I am trying to show the following as part of a bigger proof and I would like to confirm whether my reasoning is correct? Fix x a periodic point with period n of a continuous function $T$, let also f be continuous. Then obviously $f(T^n x)= f(x)$. I want to show that if $\lim_{n\to \infty} \sum_{i=0}^{n-1}\frac{f(T^i x)}{n}= L$, then $\sum_{i=0}^{n-1}\frac{f(T^i x)}{n}= L$ as well.
My proof: We have that $\lim_{m\to \infty} \sum_{i=0}^{m-1}\frac{f(T^i x)}{m}= L$. Now fix m larger than n (the period). Then m can be written as $m= kn + r$, where $r<n$, and k,r are positive integers. Then $\sum_{i=0}^{m-1}\frac{f(T^i x)}{m}= \sum_{i=0}^{nk-1}\frac{f(T^i x)}{m} + \sum_{i=nk}^{m}\frac{f(T^i x)}{m}$= $k \sum_{i=0}^{n-1}\frac{f(T^i x)}{nk+r} + \sum_{i=nk}^{m}\frac{f(T^i x)}{nk+r}$ (where the first term comes about by periodicity of $T$). Now the second sum by periodicity is independent of m, since $\sum_{i=nk}^{m}{f(T^i x)}=\sum_{i=0}^{r}{f(T^i x)}$ (and r could be chosen to be n-1 for any m, as it is bounded by n and so will be meaningless in the limit)
Now taking limit as m tends to infinity, since n is fixed and r is less than n for any m, m=kn+r implies k goes to infinity (again, r depends on m as well but since it is less than n, we can assume it is n-1 in what follows since it will be bounded by a constant and hence irrelevant in the limit). So $\lim_{m\to \infty} \sum_{i=0}^{m-1}\frac{f(T^i x)}{m}= \lim_{k\to \infty} k\sum_{i=0}^{n-1}\frac{f(T^i x)}{nk+r} + \sum_{i=0}^{r}\frac{f(T^i x)}{nk+r}= \lim_{k\to \infty} k \sum_{i=0}^{n-1}\frac{f(T^i x)}{nk+r} + \lim_{k \to \infty}\sum_{i=0}^{r}\frac{f(T^i x)}{nk+r}= \lim_{k\to \infty} k\sum_{i=0}^{n-1}\frac{f(T^i x)}{nk+r}= \lim_{k\to \infty} \frac{k}{nk+r}\sum_{i=0}^{n-1}{f(T^i x)} = \sum_{i=0}^{n-1}{f(T^i x)}$ and the claim is proved.
Is it correct? Thanks.
It is fine.
But since you assume that the limit exists, a simpler proof is the following: if $$\sum_{i=0}^{n-1}\frac{f(T^i x)}{n}= M\ne L,$$ then $$ \lim_{m\to \infty} \sum_{i=0}^{m-1}\frac{f(T^i x)}{n}=\lim_{k\to \infty} \sum_{i=0}^{kn-1}\frac{f(T^i x)}{kn}=M\ne L. $$