}{n})\right\vert \le \Vert f - f_m\Vert_\infty \tag{1}$$ for every $m,n\in\mathbb{N}$ where $f$ and $f_m$ are defined on $\mathbb{Q}\cap [0,1]$. Here is my try:

We know that $$|\sum_{k=1}^{n}a_k |\le \sum_{k=1}^{n}|a_k| $$ so we have $$\vert \frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})-\frac{1}{n}\sum_{k=1}^{n}f_m(\frac{k}{n})\vert = \vert\sum_{k=1}^n \frac{1}{n}(f(\frac{k}{n}) - f_m(\frac{k}{n}))|\le \sum_{k=1}^n \frac{1}{n}|f(\frac{k}{n}) - f_m(\frac{k}{n})|$$ If we assume $A = \sup\{|f(x) - f_m(x)| : x\in \mathbb{Q}\cap [0,1] \}$ exists, $$\sum_{k=1}^n \frac{1}{n}|f(\frac{k}{n}) - f_m(\frac{k}{n})|\le \sum_{k=1}^n \frac{1}{n}\times A = A$$ Is my proof correct? Also why $\sup$ exists in this case(maybe because it's a set of measure zero)? In addition, I couldn't prove that $(1)$ implies $$\vert S(f) -S(f_m)\vert \le \Vert f - f_m \Vert _\infty \tag{2}$$ and why we need $f_n$ converges uniformly to $f$ in order to have $S(f) \to S(f_m)$.

Answer 1

Your proof is correct. The $\sup$ exists, at least for $m$ large enough because in your previous question, you supposed that $\{f_m\}$ converges to $f$ uniformly.

Answer 2

Thanks to @mathcounterexamples.net's helpful responses, I try to write a complete answer:

Let $D = \mathbb{Q}\cap [0,1]$. Functions $f$ and $(f_m)_{m\ge1}$ are defined on $D$. Let $$H_n(f) =\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})$$ Assume $$S(f) = \lim_{n \to \infty} H_n(f) $$ and $S(f_m)$ for every $m\in \mathbb{N}$ exist. In addition, assume that $f_m \to f$ uniformly on $D$. According to the equivalent definition of uniform convergence, $$\sup_{x\in D}|f_m(x) - f(x)| \to 0$$ So we can conclude $A(m) = \sup_{x\in D}|f_m(x) - f(x)|$ is well-defined for every $m\in \mathbb{N}$ and $\lim_{m \to \infty} A(m) = 0$. Using triangle inequality, $|\sum_{k=1}^{n}a_k |\le \sum_{k=1}^{n}|a_k|$, we get $$|H_n(f_m) - H_n(f)| = |\sum_{k=1}^n \frac{1}{n}(f_m(\frac{k}{n}) - f(\frac{k}{n}))| \le \sum_{k=1}^n \frac{1}{n}|f_m(\frac{k}{n}) - f(\frac{k}{n})| \le \sum_{k=1}^n \frac{1}{n}\times A(m) = A(m)$$ Therefore we arrived at $$|H_n(f_m) - H_n(f)|\le A(m)$$ There is a well known theorem which says that $$\forall n\in \mathbb{N} :|a_n|\le M \ \text{and} \ \lim_{n \to \infty} a_n = a \implies |a|\le M$$ We use this theorem with $a_n = H_n(f_m) - H_n(f)$. Because we have assumed that $S(f)$ and $S(f_m)$ exist, $\lim_{n \to \infty} H_n(f_m) - H_n(f)$ is well-defined and $$|\lim_{n \to \infty}(H_n(f_m) - H_n(f))| = |\lim_{n \to \infty}H_n(f_m) - \lim_{n \to \infty} H_n(f)| = |S(f_m) - S(f)|\le A(m)$$ Finally we use the squeeze theorem to get $$0\le|S(f_m) - S(f)|\le A(m) \implies \lim_{m \to \infty} 0 = \lim_{m \to \infty} |S(f_m) - S(f)| = \lim_{m \to \infty} A(m) = 0$$ The last equation means that the sequence of numbers $\{S(f_m)\}$ converges to $S(f)$ which is the desired conclusion.