A:

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|x-x_0|^{ -\alpha-1}.$$Therefore, the parameter regime where the approximation works is$$\label{eq:singapprox2}
|x-x_0| \ll \tfrac12 \min\{|x-x_0|^{\alpha},|x-x_0|^{ -\alpha-1}\}.$$We have |x-x_0|^{\alpha-1} \lesssim {\lVert\rho(x)\rVert}_\infty \leq \tfrac12 {\lVert\rho(x)\rVert}_2 \leq \tfrac12 {\lVert\rho(x)\rVert}_\infty. This proves that the approximation error decays as x \to x_0 at fixed {\lVert\rho(x)\rVert}_\infty. Now we are ready to prove the theorem. Let us assume that {\lVert\rho(x)\rVert}_\infty \leq \gamma for x \in {\mathbb{R}} and prove the lower bound in. Set \beta := \alpha + 1 and \gamma := \tfrac12 {\lVert\rho(x)\rVert}_\infty. If \gamma > 1 and \beta = 2 \alpha, then |x-x_0|^\beta \gtrsim {\lVert\rho(x)\rVert}_\infty^{\alpha} and we are done. Otherwise, we use the approximation inequality and conclude that for |x-x_0| \leq \tfrac12 \min\{|x-x_0|^{\alpha},|x-x_0|^{ -\alpha-1}\} we have$$\begin{aligned} 