Upper bound on the Lambert function

Question

Is there an upper bound on the Lambert function $$W(-\frac{k}{e})$$ for $0 < k < 1$? Or should the condition be $0 < k \leq 1$?

I know that $W(-\frac{1}{e}) = -1$; would it be okay to say $W(-\frac{k}{e}) \leq -1$?

I just recently learned about Lambert functions and am having a hard time understanding all these. This came from an approximation of an expression given by:

$$\ln \phi_\infty = \mu + \lambda \phi_\infty - \mu e^{-\lambda \phi_\infty},$$

which by letting $z=\lambda \phi_\infty$, yields $$z = -\frac{1}{1+\mu}W(-\lambda(1+\mu))$$ which only exists if $$\lambda (1+\mu) \leq \frac{1}{e}.$$ If we let $\lambda = ke^{-1}\frac{1}{1+\mu}$ for $0 < k \leq 1$, we can have $$ z = -\frac{1}{1+\mu}W\bigg(-\frac{k}{e}\bigg). $$

What I need is to represent this last equation in terms of $\phi_\infty$ alone (and these parameters $\lambda, \mu$) so I was wondering if I could get an upper bound for the Lambert function and do just that.

Thank you so much for your help!

Answer 1

$W$ without a subscript usually means $W_0$.

Let $f(k)=W\left(-\frac{k}{e}\right)$ for $0<k<1$.

See Wolfram Alpha for the graph of $f$, extended by the points $(0,0)$ and $(1,-1)$.

$$\lim_{x\to 0}f(x)=0,\ \ \lim_{x\to 1}f(x)=-1$$

See at Wikipedia: Upper and lower bounds, Bounded function and Bounded set.

$f$ is a bounded function: bounded above by $0$ and bounded below by $-1$.

$$-1<f(k)<0$$