Continuous paths and limits of linear interpolations Let $C^0([a,b],\mathbb{R}^K)$ be the space of all continuous paths $\gamma : [a,b]\to \mathbb{R}^K$.
Let $(P_n)$ be a sequence of partitions of $[a,b]$ into $n$ subintervals and such that $\lim |P_n|\to 0$ in other words, the mesh of the partitions goes to zero. One chooses $P_{n+1}$ to refine $P_n$. We shall denote $P_n = \{t_n^0,\dots, t_n^n\}$.
Now for every $n$ pick a function $\mathbf{x}_n : P_n\to \mathbb{R}^K$.
We get thus a collection of functions $\{\mathbf{x}_n : n\in \mathbb{N}\}$. We further constrain these so that $$\mathbf{x}_n |_ {P_m} = \mathbf{x}_m.$$
Now define the functions $\gamma_n : [a,b]\to \mathbb{R}^K$ by 
$$\gamma_n(t)=\mathbf{x}_n(t_n^k)+\frac{\mathbf{x}_n(t_n^{k+1})-\mathbf{x}_n(t_n^k)}{t_n^{k+1}-t_n^k}(t-t_n^k),\quad t\in [t_n^k,t_n^{k+1}],\quad t_n^k \in P_n,\quad \forall k\in \{0,\dots, n\}.$$
We thus have a sequence in $C^0([a,b],\mathbb{R}^K)$.
In Physics it is usual to assume that this kind of sequence converges in a sense to a continuous path, and that every continuous path is the limit of such a sequence.
Now is this true? If so, is this convergence pointwise or is it convergence in some norm defined on the space of paths?
My first guess was to work with pointwise convergence and try to show that $(\gamma_n(t))$ is a Cauchy sequence for all $t$ but I couldn't work this out (maybe it is even wrong).
How the correct statement can be shown?
 A: It's certainly not true that $(\gamma_n)$ always converges: you could choose the $\mathbf{x}_n$ to be very bad.  For instance, for $K=1$ if you define $\mathbf{x}_n(t^k_n)=n$ for all $n$ where $t^k_n$ is the new point added in $P_n$, then it is easy to see that $\gamma_n(t)\to \infty$ for any $t$ that is not in any $P_n$ (since it is eventually surrounded by points of the partition that get mapped to arbitrarily large numbers.  More generally, $P=\bigcup_n P_n$ is some countable dense subset of $[a,b]$, and the pointwise limit of the $\gamma_n$ must agree with the $\mathbf{x}_n$ on $P$.  So if you take any function on $P$ which is not continuous (or which does not extend continuously to $[a,b]$) and use it to define the $\mathbf{x}_n$, then the $\gamma_n$ cannot possibly converge pointwise to a continuous function on $[a,b]$.
On the other hand, it is true that every continuous path can be obtained as a uniform limit in this way.  Indeed, given any continuous path $f$, just let $\mathbf{x}_n=f|_{P_n}$ for all $n$, and then $\mathbf{x}_n$ converges uniformly to $f$.  This follows easily from uniform continuity of $f$: as the mesh of $P_n$ gets small, $f$ becomes approximately constant on each interval formed by the partition (with the error in the approximation being uniform across all the intervals) so the piecewise linear approximation by $\mathbf{x}_n$ gets arbitrarily close to $f$.  (This argument is closely related to the proof that a continuous function on a compact interval is Riemann integrable.)
Combining the two parts, we see that in fact $(\gamma_n)$ converges pointwise to a continuous function iff the function on $P=\bigcup_n P_n$ obtained by gluing together the $\mathbf{x}_n$ extends to a continuous function on $[a,b]$, in which case the convergence is in fact uniform.  Note moreover that a function on $P$ extends to a continuous function on $[a,b]$ iff it is uniformly continuous.
