Polygonal functions with sufficiently steep nonvertex slopes are dense in $C([0,1])$

By [L1], choose a finite partition $0=x_0<\cdots <x_r=1$ so that $\mathrm{\mid }f(s)-f(t)\mathrm{\mid }<\epsilon \mathrm{/}4$ whenever $s,t$ are in one partition interval. Let $g$ be the affine interpolant through $(x_i,f(x_i))$.

The affine-interpolation formula makes $\mathrm{\mid }g(t)-f(t)\mathrm{\mid }<\epsilon \mathrm{/}4$ on every partition interval. Let $S$ be the maximum of the finitely many absolute slopes of $g$.

Choose $0<\eta <\epsilon \mathrm{/}4$. By [L2], subdivide every partition interval evenly enough to support a continuous triangular sawtooth $w$, zero at the old vertices, with $\parallel w{\parallel }_{\mathrm{\infty }}\le \eta$ and every nonvertex slope of absolute value greater than $S+M$.

Put $h=g+w$. On each new affine piece, the reverse triangle inequality gives $\mathrm{\mid }{h}^{\mathrm{\prime }}\mathrm{\mid }\ge \mathrm{\mid }{w}^{\mathrm{\prime }}\mathrm{\mid }-\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }>M$, while $\parallel h-f{\parallel }_{\mathrm{\infty }}<\epsilon \mathrm{/}2<\epsilon$.

This $h$ has the required finite polygonal structure, approximation, and slope bound.