= Solution
The <Butcher contractivity theorem> states that an <algebraically stable> <Runge-Kutta method> is <B-stable>: for a <dissipative vector field> and any positive step for which the stage problems are solved, its step does not increase the distance between two numerical solutions. The required condition on the <dissipative vector field> is
$$
\operatorname{Re}\langle f(t,u)-f(t,v),u-v\rangle\leq0
$$
for all compared states at each common time. The <inner product> can be real Euclidean or complex Hermitian. The <stage solvability of an implicit Runge-Kutta method> is assumed in this comparison, not a conclusion from a distance estimate alone.
Let $d$ be the difference of starting values, $D_i$ the differences of corresponding stages and $F_i$ the differences of their vector-field values. Then
$$
D_i=d+h\sum_ja_{ij}F_j,\qquad
d_+=d+h\sum_i b_iF_i.
$$
Expanding the squared <norm> of the update gives
$$
\|d_+\|^2=\|d\|^2+2h\sum_i b_i\operatorname{Re}\langle d,F_i\rangle
+h^2\sum_{i,j}b_ib_j\operatorname{Re}\langle F_i,F_j\rangle.
$$
Substitute $d=D_i-h\sum_ja_{ij}F_j$ in the linear term. Symmetrizing the double sum proves the <Runge-Kutta contractivity identity>
$$
\boxed{\|d_+\|^2=\|d\|^2
+2h\sum_i b_i\operatorname{Re}\langle D_i,F_i\rangle
-h^2\sum_{i,j}m_{ij}\operatorname{Re}\langle F_i,F_j\rangle.}
$$
Each <inner product> in the first sum is nonpositive by the defining inequality for a <dissipative vector field>, and its weight is nonnegative. The final <quadratic form> is nonnegative: resolve the <vectors> $F_i$ into components and apply <positive semidefiniteness> of $M$ to each component <vector>. Hence $\boxed{\|d_+\|\leq\|d\|}$, which is exactly <B-stability>. No <linearization> of the <vector field> has been used.
For the <Dahlquist test equation> $y'=\lambda y$ with $\operatorname{Re}\lambda\leq0$, this also bounds the amplification factor by one wherever the stages are well defined. With strictly positive weights, no stage pole can occur in the open left half-plane. Indeed, if $(I-zA)v=0$ for nonzero $v$, then
$$
v^*Mv=2\operatorname{Re}(1/z)v^*Bv-|b^Tv|^2<0
$$
when $\operatorname{Re}z<0$, contradicting $M\succeq0$. This proves the usual <A-stability> corollary for the scalar <stability function>; boundedness from the open half-plane makes any boundary pole in that <rational function> removable. Zero-weight redundant stages require their own solvability treatment. The nonlinear <B-stability> statement is stronger than <A-stability>, and the two notions should not be conflated.
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