Let $X$ be a Banach space and assume that $a0$ there is a $\delta>0$ such that for $a=t_0 One can show that contiuous functions are integrable in the sense of the definition above. My question is how to determine the integral of the function $f:\left[0,1\right]\rightarrow c_0$ given by
$f(x)=\frac{1}{2}\left(x-\frac{1}{2^n}\right)e_n+\frac{1}{4}\left(\frac{1}{2^{n-1}}-x\right)e_{n+1}$, for $x\in\left[\frac{1}{2^n}, \frac{1}{2^{n-1}}\right]$, $n\in\mathbb{N}$ and $f(0)=0$, where $e_n=(0,...,0,1,0,...,0)$ with $1$ being a the $n$-th place? Notice that $f$ is continuous so it is integrable. I guess that the integral may be of the form $\sum_{n=1}^{\infty} \frac{1}{2^{n+1}}e_n$ or something like this but can not deal with it, so I will be very grateful for your help.
Integral of a function taking values in $c_0$
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functional-analysis
integration
normed-spaces
1 Answers
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Your function $f$ can be written as the sum of a series $f(x)=\sum_{n\geq1}\chi_{I_n}(x)f(x)$, with $I_n=\left[\frac{1}{2^n}, \frac{1}{2^{n-1}}\right)$ and $\chi_{I_n}$ the characteristic function of $I_n$. This series converges uniformly, and presumably you can show that your Riemann integrals behave well with such series. Then you can interchange sums and integrals.
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0Thanks for the hint, I will try to use it – 2011-10-12