I want to prove following statement: Let $f_n:[0,1]\to\mathbb{R}$ be sequence of continuous functions uniformly convergent to $f\left(x\right)$. If $\left(f_n\left(x\right)-f_n\left(x\frac{n}{n+1}\right)\right)n\to g$ uniformly on $[-1,1]$ for some continuous function $g$ then $f'=g$.
My attempt to prove it is following:
By uniform convergence of $f_n$ and $\left(f_n\left(x\right)-f_n\left(x\frac{n}{n+1}\right)\right)n$ we know that there exists sequence $\epsilon_n$ convergent to $0$ such that:$$|f\left(x\right)-f_n\left(x\right)|<\epsilon_n,~x\in[0,1]~n\in\mathbb{N}$$$$\left|\left(f_n\left(x\right)-f_n\left(x\frac{n}{n+1}\right)\right)n-g\left(x\right)\right|<\epsilon_n,~x\in[0,1]~n\in\mathbb{N}$$Let's fix $h\in (0,1]$ and set $k_n=\lfloor nh\rfloor$. Then from inequality above we know that$$\left|\left(f_n\left(x\right)-f_n\left(x\left(\frac{n}{n+1}\right)^{k_n+1}\right)\right)n-\left(g\left(x\right)+g\left(x\left(\frac{n}{n+1}\right)\right)+\dots+g\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)\right)\right|\leq$$$$\leq \left|\left(f_n\left(x\right)-f_n\left(x\left(\frac{n}{n+1}\right)\right)\right)n-g\left(x\right)\right|+\left|\left(f_n\left(x\left(\frac{n}{n+1}\right)\right)-f_n\left(x\left(\frac{n}{n+1}\right)^2\right)\right)n-g\left(x\left(\frac{n}{n+1}\right)\right)\right|+\dots+$$$$+\left|\left(f_n\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)-f_n\left(x\left(\frac{n}{n+1}\right)^{k_n+1}\right)\right)n-g\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)\right|\leq k_n\epsilon_n$$Diving by $k_n$ gives:$$\left|\left(f_n\left(x\right)-f_n\left(x\left(\frac{n}{n+1}\right)^{k_n+1}\right)\right)\frac{n}{k_n}-\frac{g\left(x\right)+g\left(x\left(\frac{n}{n+1}\right)\right)+\dots+g\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)}{k_n}\right|\leq \epsilon_n$$Now we have:$$\left|\left(f\left(x\right)-f\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)\right)\frac{n}{k_n}-g\left(x\right)\right|\leq \left|f\left(x\right)-f_n\left(x\right)\right|\frac{n}{k_n}+\left|f\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)-f\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)\right|\frac{n}{k_n}+$$$$+\left|g\left(x\right)-\frac{g\left(x\right)+g\left(x\left(\frac{n}{n+1}\right)\right)+\dots+g\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)}{k_n}\right|+$$$$+\left|\left(f_n\left(x\right)-f_n\left(x\left(\frac{n}{n+1}\right)^{k_n+1}\right)\right)\frac{n}{k_n}-\frac{g\left(x\right)+g\left(x\left(\frac{n}{n+1}\right)\right)+\dots+g\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)}{k_n}\right|\leq 2\epsilon_n\frac{n}{k}+$$$$+\sup_{t\in[x\left(\frac{n}{n+1}\right)^{k_n},x]}\left|g\left(x\right)-g\left(t\right)\right|+\epsilon_n\leq \left(1+2\frac{n}{k}\right)\epsilon_n+\sup_{t\in[xe^{-h},x]}\left|g\left(x\right)-g\left(t\right)\right|$$We obtained $\left|\left(f\left(x\right)-f\left(x\left(\frac{n}{n+1}\right)^{k_n}\right)\right)\frac{n}{k_n}-g\left(x\right)\right|\leq\left(1+2\frac{n}{k}\right)\epsilon_n+\sup_{t\in[xe^{-h},x]}|g\left(x\right)-g\left(t\right)|$ taking $n\to\infty$ by continuity of $f$ we get:$$|\left(f\left(x\right)-f\left(xe^{-h}\right)\right)\frac{1}{h}-g\left(x\right)|\leq\sup_{t\in[xe^{-h},x]}|g\left(x\right)-g\left(t\right)|$$Substituting $h=-\ln(1+\frac{u}{x})$ we get $|x\frac{f\left(x\right)-f\left(x+u\right)}{u}\frac{u}{-\ln(1+\frac{u}{x})}-g\left(x\right)|\leq\sup_{t\in[x+u,x]}|g\left(x\right)-g\left(t\right)|$ now by taking $u\to 0$ we have$$|xf'\left(x\right)-g\left(x\right)|\leq 0$$My question is: Is this proof (or rather idea of a proof) correct? I'm also looking for some simpler proofs.