The L2 mass of the trapezoidal cutoff is 2/3

The trapezoidal cutoff $\eta_1$ of Section 8 has $L^2$ mass exactly $2/3$:

Here $\eta_1(t) = \max\bigl(0,\; 1 - 10\,\mathrm{dist}(t, [\tfrac15, \tfrac45])\bigr)$, so $\eta_1$ is the trapezoid that equals $1$ on $[0.2, 0.8]$, falls linearly to $0$ over $[0.1, 0.2]$ and over $[0.8, 0.9]$, and vanishes outside $[0.1, 0.9]$.

The computation. The plateau contributes $0.8 - 0.2 = 3/5$. Each ramp contributes, by the substitution $u = 10t - 1$ on $[0.1, 0.2]$,

and the two ramps are symmetric about $t = 1/2$, so together they give $1/15$. Hence

Why the exact value is wanted. TaoFivePrimes.S1_major_arc_L2_mass_corollary49_limit bounds the major-arc mass of $S_1$ below by $0.94\left(\int\eta_1^2\right)x$, and the form the minor-arc argument consumes is TaoFivePrimes.S1_major_arc_L2_mass_626, namely $0.626\,x$. The two match precisely because

with a margin of about one part in a thousand. So $2/3$ is not merely a convenient value: any estimate of $\int\eta_1^2$ accurate to worse than $0.1\%$ would fail to bridge the two nodes, and the exact evaluation is what makes the chain close.

The same $2/3$ appears in TaoFivePrimes.eta_cutoff_model_mass and in TaoFivePrimes.eta1_quadratic_prime_mass, where the sifted prime sum weighted by $\eta_1^2$ is $\frac{2}{3}(1+\varepsilon)x$ — the arithmetic factor is $1$ on average, so the whole constant comes from this integral.

Note that no smoothness is needed: $\eta_1$ is piecewise linear and compactly supported, so the integral is a finite sum of elementary pieces, and the only care required is in handling Metric.infDist to the interval $[\tfrac15,\tfrac45]$ on each of the five regions.