Continuity does not imply finite quadratic variation

For $m\ge 1$ put $I_m:=[2^{-(m+1)},2^{-m}]$, $n_m:=2m^4$, ${\delta }_m:=2^{-(m+2)}\mathrm{/}{m}^4=2^{-(m+1)}\mathrm{/}{n}_m$, and let $s_{m,j}:=2^{-(m+1)}+j{\delta }_m$ for $0\le j\le n_m$; define $x(s_{m,j}):=1\mathrm{/}m$ for odd $j$ and $x(s_{m,j}):=0$ for even $j$, interpolate $x$ linearly between consecutive vertices of each block, and set $x:=0$ on $\{0\}\cup [1\mathrm{/}2,1]$.

The function $x$ is well defined and continuous: on each block it is piecewise linear hence continuous, the last vertex of $I_m$ has even index and value $0$, matching the value $0$ at the shared endpoints of consecutive blocks and on $[1\mathrm{/}2,1]$, and for $t\in (0,2^{-m}]$ one has $\mathrm{\mid }x(t)\mathrm{\mid }\le 1\mathrm{/}m$, so $x(t)\to 0=x(0)$ as $t\downarrow 0$.

For $n\ge 1$ let ${\pi }_n$ be the partition of $[0,1]$ whose point set is the dyadic grid $\{k{2}^{-n}:0\le k\le 2^n\}$ together with every vertex $s_{m,j}$ with $1\le m\le n$; each ${\pi }_n$ is a finite partition in the sense of Partition of $[a,b]$ as a finite strictly increasing list $a=t_0<t_1<\cdots <t_n=b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, the sequence is refining, and $\mathrm{mesh}({\pi }_n)\le 2^{-n}$.

No dyadic point of level $n$ lies in the interior of $I_n$: such a point would be $k{2}^{-n}$ with $1\mathrm{/}2<k<1$, and no integer satisfies this; hence the points of ${\pi }_n$ inside $I_n$ are exactly the vertices $s_{n,0}<\cdots <s_{n,n_n}$, and consecutive points of ${\pi }_n$ inside that block are consecutive vertices.

The quadratic sum along ${\pi }_n$ therefore contains the $n_n$ vertex increments of the block $I_n$, each of absolute value $1\mathrm{/}n$, so it is at least $n_n\cdot (1\mathrm{/}n{)}^2=2n^4\mathrm{/}{n}^2=2n^2$; since $2n^2\to \mathrm{\infty }$, the quadratic sums along the refining sequence $({\pi }_n)$ diverge to $+\mathrm{\infty }$ for this continuous function.

Consequently continuity alone does not force finite quadratic variation along a prescribed refining sequence with vanishing mesh: the witness is the explicit sawtooth function above, whose block $I_n$ alone contributes $2n^2$ to the $n$-th quadratic sum; the example also shows that the mesh condition of Quadratic variation along a partition sequence is not sufficient by itself, and the construction uses no choice principle.