I'm trying to define integrals of Banach space-valued functions, following the approach of the book by Bühler and Salamon. I'm facing the following problem.
Let $X$ be a (real or complex) Banach space and let $x:[a,b]\to X$ be a continuous function. For each $n\in \mathbb{N}$, define$$\xi _n:=\sum \limits _{k=0}^{2^n-1}\frac{b-a}{2^n}x\left (a+k\frac{b-a}{2^n}\right ),\qquad \delta _n:=\sup \limits _{|s-t|\leq 2^{-n}(b-a)}||x(s)-x(t)||.$$I'm trying to prove that$$||\xi _{m+n}-\xi _n||\leq (b-a)\delta _n$$for all $m,n\in \mathbb{N}$. What I got so far is $$||\xi _{n+m}-\xi _n||=\left |\left |\sum \limits _{k=0}^{2^{n+m}-1}\frac{b-a}{2^{n+m}}x\left (a+k\frac{b-a}{2^{n+m}}\right )-\sum \limits _{k=0}^{2^n-1}\frac{b-a}{2^n}x\left (a+k\frac{b-a}{2^n}\right )\right |\right | \\ =(b-a)\left |\left |\sum \limits _{k=0}^{2^{n+m}-1}\frac{1}{2^{n+m}}x\left (a+k\frac{b-a}{2^{n+m}}\right )-\sum \limits _{k=0}^{2^n-1}\frac{1}{2^n}x\left (a+k\frac{b-a}{2^n}\right )\right |\right | \\ =(b-a)\left |\left |\frac{1}{2^m}\sum \limits _{k=0}^{2^{n+m}-1}\frac{1}{2^n}x\left (a+k\frac{b-a}{2^{n+m}}\right )-\sum \limits _{k=0}^{2^n-1}\frac{1}{2^n}x\left (a+k\frac{b-a}{2^n}\right )\right |\right | \\ =(b-a)\left |\left |\frac{1}{2^m}\sum \limits _{k=2^n}^{2^{n+m}-1}\frac{1}{2^n}x\left (a+k\frac{b-a}{2^{n+m}}\right )+\frac{1}{2^m}\sum \limits _{k=0}^{2^n-1}\frac{1}{2^n}x\left (a+k\frac{b-a}{2^{n+m}}\right )-\sum \limits _{k=0}^{2^n-1}\frac{1}{2^n}x\left (a+k\frac{b-a}{2^n}\right )\right |\right |.$$ Clearly I should prove that the vector sum has norm $\leq \delta _n$, but I'm not sure if breaking the sum in this way leads to the desired result or if there is a simpler/shorter way of solving this.
Start by considering a single subinterval of a partition $P$ of $[a,b]$.
Then, if $\{ y_r,y_{r+1},\dots,y_s\}$ is the partition of it induced by a refinement of $P$, prove that $\omega \, (y_s-y_r)$ is a bound of $$\left \| \sum_{j=r+1}^s x(y_{j-1})\,(y_j-y_{j-1})-x(y_r)\,(y_s-y_r)\right \|$$ being $$\omega = \sup _{|u-v| \le \|P\|} \|x(u)-x(v)\|$$ Finally prove that $\,\omega \, (b-a)$ is a bound of the norm of the difference of the associated (left) Riemann sums.