**Proof.**

1. By the proof of Theorem 2.1 in Bayraktar et al. (2014a), we have that for 0 < *x* < *c*Λ2,

$$(\mathcal{L} - q) \left( \overline{Z}^{(q)}(\mathbf{x}) + \frac{\psi'(0+)}{q} \right) = 0 \quad \text{and} \quad (\mathcal{L} - q) Z^{(q)}(\mathbf{x}) = 0.$$

This implies that for 0 < *x* < *c*Λ2 ,

$$(\mathcal{L} - q)v\_{\delta,\Lambda}(\mathfrak{x}) = 0.$$

	- (i) If we take *y* ≤ *x*, and *c*Λ2≤ *x*, we obtain

$$
\mu^x\_\Lambda(y) = Z^{(q)}(y)\zeta\_\Lambda(x) + \Lambda \left( \overline{Z}^{(q)}(y) + \frac{\psi'(0+)}{q} \right).
$$

Recall the functions *ζ*Λ and *H* are as in Equations (7) and (14), respectively. Then,

$$\begin{split} \lim\_{y \uparrow \mathbf{x}} \frac{\mathbf{d}^2 \mu^x\_{\mathbf{A}}}{\mathbf{d} y^2}(y) &= \Lambda q \mathcal{W}^{(q)}(\mathbf{x}) + q \mathcal{W}^{(q) \prime}(\mathbf{x}) \xi\_{\mathbf{A}}(\mathbf{x}) \\ &= \frac{\mathcal{W}^{(q) \prime}(\mathbf{x})}{\mathcal{W}^{(q)}(\mathbf{x})} \left( 1 - \Lambda \left( Z^{(q)}(\mathbf{x}) - q \mathcal{W}^{(q)}(\mathbf{x}) \frac{\mathcal{W}^{(q)}(\mathbf{x})}{\mathcal{W}^{(q) \prime}(\mathbf{x})} \right) \right) \\ &= \frac{\mathcal{W}^{(q) \prime}(\mathbf{x})}{\mathcal{W}^{(q)}(\mathbf{x})} (1 - \Lambda H(\mathbf{x})) \\ &= -q \mathcal{W}^{(q)}(\mathbf{x}) \tilde{\xi}\_{\mathbf{A}}^{\prime}(\mathbf{x}). \end{split}$$

By Proposition 2, we know that *a*Λ < *c*Λ2 ≤ *x*. Then, lim *y*↑*x* <sup>d</sup>2*ux*Λ <sup>d</sup>*y*<sup>2</sup> (*y*) ≥ 0 = <sup>d</sup>2*uc*Λ2Λ d*x*<sup>2</sup> (*x*), since *ζ*Λ(*x*) < 0 by Remark 6.

(ii) We have for *y* ∈ [0, *c*Λ2 ],

$$\frac{\mathrm{d}u\_{\Lambda}^{c\_2^{\Lambda}}}{\mathrm{d}y}(y) = \Lambda Z^{(q)}(y) + q\mathcal{W}^{(q)}(y)\zeta\_{\Lambda}(c\_2^{\Lambda}) \geq \Lambda Z^{(q)}(y) + q\mathcal{W}^{(q)}(y)\zeta\_{\Lambda}(x) = \frac{\mathrm{d}u\_{\Lambda}^{x}}{\mathrm{d}y}(y),$$

which comes from the fact that for *x* ≥ *c*Λ2 > *a*Λ, then *ζ*Λ(*c*Λ2 ) ≥ *ζ*Λ(*x*) by Remark 2. On the other hand, for *y* ∈ (*c*Λ2 , *<sup>x</sup>*], we have, using the fact that *ζ*Λ(*y*) ≥ *ζ*Λ(*x*),

$$\begin{split} \frac{\mathbf{d}u^{\boldsymbol{\Lambda}}\_{\Lambda}}{\mathbf{d}y}(\boldsymbol{y}) &= \Lambda \boldsymbol{Z}^{(q)}(\boldsymbol{y}) + q\mathcal{W}^{(q)}(\boldsymbol{y})\tilde{\boldsymbol{\zeta}}\_{\Lambda}(\boldsymbol{x}) \\ &\leq \Lambda \boldsymbol{Z}^{(q)}(\boldsymbol{y}) + q\mathcal{W}^{(q)}(\boldsymbol{y})\tilde{\boldsymbol{\zeta}}\_{\Lambda}(\boldsymbol{y}) \\ &= \Lambda \boldsymbol{Z}^{(q)}(\boldsymbol{y}) + q\mathcal{W}^{(q)}(\boldsymbol{y}) \frac{1 - \Lambda \boldsymbol{Z}^{(q)}(\boldsymbol{y})}{q\mathcal{W}^{(q)}(\boldsymbol{y})} = 1 = \frac{\mathbf{d}u^{\boldsymbol{\zeta}^{\Lambda}\_{\Lambda}}\_{\Lambda}}{\mathbf{d}y}(\boldsymbol{y}). \end{split}$$

(iii) We note that

$$\begin{aligned} u\_{\Lambda}^{\chi}(c\_2^{\Lambda}) &= \Lambda \left( \overline{Z}^{(q)}(c\_2^{\Lambda}) + \frac{\psi'(0+)}{q} \right) + Z^{(q)}(c\_2^{\Lambda}) \zeta\_{\Lambda}(\mathfrak{x}) \\ &\leq \Lambda \left( \overline{Z}^{(q)}(c\_2^{\Lambda}) + \frac{\psi'(0+)}{q} \right) + Z^{(q)}(c\_2^{\Lambda}) \zeta\_{\Lambda}(c\_2^{\Lambda}) = u\_{\Lambda}^{\chi \natural}(c\_2^{\Lambda}). \end{aligned}$$

This and Point (ii) imply that (*uc*Λ2Λ − *<sup>u</sup>x*Λ)(*x*) ≥ 0. (iv) We have

$$\frac{\mathrm{d}u\_{\Lambda}^{c\_2^{\Lambda}}}{\mathrm{d}x}(x) = 1 = \lim\_{y \to x} \frac{\mathrm{d}u\_{\Lambda}^{x}}{\mathrm{d}y}(y).$$

Thus, by similar arguments to those in the proof of Theorem 2 in Loeffen (2008), we obtain the result.
