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For prescribed values $F(x_i)$ at ordered points $x_0<\cdots <x_n$, the connect-the-dots interval map is the unique piecewise-linear function obtained by linear interpolation between consecutive data points.

Between two prescribed values, linear interpolation assigns $(1-t)f(a)+tf(b)$ at $(1-t)a+tb$, for $0\le t\le 1$. Joining successive samples produces a piecewise-linear function. For a continuous function on a compact interval, interpolants along grids with mesh tending to zero converge in the uniform norm; their error is bounded by the modulus of continuity at the mesh size.

Work in the complete Banach space $C([0,1])$ with the uniform norm. Each $E_m$ is closed. Indeed if $f_j\to f$ uniformly and $x_j$ witnesses membership in $E_m$, pass to a subsequence with $x_j\to x$. For each $y\in [0,1]$, uniform convergence and continuity giveThus $x$ witnesses membership for $f$.

Each $E_m$ has empty interior. Approximate an arbitrary $f$ uniformly by a piecewise-linear function $q$ on a uniform mesh, and let $D$ bound the absolute slopes of $q$. On a sufficiently fine uniform refinement of mesh $\delta$, add a triangular oscillation of fixed small amplitude $b$, alternating between $b$ and $-b$ at successive vertices. The resulting function $g$ is still as close to $f$ as desired, and every segment has absolute slope at least $2b/\delta -D$. For any $x$, choose the farther endpoint of a segment containing $x$. Its distance $h$ is between $\delta /2$ and $\delta$, sowhen $\delta$ is small. Therefore $g\notin E_m$.

By the Baire category theorem, the complement of ${\bigcup }_m{E}_m$ is dense and nonempty. If the local limsup in the question were finite at some $x$, its nearby differences would satisfy a finite bound. The remaining differences, with $|h|\ge \delta >0$, are bounded by $2\Vert f{\Vert }_{\infty }{\delta }^{-\alpha }$. Combining the bounds would put $f$ in some $E_m$, a contradiction. Hence the required limsup is infinite at every point, including one-sided approaches at the endpoints.

TakeNowsoThe recurrent pieces of this covering graph are the two-cycle $I_0\leftrightarrow I_2$ and the loop at $I_1$; neither contains two competing closed routes. Alsoso this ordering forces no 3-cycle.

The bound is attained by the piecewise-linear functionIndeed,is a 4-cycle with the required spatial order. On $I_0\cup I_2$, the fourth iterate is the identity function. Within $I_1$, the fixed point is $5/9$ andso every other point eventually leaves $I_1$ and enters $I_0\cup I_2$. The map therefore has no horseshoe in any iterate and is not chaotic. The minimum number of distinct 3-cycles is consequently $\boxed{0}$.