= Solution
For the level-$N$ dyadic partition, write
$$
V_N(t)=\sum_{k=1}^{2^N}
|f(k2^{-N}\wedge t)-f((k-1)2^{-N}\wedge t)|.
$$
The dyadic partitions are nested, so the <triangle inequality> makes $V_N(t)$ nondecreasing in $N$, and $\lVert f_t\rVert=\lim_NV_N(t)$.
Fix $t_1<t_2$. As the mesh tends to zero, the last dyadic point before $t_1$ approaches $t_1$. Refining from there to $t_2$, the triangle inequality says that the added variation is at least $|f(t_2)-f(t_1)|$ minus the two endpoint errors, which tend to zero by <continuous function>[continuity]. Therefore
$$
\lim_N\bigl(V_N(t_2)-V_N(t_1)\bigr)
\geq|f(t_2)-f(t_1)|.
$$
Since $\lVert f\rVert<\infty$, both limits are finite and may be subtracted, giving
$$
\lVert f_{t_2}\rVert-\lVert f_{t_1}\rVert
\geq|f(t_2)-f(t_1)|.
$$
Solved by gpt-5.6-sol high.
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