Let $X=C[0,1]$. For $n\in\mathbb{N}$ and $f\in X$, define the Riemann sum with uniform partition $$ R_n(f)=\frac1n\sum_{k=1}^n f\left(\frac kn\right).$$ Show that if $f\in X$ is Lipschitz, then the sequence $$ \left\{n \left( \int_0^1f(t)dt-R_n(f)\right):n\in\mathbb{N}\right\}$$ is bounded.
My attempt: We need to find $K>0$ (independent of $n$) such that $\forall n: n\left|\int_0^1f(t)dt-R_n(f) \right|\le K$. We have $$ \left| \int_0^1f(t)dt-R_n(f)\right|=\left| \int_0^1 \left( f(t)-\frac 1n\sum_{k=1}^n f\left(\frac kn\right)\right)dt\right|\le \int_0^1\left| f(t)-\frac 1n\sum_{k=1}^n f\left(\frac kn\right)\right|dt.$$ The function $f$ is Lipschitz, meaning that $\exists C>0:|f(t)-f(s)|\le C|t-s|$ for all $t,s\in [0,1]$. I was thinking of considering $$ |f(t)-f(0)+f(0)-f(1/n)+f(1/n)-\dots-f(1)+f(1)-\frac 1n\sum_k f(k/n)|\\\le C_0|t|+C_1\frac 1n + C_2\frac 1n+\dots+C_n\frac1n+\frac1n\sum_k |f(\frac kn)|\\ \le C_0|t|+\frac 1n(C+\sum_k |f(\frac kn)|).$$ Integrating this final term does not yield an upperbound independent of $n$.
I'm probably missing results from integration theory regarding partition and such. Can anyone provide hints?
Thanks.
Hint: $$\int_0^{1} f(t)dt= \sum\limits_{k=1}^{n}\int_{(k-1)/n}^{k/n} f(t)dt$$$$=\sum\limits_{k=1}^{n}\int_{(k-1)/n}^{k/n} f(\frac k n)dt+\sum\limits_{k=1}^{n}\int_{(k-1)/n}^{k/n} [f(t)-f(\frac k n)]dt.$$ The first term is exactly $R_n(f)$. Use Lipschitz condition to show that $n$ times the second term is bounded.