# For which $\alpha>0$ does $x\le|W(-cx^2)|^{-\alpha}$

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

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

• 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? – Sherwin Lott Jul 12 '20 at 19:50
• 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? – Sherwin Lott Jul 12 '20 at 20:06
• @SherwinLott I'd assume $W$ and $x$ must be real, because I don't think inequalities are defined for complex numbers. – Polygon Jul 12 '20 at 20:14
• 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 $x$? – Polygon Jul 12 '20 at 20:21
• 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. – River Li Jul 13 '20 at 4:09

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.)

• 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. – Polygon Jul 13 '20 at 15:45
• @Polygon: I've now extended the argument to your case of the lower branch, thanks! Both seemed interesting regardless of which OP intended. – Sherwin Lott Jul 13 '20 at 18:48