Is sum of differences bounded for $C^0[a,b]$ functions? Let $a < b$ and suppose that $f \in C^{0}([a,b]).$ Is it true that
$$ \limsup_{h \rightarrow 0} \sum_{i}| f(t_{i+1})-f(t_i)|^2 = 0 $$
where $a < \{ t_i\} < b$ and $h = t_{i+1} - t_i$?
This is certainly true for $C^1([a,b])$ functions since they are Lipschitz but I cannot prove anything for functions in $C^0([a,b])$.
 A: Do you consider one partition of $[a,b]$ with shrinking width or the strong quadratic variation, i.e.
$$
\limsup_{ h \downarrow 0} \, \sup \{ \sum_{i=1}^n |f(t_{i+1})-f(t_i)|^2 :  a \le t_1 < \ldots t_n \le b, \, \Delta t_i \le h \}?
$$
One famous example is the Browian motion which has $\mathbb{P}$-a.s. $\infty$ strong quadratic variation - that's was proven by Levy in  1948. However, the quadratic variation in the sense of stochastic analysis is $\mathbb{P}$-a.s. equal to $t$ and for any partition with $\Delta t_i = o(1/\log(n))$ [result of Dudley] one has
$$
\sum_{i=1}^n |f(t_{i+1})-f(t_i)|^2 \rightarrow t
$$
$\mathbb{P}$-almost sure.
We can also construct an explicit example as follows: Take $f(x) = x^{\alpha} \sin(1/x)$ for $\alpha \in (0,1/2)$. Then for $x_k = ( \pi k + \pi + 2 \pi n )^{-1}$ with $k=1,\ldots, n-1$ we have $0 \le x_{k}-x_{k+1} \le 1/n$ and
\begin{align}
  \sum_{k=1}^{n-1} |f(x_k) - f(x_{k+1})|^2 &= \sum_{k=1}^{n-1} |x_k^\alpha + x_{k+1}^\alpha|^2 \\
& \ge  \sum_{k=1}^n x_k^{2\alpha} \\
& \ge \int_1^n \frac{1}{(\pi(x+1) + 2 \pi n)^{2 \alpha}} \mathrm{d} x \\
& \gg_{\alpha} n^{-2\alpha+1}.
\end{align}
This diverges if $\alpha \in (0,1/2)$.
