A mistake on the Borwein - Preiss's Variational Principle statement?? I know the following generalization of Borwein - Preiss's  Variational  Principle(BPVP), known as Loewen-Wang's Variational principle (LWVP)

$\textbf{Loewen- Wang Variational Principle}$
Let $f:X\to \mathbb{\bar{R}}$ be proper, l.s.c and bounded below. Let $\epsilon >0$ and consider a point $\bar{x}$ such that ${f(\bar{x})\leq \inf_X f +\epsilon}.$  Let $\rho:X\to \mathbb{R}$ be continuous and such that
$$\sup\{\|x\|: \rho(x)>1\}< +\infty.$$ Then, for any decreasing sequence $\{\mu_n\}\subseteq (0,1)$ such that
$$\sum_{n=0}^\infty \mu_n <+\infty$$ there exists a sequence $\{z_n\}\subseteq X$ convergent to some $z\in X$ such that

*

*$\rho(z-\bar{x})<1,$

*$z$ is a strong minimizer of the function

$$f(x)+ \epsilon\sum_{n=0}^\infty \mu_n \rho((n+1)(x-z_n)).$$

In the proof of this result they don't use the condition $$\sum_{n=0}^\infty \mu_n <+\infty,$$ so here is my first question:

Question 1: Why do we need this condition? Is it for somehow guarantee that
$$\sum_{n=0}^\infty \mu_n \rho((n+1)(x-z_n))< +\infty ?$$

Now, I tried to deduce BPVP from the previous statement.

$\textbf{Borwein - Preiss's  Variational  Principle}$
Let $f:X\to \mathbb{\bar{R}}$ be proper, l.s.c and bounded below. Let $p\geq 1,\epsilon >0$ and consider a point $\bar{x}$ such that ${f(\bar{x})\leq \inf_X f +\epsilon}.$  Then, for each $\lambda >0$ there exists a sequence $\{\nu_n\}\subseteq (0,1)$ such that
$$\sum_{n=0}^\infty \nu_n =1,$$ and there exists a sequence $\{z_n\}\subseteq X$ convergent to some $z\in X$ such that

*

*$\|z-\bar{x}\|\leq \lambda,$

*$z$ is a strong minimizer of the function

$$f(x)+ \frac{\epsilon}{\lambda ^p}\sum_{n=0}^\infty \nu_n \|(x-z_n)\|^p.$$

Here is my proof of this statement: In order to apply LWVP, we take
$$\rho(x)=\frac{1}{\lambda^p}\|x\|^p,\;\mu_n=\frac{1}{2^{n+1}(n+1)^p}.$$
It is easy to see that all the conditions of LWVP are satisfied and then it follows the existence of $z_n$ convergent to some $z$ such that $z$ is a strong minimizer of $$f(x)+ \epsilon\sum_{n=0}^\infty \mu_n \rho((n+1)(x-z_n))=$$
$$f(x)+ \frac{\epsilon}{\lambda^p}\sum_{n=0}^\infty \frac{1}{2^{n+1}(n+1)^p} \|(n+1)(x-z_n)\|^p= $$
$$f(x)+ \frac{\epsilon}{\lambda ^p}\sum_{n=0}^\infty \nu_n \|(x-z_n)\|^p,$$ with $$\nu_n= \frac{1}{2^{n+1}}.$$ $\Box$
Now, in the book I am reading (Schirotzek, Nonsmooth Analysis), instead of taking
$$\mu_n=\frac{1}{2^{n+1}(n+1)^p},$$ they take $$\mu_n=\frac{1}{2^{n+1}(n+1)\sigma},$$ where $$\sigma= \sum_{n=0}^\infty \frac{(n+1)^{p-1}}{2^{n+1}} $$ and proceed as I did. Now, although I find this last proof to be correct, my question is

Question 2: Is my reasoning correct, so that in BPVP we can remove the assumption of the existence of $\{\nu_n\}$ since it can be chosen to be $\nu_n=\frac{1}{2^{n+1}} ?$

I have searched other papers and all of them state the theorem without recognizing this, which makes me think that I am wrong somewhere. Any help is appreciated. Thanks in advance to all.

Regarding your first question, you gave the answer yourself. The convergence of the series $\sum{\mu_n}$ is used in order to make sure that the series 
$\rho_{\infty}(x)=\sum_{n=0}^\infty \mu_n \rho((n+1)(x-z_n))$ is convergent. Indeed, in the course of the proof they show that $\rho((n+1)(x-z_n))<1$ for all $n$, which implies that the function $\rho_{\infty}$ is well defined.
As for your second question, please note that the existence of $\nu_n$ is not an assumption of the BPVP, but it is part of the assertion. The fact that your choice $\nu_n=\frac{1}{2^{n+1}}$ always works can also be read off the original proof of Borwein and Price. Indeed, they show that whenever you choose $\varepsilon_1$ such that 
$$f(\bar{x})-\inf_X f<\varepsilon_1<\varepsilon\enspace\enspace\enspace      (1)$$
then for every number $\mu$ satisfying
$$0<\mu<1-\frac{\varepsilon_1}{\varepsilon}$$
the result of their theorem holds with $\nu_n=\mu^n(1-\mu)$. Since we can always adjust $\varepsilon,\varepsilon_1$ so that $\varepsilon>2\varepsilon_1$ will hold, hence $1-\frac{\varepsilon_1}{\varepsilon}>\frac{1}{2}$ will hold, (observe that we can increase $\varepsilon$ if needed while keeping inequality ($1$)), we can therefore always choose $\mu=\frac{1}{2}$, which gives your $\nu_n=\frac{1}{2^{n+1}}$.
