Limit of a sequence of functionals $F_n : C([0,1]) \to \mathbb{R}$ Consider the Banach space $C([0,1])$ with $\| . \|_{\infty}$ maximum norm. Consider the linear functional $F_{n} : C([0,1]) \to \mathbb{R}$, $ F_{n} (f) = f(1) - f\big(\frac{2n - 2}{2n - 1}\big) + f\big(\frac{2n-3}{2n-1}\big) - f\big(\frac{2n-4}{2n-1}\big) + ... - f(0)$
Show that $\displaystyle \lim_{n \to \infty} F_n(f) = \frac{f(0)-f(1)}{2}$.
I already proved that $C^1([0,1])$ is dense in $C([0,1])$ using Weierstrass Stone theorem for a previous item in this same exercise. But I don't see how can I prove this limit. 
Any help would be appreciated.
Thanks.
 A: Assuming that
$$\lim_{n\to\infty} F_n( \color{red}{f} ) = \frac{f(1) \color{red}{-} f(0)}{2} \tag{*}$$
is what really you mean, it is easy to prove the convergence for $f \in C^1([0, 1])$. Indeed, by the mean value theorem, for each $n$ and $k$, there exists $\xi_k = \xi_{f,n,k} \in (\frac{2k}{2n-1}, \frac{2k+1}{2n-1})$ such that
$$ f\left(\frac{2k+1}{2n-1}\right) - f\left(\frac{2k}{2n-1}\right) = f'(\xi_k) \frac{1}{2n-1}. $$
Thus it follows that
\begin{align*}
\sum_{k=0}^{n-1} \left( f\left(\frac{2k+1}{2n-1}\right) - f\left(\frac{2k}{2n-1}\right) \right)
&= \sum_{k=0}^{n-1} f'(\xi_k) \frac{1}{2n-1} \\
&\xrightarrow[\ n\to\infty \ ]{}
\frac{1}{2} \int_{0}^{1} f'(x) \, dx
= \frac{f(1) - f(0)}{2}.
\end{align*}
On the other hand, there are continuous functions for which $\text{(*)}$ is not true. For instance, let $f$ be the Cantor-Lebesgue function. Then
$$ F_{(3^n+1)/2}(f) = \sum_{k=0}^{(3^n-1)/2} \left( f\left(\frac{2k+1}{3^n}\right) - f\left(\frac{2k}{3^n}\right) \right) = 1 $$
since the Cantor set $C$ is contained in $\bigcup_{k=0}^{(3^n-1)/2} [ \frac{2k+1}{3^n}, \frac{2k}{3^n} ]$ and $f$ increases only on $C$. So we have
$$ \limsup_{n\to\infty} F_n(f) = 1 \neq \frac{1}{2} = \frac{f(1) - f(0)}{2}. $$
