I was reading the section of Pseudoconvexity and plurisubharmonicity of Hörmander's book (Introduction to complex analysis in several variables), and I have a doubt regarding the proof of the following theorem:
Theorem 2.6.7. If $\Omega$ is an open set in $\mathbb{C}^{n}$, the following conditions are equivalent:
(i) $-\log\delta_{\Omega}$ is plurisubharmonic in $\Omega$, where $\delta_{\Omega}(z)=\mbox{dist}(z,\mathbb{C}^{n}\setminus \Omega)=\inf\{\|z-w\|:w\in \mathbb{C}^{n}\setminus \Omega\}$.
(ii) There exist a continuous plurisubharmonic function $u$ in $\Omega$ such that $$\Omega_{c}=\{z\in\Omega : u(z)<c\}$$ is relatively compact in $\Omega$
(iii) $\hat{K}^{P}_{\Omega}$ is relatively compact in $\Omega$ if $K$ is relatively compact in $\Omega$, where $\hat{K}^{P}_{\Omega}=\{z\in\Omega : u(z)\leq\sup_{K}u\hspace{0.2cm}\mbox{for all}\hspace{0.2cm}u\hspace{0.2cm}\mbox{plurisubharmonic in}\hspace{0.2cm}\Omega\}$.
My doubt is in the proof of (iii) implies (i).
For this the author takes $z_{0}\in\Omega$, $w\in\mathbb{C}^{n}\setminus\{0\}$, and chooses $r>0$ so that $D=\{z_{0}+\tau w : |\tau|\leq r\}\subseteq \Omega$ then he considers $f(\tau)$ analytic polynomial such that $$-\log\delta_{\Omega}(z_{0}+\tau w, \mathbb{C}^{n}\setminus\Omega)\leq \mbox{Re}f(\tau),\hspace{0.2cm}|\tau|=r,$$ equivalently $$|e^{-f(\tau)}|\leq \delta_{\Omega}(z_{0}+\tau w, \mathbb{C}^{n}\setminus\Omega),\hspace{0.2cm}|\tau|=r.$$ Then he wants prove the same last inequality when $|\tau|\leq r$. To do this, he takes any vector $a\in\mathbb{C}^{n}$ with $\|a\|<1$ and for $0\leq\lambda\leq 1$ considers the mappings $$\tau \longmapsto z_{0}+\tau w+\lambda e^{-f(\tau)} a ,\hspace{0.4cm}|\tau|\leq r.$$ He denotes the image of this mappings by $D_{\lambda}$.
Then he puts $$\Lambda=\{\lambda\in [0,1] : D_{\lambda}\subseteq\Omega\}.$$ From (iii) and that $[0,1]$ is connected, it follows that $\Lambda=[0,1]$. In particular $1\in \Lambda$, then $D_{1}=\{z_{0}+\tau w+e^{-f(\tau)}a : |\tau|\leq r \}\subseteq\Omega$.
My specific doubt is the following, how from here it is concluded that $$|e^{-f(\tau)}|\leq \delta_{\Omega}(z_{0}+\tau w, \mathbb{C}^{n}\setminus\Omega),\hspace{0.2cm}\mbox{for all}\hspace{0.2cm}|\tau|\leq r.$$ ?
I tried it by contradiction, that is, there exists $|\tau_0|\leq r$ with $$|\delta_{\Omega}(z_0 +\tau_0 w, \mathbb{C}^{n}\setminus\Omega)|<|e^{-f(\tau_0)}|$$ but unfortunately I have not achieved anything.
Thanks
Since $D_1 \subset \Omega$, we have $$ z_0+\tau \omega + e^{-f(\tau)}a \in \Omega $$ for all $a\in \mathbb{C}^n$ with $\|a\|<1$ and all $|\tau| \leq r$. Therefore, for every fixed $|\tau|\leq r$ and for every $\varepsilon>0$, the closed ball $$ B(z_0+\tau \omega,(1-\varepsilon)|e^{-f(\tau)}|) $$ is contained in $\Omega$. This implies that $$ \delta_{\Omega}(z_0+\tau \omega) \geq (1-\varepsilon)|e^{-f(\tau)}| $$ for all $|\tau|\leq r$ and all $\varepsilon>0$. Since $\varepsilon>0$ is arbitrary, you get the result.