The commutator has entries \([D,f]_{ij}=D_{ij}(f_j-f_i)\). Adding a
constant to \(f\) does not change it, so set \(f_1=0\), \(f_2=x\) and
\(f_3=y\). The entries above the diagonal are \(a=D_{12}x\),
\(b=D_{13}y\) and \(c=D_{23}(y-x)\). Let
\(\omega =(-c,b,-a)\) and let \(\sigma ^{2}=a^{2}+b^{2}+c^{2}\). Write
\(M\) for the matrix of \([D,f]\). It is skew-symmetric, so its square
is the negative of \(M^{\mathrm T}M\). Expanding that product gives
\(M^{\mathrm T}M=\sigma ^{2}I-\omega \omega ^{\mathrm T}\). Thus
\(M^{\mathrm T}M\) fixes the line through \(\omega \) by \(0\) and acts
as \(\sigma ^{2}\) on the orthogonal plane. The operator norm of
\([D,f]\) is therefore \(\sigma \). The constraint \(\sigma \le 1\) reads
\[(D_{12}^{2}+D_{23}^{2})x^{2}-2D_{23}^{2}xy+(D_{13}^{2}+D_{23}^{2})y^{2}\le 1.\]
Write \(A=D_{13}^{2}+D_{23}^{2}\). If \(A>0\), the left side is a
quadratic polynomial in \(y\) opening upwards. The discriminant of
this polynomial set equal to \(1\) is \(4A-4\Delta x^{2}\). A real
\(y\) exists exactly when this discriminant
is nonnegative. If \(\Delta >0\), that condition is
\(x^{2}\le A/\Delta \), so the largest value of \(|x|\) is the displayed square
root. If \(\Delta =0\), the discriminant is positive for every \(x\),
and there is no finite upper bound.
If \(A=0\), then \(D_{13}=D_{23}=0\) and the constraint reduces to
\(D_{12}^{2}x^{2}\le 1\). A nonzero \(D_{12}\) gives the reciprocal
bound. If \(D_{12}\) vanishes as well, every \(x\) is allowed. Since
\(A=0\) forces \(\Delta =0\), this case and the case \(A>0\) with
\(\Delta =0\) cover every vanishing \(\Delta \).
On the path with vertices \(0,1,2,3\) and the adjacency matrix as
\(D\), the same commutator norm gives distance \(2\) from \(0\) to
\(3\). The graph distance along the three edges is \(3\). Let
\(a=f_1-f_0\), \(b=f_2-f_1\) and \(c=f_3-f_2\), so
\(f_3-f_0=a+b+c\). The commutator is zero except for these three
adjacent pairs. Replacing each difference by its absolute value does
not change the operator norm: the signs are absorbed by changing the
signs of the last three basis vectors. Thus it is enough to take
\(a,b,c\ge 0\).
Suppose \(a+b+c>2\). If \(a\ge 2\), the first basis vector is sent to
a vector of length \(a\), so the norm exceeds \(1\). If \(a<2\), then
\(b+c>2-a\). The vector \((0,1,0,-1)\) has length \(\sqrt {2}\) and is
sent to \((a,0,-(b+c),0)\), whose square length is
\(a^{2}+(b+c)^{2}\). That square exceeds
\(a^{2}+(2-a)^{2}=2(a-1)^{2}+2\ge 2\), so the operator norm exceeds
\(1\). Therefore every admissible triple has \(a+b+c\le 2\).
The values \(a=c=1\) and \(b=0\) give \(f=(0,1,1,2)\). The commutator
is the orthogonal sum of two copies of the block with entries
\(0,1,-1,0\), and each block has norm \(1\). The difference of the
endpoint values is \(2\), so the distance is exactly \(2\).