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Let $x\in(0,1)$. I want to know for which $\alpha>0$ it's true that $$ x\le|W(-cx^2)|^{-\alpha},\label{1}\tag{$\ast$} $$

where $W$ is the Lambert W-function and $c>0$ is some constant.

In my numerical tests, the value of $c$ didn't really seem to matter, but \eqref{1} seemed to hold for very small $\alpha$, for example $\alpha\approx 0.001$. It seems difficult to prove analytically because of the non-elementary nature of the Lambert W function.

For negative $y<0$, it seems to be true that $W(-y)<0$. So we can rewrite \eqref{1} as

$$ x(-W(-cx^2))^{-\alpha}\le 1. $$

We can define a function $f(x)=x(-W(-cx^2))^{-\alpha}$. Then $f(0)=0$, $f>0$ on $(0,1)$ and $f\in C^1$ since $W$ is differentiable on $(0,1)$ as it does not include the points $\{0,\frac{1}{e}\}$.

So the maximum of $f$ reached at $x_0$ should satisfy

$$f'(x_0)=0\label{2}\tag{$\ast\ast$}$$

where

$$ f'(x)=\left(-W(-cx^2)\right)^\alpha\left(1-\frac{2 c\alpha x^2 W'(-cx^2)}{W(-cx^2)} \right), $$

So \eqref{2} is

\begin{align*} &\left(-W(-cx_0^2) \right)^\alpha-2c\alpha x_0^2\left(-W(cx_0^2) \right)^{\alpha-1}W'(-cx_0^2)=0 \\ \iff& \alpha=-c'\frac{W(-cx_0^2)}{x_0^2 W'(-cx_0^2)} \\ \iff& x^2_0\frac{\mathrm{d}}{\mathrm{d}x_0}\log\left( W(-cx_0^2)\right)=-c''\frac{1}{\alpha}, \end{align*}

But I don't see how to go from here, i.e., how to invert the function

$$ \psi(x)=x^2\frac{\mathrm{d}}{\mathrm{d}x}\log\left(W(-cx^2)\right) $$ to recover $x_0$ as

$$ x_0=\psi^{-1}\left(-c''\frac{1}{\alpha}\right), $$

and plug that back into \eqref{1}.

But in Mathematica, it gives

$$ c'''\psi^{-1}\left(-c''\frac{1}{\alpha}\right)=\pm c'''\left(\alpha W\left(\mp c'' i\frac{1}{\sqrt{\alpha}}\right)\right)^{-\frac{1}{2}}, $$

which isn't very helpful!

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  • $\begingroup$ I'm confused why you say the value of $c$ doesn't seem to matter much? It can scale up/down $|W(-cx^2)|$ arbitrarily? $\endgroup$ – Sherwin Lott Jul 12 '20 at 19:50
  • $\begingroup$ I believe your second equation should be: $x(-W(-cx^2))^{\alpha}\le 1$? Also, you seem to be assuming $W$ is not a complex number there? Are you constraining $c$ so that $W(-cx^{2})$ is real? $\endgroup$ – Sherwin Lott Jul 12 '20 at 20:06
  • $\begingroup$ @SherwinLott I'd assume $W$ and $x$ must be real, because I don't think inequalities are defined for complex numbers. $\endgroup$ – Polygon Jul 12 '20 at 20:14
  • $\begingroup$ Does this only consider the upper branch of the $W$ function? I can see it happening where $x$ is between the two branches - greater than one, but less than the other. In that case how can we say if the function is greater than or equal to $x$? $\endgroup$ – Polygon Jul 12 '20 at 20:21
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    $\begingroup$ If so, I think that the problem is: Let $c > 0$ be a constant. For which $\alpha > 0$, it is true that $$x \le |W(-cx^2)|^\alpha$$ for all $x$ in $(0, 1/\sqrt{c\mathrm{e}}]\cap (0, 1)$, where $W(\cdot)$ is the principal branch of the Lambert W function. $\endgroup$ – River Li Jul 13 '20 at 4:09
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There are a few ambiguities:

  • By treating $W(-cx^{2})$ as real-valued, you're implicitly assuming $cx^{2} \leq e^{-1}$.
  • There are two different values for $W(-cx^{2})$, the principal branch in $(-1, 0)$ and the lower branch in $(-\infty, -1)$, which are you using?
  • On the principal branch, your inequality trivially holds for all $\alpha > 0$ since $x <1 < |W(-cx^2)|^{-\alpha}$.

Thus, there are two possible nontrivial questions:

  1. For which $\alpha > 0$ does $x \leq |W(-cx^{2})|^{\alpha}$ hold for all $x \in (0, \min\{1, 1/\sqrt{ce}\})$ on the primary branch of $W$.
  2. For which $\alpha > 0$ does $x \leq |W(-cx^{2})|^{-\alpha}$ hold for all $x \in (0, \min\{1, 1/\sqrt{ce}\})$ on the lower branch of $W$.

My conclusions are as follow:

  1. The inequality holds if $\alpha \leq 1/2$ and $c$ is not too small.
  2. The inequality holds for any given $\alpha$ if $c$ is not too small.

Question 1

$W(-cx^{2})$ is defined to be the solution on $(-1,0)$ to the equation: $$-c x^{2} = we^{w}$$ But what matters is $|W(-cx^2)|^{\alpha}$, so let's rewrite the equation in terms of $\hat{w}$ where: $$\hat{w} = (-w)^{\alpha} \in (0,1)$$ $$ \Rightarrow w = -\hat{w}^{1/\alpha}$$ Thus, $|W(-cx^2)|^{\alpha}$ is the solution on $(0,1)$ to the equation: $$c x^{2} = \hat{w}^{1/\alpha}e^{-\hat{w}^{1/\alpha}}$$ The right hand side is increasing for $\hat{w} \in (0,1)$, since its derivative is: $$(1/\alpha)\hat{w}^{1/\alpha-1}e^{-\hat{w}^{1/\alpha}}(1-\hat{w}^{1/\alpha})$$ Therefore, the solution is at least $x$ if and only if: $$cx^{2} \geq x^{1/\alpha}e^{-x^{1/\alpha}}$$ $$\Leftrightarrow \log(c) + (2-1/\alpha)\log(x) \geq - x^{1/\alpha}$$ $$\Leftrightarrow \log(c) \geq (1/\alpha - 2)\log(x) - x^{1/\alpha} \equiv \gamma(x)$$


If $\alpha > 1/2$, then the right hand is arbitrarily large for small $x$, hence the inequality is violated.

If $\alpha = 1/2$, then $\log(c) \geq -x^{2}$ only holds for all $x \in (0,1)$ if $c \geq 1$.

If $\alpha < 1/2$, then let's maximize $\gamma(x)$: $$\gamma'(x) = \frac{1/\alpha - 2}{x} - (1/\alpha)x^{1/\alpha - 1}$$ $$\gamma''(x) < 0$$ The first order condition is $\gamma'(\tilde{x})=0 \Rightarrow \tilde{x} = (1 - 2\alpha)^{\alpha}$, so: $$\begin{align} \gamma(\tilde{x}) &= (1 - 2\alpha)\log(1-2\alpha) - (1 - 2\alpha) \\ &=(1-2\alpha)(\log(1-2\alpha)-1) \end{align}$$

Thus, the inequality holds if: $$\boxed{\log(c) \geq (1-2\alpha)(\log(1-2\alpha)-1)}$$ This is sufficient but not quite necessary since $\tilde{x}$ may lie outside of $(0, 1/\sqrt{ce})$. Instead, let $\tilde{x}(c) = \min\left\{(1 - 2\alpha)^{\alpha}, \, 1/\sqrt{ce}\right\}$, then the inequality holds for any $\alpha$ and $c$ that satisfy:

$$\log(c) \geq \gamma(\tilde{x}(c))$$

(We can solve for the exact bound on $c$ as a function of $\alpha$ by setting it equal.)


Question 2

$W(-cx^{2})$ is defined to be the solution on $(-\infty,-1)$ to the equation: $$-c x^{2} = we^{w}$$ But what matters is $|W(-cx^2)|^{-\alpha}$, so let's rewrite the equation in terms of $\hat{w}$ where: $$\hat{w} = (-w)^{-\alpha} \in (0,1)$$ $$ \Rightarrow w = -\hat{w}^{-1/\alpha}$$ Thus, $|W(-cx^2)|^{-\alpha}$ is the solution on $(0,1)$ to the equation: $$c x^{2} = \hat{w}^{-1/\alpha}e^{-\hat{w}^{-1/\alpha}}$$ The right hand side is increasing for $\hat{w} \in (0,1)$, since its derivative is: $$(1/\alpha)\hat{w}^{-1/\alpha-1}e^{-\hat{w}^{-1/\alpha}}(\hat{w}^{-1/\alpha}-1)$$ Therefore, the solution is at least $x$ if and only if: $$cx^{2} \geq x^{-1/\alpha}e^{-x^{-1/\alpha}}$$ $$\Leftrightarrow \log(c) + (2+1/\alpha)\log(x) \geq - x^{-1/\alpha}$$ $$\Leftrightarrow \log(c) \geq -(1/\alpha + 2)\log(x) - x^{-1/\alpha} \equiv \hat{\gamma}(x)$$


There are no obvious cases here, so let's just maximize $\hat{\gamma}(x)$. The first order condition is characterized by: $$\hat{\gamma}'(x) = -\frac{1/\alpha + 2}{x} + (1/\alpha)x^{-1/\alpha - 1}$$ $$\hat{\gamma}'(\tilde{x}) = 0 \Rightarrow \tilde{x} = (1+2\alpha)^{-\alpha}$$

$$\hat{\gamma}''(x) = \frac{1/\alpha + 2}{x^{2}} - (1/\alpha)(1/\alpha + 1)x^{-1/\alpha - 2}$$ $$\hat{\gamma}''(\tilde{x})<0$$

Since the first-order condition is only satisfied at $\tilde{x}$, and $\hat{\gamma}$ is concave at that point, $\hat{\gamma}$ is increasing on $(0,\tilde{x})$ and decreasing on $(\tilde{x}, 1)$.

$$\hat{\gamma}(\tilde{x}) = (1+2\alpha)(\log(1+2\alpha)-1)$$

Thus, the inequality holds if: $$\boxed{\log(c) \geq (1+2\alpha)(\log(1+2\alpha)-1)}$$

This is sufficient but not quite necessary since $\tilde{x}$ may lie outside of $(0, 1/\sqrt{ce})$. Instead, let $\tilde{x}(c) = \min\left\{(1 + 2\alpha)^{-\alpha}, \, 1/\sqrt{ce}\right\}$, then the inequality holds for any $\alpha$ and $c$ that satisfy:

$$\log(c) \geq \hat{\gamma}(\tilde{x}(c))$$

(We can solve for the exact bound on $c$ as a function of $\alpha$ by setting it equal.)

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  • $\begingroup$ About $-\alpha$ vs $\alpha$, if OP is referring to the lower branch, then only $-\alpha$ would make sense. This leads me to think that OP is only considering the lower branch, but some clarification by OP would be nice. $\endgroup$ – Polygon Jul 13 '20 at 15:45
  • $\begingroup$ @Polygon: I've now extended the argument to your case of the lower branch, thanks! Both seemed interesting regardless of which OP intended. $\endgroup$ – Sherwin Lott Jul 13 '20 at 18:48

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