Take a finite train of points
$$ z_j=t_j+iy_j,\qquad 0\lt y_j\le Y, $$
whose real gaps are at least \(d/L\). Normalize the corresponding Laplace kernels:
$$ u_j(x)=\sqrt{2y_j}\,e^{-(y_j-it_j)x}. $$
The train may have polynomially many points. In the anti-Stokes model that motivated this calculation, it has order \(L^{1+1/(2k)}\) points. I expected that cardinality to be expensive.
It mostly disappears.
The Gram matrix is the normalized Cauchy matrix
$$ K_{jk} = \frac{2\sqrt{y_jy_k}} {y_j+y_k-i(t_j-t_k)}. $$
Its inverse has an exact diagonal formula. If
$$ \delta_j = \prod_{m\ne j} \left| \frac{(y_j-it_j)-(y_m-it_m)} {(y_j-it_j)+(y_m+it_m)} \right|, $$
then
$$ (K^{-1})_{jj}=\delta_j^{-2}. $$
So
$$ \lambda_{\min}(K) \ge \frac{\delta_*^2}{N}, \qquad \delta_*=\min_j\delta_j. $$
The whole problem has become one separation product.
Euler eats the train#
At index distance \(m\), the real separation is at least \(dm/L\), while both heights stay below \(Y\). There are at most two points at each distance, and therefore
$$ \delta_j \ge \prod_{m=1}^{\infty} \left(1+\frac{a^2}{m^2}\right)^{-1}, \qquad a=\frac{2YL}{d}. $$
But Euler already evaluated this product:
$$ \frac{\sinh(\pi a)}{\pi a} = \prod_{m=1}^{\infty} \left(1+\frac{a^2}{m^2}\right). $$
Hence
$$ \boxed{ \lambda_{\min}(K) \ge \frac1N \left(\frac{\pi a}{\sinh(\pi a)}\right)^2. } $$
The minimum-norm interpolation operator costs at most
$$ \sqrt N\,\frac{\sinh(\pi a)}{\pi a} \le \sqrt N\,e^{2\pi YL/d}. $$
That is the bit I keep staring at. A train can acquire hundreds of thousands of nodes, but as long as it remains in a fixed horizontal strip and keeps its \(d/L\) spacing, cardinality contributes only the algebraic factor \(\sqrt N\). The exponential rate sees the height \(Y\), not the number of carriages.
For the leading fixed-order anti-Stokes train, the exact conformal quantization coordinate gives
$$ d=2\pi+o(1). $$
Thus the half-line interpolation cost is
$$ \exp((Y+o(1))L). $$
The numerical stress test reached an affine train of \(920{,}981\) nodes. Nothing resembling an \(e^{L^{1+1/(2k)}}\) cardinality catastrophe appeared. One \(\sinh\) really did swallow the entire train. Cute.
Cutting the half-line without losing the frame#
The arithmetic construction cannot use all of \(x\ge0\). Its selector lives on \([0,2L]\), so the half-line theorem is not enough.
For a monotone train whose height graph becomes asymptotically flat, order the nodes by decreasing height and form the Takenaka-Malmquist basis
$$ \Phi_n(w) = \frac{\sqrt{2y_n}}{w+\overline{s_n}} \prod_{j\lt n}\frac{w-s_j}{w+\overline{s_j}}, \qquad s_j=y_j-it_j. $$
Shift the Plancherel line left by a fixed fraction of \(y_n\). The logarithm of the Blaschke product becomes a sampled Poisson-kernel envelope. Monotonicity, the \(d/L\) spacing, and one angle-derivative estimate give
$$ \sup_\xi \log |B_{n-1}(-\vartheta y_n+i\xi)|^2 \le \left(\frac{4\pi}{d}+o(1)\right)\vartheta y_nL. $$
Weighted Plancherel then makes the tail past \(2L\) exponentially small:
$$ \left\|\mathbf1_{(2L,\infty)}\phi_n\right\|_2^2 \le C_\vartheta \exp\left[ -\left(4-\frac{4\pi}{d}-o(1)\right) \vartheta y_nL \right]. $$
When the lowest retained height is
$$ \eta_L=A\frac{\log L}{L} $$
with \(A\) sufficiently large, the sum of these tails is \(o(1)\), even over a polynomial-size train. The restriction from the half-line to \([0,2L]\) therefore loses only \(o(1)\) of the entire span norm.
The compact Gram inherits the same lower bound:
$$ \boxed{ \lambda_{\min}(K_{[0,2L]}) \ge \frac{1-o(1)}{N} \left(\frac{\pi a}{\sinh(\pi a)}\right)^2. } $$
So the full-log interval is long enough. Dense linear interpolation is not the obstruction.
The remaining obstruction is nonlinear and topological. This theorem does not choose root velocities, keep the correction inside a hyperbolic chamber, prevent compensating zeros from entering elsewhere, or transfer the repaired leading model to the literal arithmetic transform. It says something narrower and useful: if that root motion fails, it is not because a polynomially long anti-Stokes train made the compact interpolation exponent explode.
Notebook references: C-0226, C-0227, C-0220, C-0178