An illustration of convex integration theory: uniform approximation by a function with restricted values of the derivative I'm reading Vincent Borelli's lectures on Convex Integration Theory and he begins by introducing two small examples. I am a little puzzled by his explanation, and I'm sure it's just something elementary that I'm not "seeing".  
Here is the problem. Let $\mathcal{R} \subset \mathbb{R}^3$ be a path-connected subset and $f_0: [0,1] \to \mathbb{R}^3$ a given differentiable map such that $f_0'(t)$ lies in the interior of the convex hull of $\mathcal{R}$ for all $t \in [0,1]$. We want to find a $C^1$ map $f: [0,1] \to \mathbb{R}^3$ such that:
(i) for all $t \in [0,1]$, $f'(t) \in \mathcal{R}$
(ii) $||f - f_0||_{C^0} \leq \delta$ for given $\delta$
Borrelli constructs $f'$ to resemble a kind of spring and claims that this ensures that, when integrating, the resulting map will be close to the initial map (i.e. condition (ii) will be satisfied). And that is exactly what I'm not understanding. To be precise, he chooses a continuous family of loops of $\mathcal{R}$, 
$h: [0,1] \to C^0(\mathbb{R}/ \mathbb{Z}, \mathcal{R}), u \mapsto h_u$ such that 
$\int_{[0,1]} h_u(s) ds = f_0'(u)$. Then he defines $f'(t) = h_t(\{Nt\})$ where he says that $N \in \mathbb{N}^*$ and ${Nu}$ is the fractional part of $Nt$.
I don't see how this ensures that condition $(ii)$ is satisfied! I'd be grateful if someone could explain why it is the case.
 A: The definition $f'(t) = h_t(\{Nt\})$ ensures $f'$ is continuous, but let's consider another, piecewise smooth, function $g$ defined by $g'(t) = h_{\lfloor Nt\rfloor /N}(\{Nt\})$. Since $ |t - \lfloor Nt\rfloor /N|\le 1/N$, it follows that $\sup|g' - f'|$ is small when $N$ is large, hence $\sup|g-f|$ is small as well. So it suffices to estimate $\sup|g-f_0|$. 
By the definition of $g$,
$$\int_{k/N}^{(k+1)/N} g'(t) \,dt = \frac{1}{N} f_0'(k/N),\quad k=0, \dots, N-1 \tag1$$
By the Mean Value Theorem,
$$
f_0((k+1)/N) - f_0(k/N) = \frac{1}{N} f_0'(k/N) + o(1/N)
\tag2$$ 
Thus, both $g$ and $f_0$ change by about the same amount on $[k/N, (k+1)/N]$:
$$
|(g((k+1)/N) - g(k/N)) - (f_0((k+1)/N) - f_0(k/N))| = o(1/N)
\tag3$$
Summing this over $k$ and using the normalization $g(0)=f_0(0)$, we obtain 
$$ 
\sup_{k} |g(k/N)-f_0(k/N)| \to 0 \quad \text{as } \ N\to\infty
\tag4$$ 
Finally, since $h$ is a bounded function (by compactness), we have $g'$ bounded by a constant $M$ independent of $N$. Hence
$$
\sup_{ k/N \le x \le (k+1)/N} |g(x) - g(k/N)| \le  M/N  
\tag5$$ 
and similarly for $f_0$ (since $f_0'$ is bounded). From (4) and (5) it follows that $\sup|g-f_0|\to 0$ as $N\to\infty$.
