PCHIP endpoint slopes lie in the Hermite monotonicity square

At an interior node $k$ whose adjacent secants have nonpositive product, [F1] sets $m_k=0$. If their product is positive, write $r=d_k\mathrm{/}{d}_{k-1}>0$ and $W=w_1+w_2=3(h_{k-1}+h_k)$. Dividing the harmonic formula in [F1] by $d_k$ gives $m_k\mathrm{/}{d}_k=W\mathrm{/}(w_{1}r+w_2)$, which is positive and at most $W\mathrm{/}{w}_2\le 3$, because $w_2=h_k+2h_{k-1}\ge h_k+h_{k-1}$. Dividing instead by $d_{k-1}$ gives $m_k\mathrm{/}{d}_{k-1}=W\mathrm{/}(w_1+w_2\mathrm{/}r)\le W\mathrm{/}{w}_1\le 3$, since $w_1=2h_k+h_{k-1}\ge h_k+h_{k-1}$.

For a nonzero first secant, put $q=d_1\mathrm{/}{d}_0$. The raw endpoint ratio is ${\widehat{m}}_0\mathrm{/}{d}_0=1+\frac{h_0}{h_0+h_1}(1-q)$. If $q\ge 0$, this ratio is at most $1+h_0\mathrm{/}(h_0+h_1)<2$; clipping makes it zero if it is nonpositive. If $q<0$, the ratio is greater than $1$, and the cap branch reduces it to $3$ exactly when it exceeds $3$. Thus the final ratio is in $[0,3]$. If $d_0=0$, the raw endpoint value is either zero or has sign different from $d_0$, so the clip sets it to zero. The reflected argument proves the same bound at the right endpoint.

For two nodes both selected slopes equal $d_0$, so they are zero when $d_0=0$ and their ratios to $d_0$ are $1$ otherwise. For $n\ge 3$, if $d_i=0$, its interior endpoint slopes are set to zero by the nonpositive-product branch, and a first or last endpoint slope is zero by the clip rule. If $d_i\mathrm{\ne }0$, each of its endpoint slopes is either an interior slope covered by step 1.1 or an endpoint slope covered by step 1.2. These cases prove both alternatives in the statement.