ScalingStacks

Proof. [040F]

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Proof.

It suffices to prove the growth estimate for z1,z2,z0z_{1},z_{2},z_{0}. By elementary calculation |∂α1∂η|≤C​a22​|η|(μ12+a22​|η|2)3/2|\frac{\partial\alpha_{1}}{\partial\eta}|\leq\frac{Ca_{22}|\eta|}{(\mu_{1}^{2}+a_{22}|\eta|^{2})^{3/2}}, so upon integration

∫μ1∞|∂α1∂η|​d​μ1≤C|η|​(μ1μ12+a22​|η|2−1)≤C|η|,\int_{\mu_{1}}^{\infty}|\frac{\partial\alpha_{1}}{\partial\eta}|d\mu_{1}\leq\frac{C}{|\eta|}(\frac{\mu_{1}}{\sqrt{\mu_{1}^{2}+a_{22}|\eta|^{2}}}-1)\leq\frac{C}{|\eta|},

hence |βi|≤C|η||\beta_{i}|\leq\frac{C}{|\eta|} by the integral definition of βi\beta_{i}. From

d​log⁡|z1|=d⁡(a11​μ1+a12​μ2)+α1​d​μ1+α3​d​(μ1−μ2)+Re​(β1​d​η),d\log|z_{1}|=d(a_{11}\mu_{1}+a_{12}\mu_{2})+\alpha_{1}d\mu_{1}+\alpha_{3}d(\mu_{1}-\mu_{2})+\text{Re}(\beta_{1}d\eta),

we integrate to obtain the growth bound on z1z_{1} for A1/4​|μ→|a≥1A^{1/4}|\vec{\mu}|_{a}\geq 1. But |z1||z_{1}| is a continuous function, so the bound holds also near the origin. Similarly we can bound |z0||z_{0}| and |z2||z_{2}|. ∎

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