be useful later on.

\[lem:Ideltabound\] For all $\delta \geq 2$ and $0<x\leq\tfrac12$, $I_{\delta}\left(x\right) \leq \left(1-x\right)
-\left(1-x\right)^{2\left(1-1/\delta\right)}$.

The claim can be verified numerically. For $w \in \left(0,\frac{1}{2}\right]$, let $\tilde{y}
=\left(1-x\right)-\left(1-x\right)^{2\left(1-w\right)}$. Fig. \[fig:cplot\] gives a contour plot of the function $$\tilde{I}\left(x,w\right) = x^{x\left(1-w\right)}\left(1-x\right)^{\left(1-x\right)
\left(1-w\right)}\left(1-x-\tilde{y}\right)^{-\left(1-x-\tilde{y}\right)/2}
\left(x-\tilde{y}\right)^{-\left(x-\tilde{y}\right)/2}\tilde{y}^{-\tilde{y}}.$$

\[htbp\]

Note that $\tilde{I}\left(x,w\right)\geq 1$, which immediately implies Lemma \[lem:Ideltabound\] when we take $w=1/\delta$. It is easy to verify that the boundary values corresponding to $x \downarrow 0$ and $x \uparrow 1$ result in $\tilde{I}\left(x,w\right) \downarrow 1$.

\[lem:Ideltamonotone\] The function $x\rightarrow\frac{I_{\delta}\left(x\right)}{x}$ is non-increasing on $\left(0,\frac{1}{2}\right]$.

By definition of $I_{\delta}\left(x\right)$, the function $$\hat{I}\left(x,z\right) = x^{x\left(1-1/\delta\right)}\left(1-x\right)^{\left(1-x\right)
\left(1-1/\delta\right)}\left(1-x-xz\right)^{-\left(1-x-xz\right)/2}
\left(x-xz\right)^{-\left(x-xz\right)/2}\left(xz\right)^{-xz}$$ satisfies $\hat{I}(x,\frac{I_{\delta}\left(x\right)}{x})=1$ for all $x\in\left(0,\frac{1}{2}\right]$. It will therefore suffice to show that $$\label{Itask}
\frac{\partial}{\partial x}\vert_{z=\frac{I_{\delta}\left(x\right)}{x}} \hat{I}\left(x,z\right) \geq 0,$$ since this implies that, for $\epsilon$ sufficiently small, $\hat{I}(x+\epsilon,\frac{I_{\delta}
\left(x\right)}{x})\geq1$, and hence that $$\inf\left\{w\colon\,\hat{I}\left(x+\epsilon,w\right)\geq1\right\} \leq\frac{I_{\delta}\left(x\right)}{x},$$ and thus $\frac{I_{\delta}(x+\epsilon)}{x+\epsilon} \leq \frac{I_{\delta}\left(x\right)}{x}$. Observe that $$\begin{aligned}
&\frac{\partial}{\partial x}\hat{I}\left(x,z\right)\\
&= \hat{I}\left(x,z\right)
\left\{ \log\left(\left(\frac{x}{1-x}\right)^{\left(1-1/\delta\right)}
\left(1-x-xz\right)^{\left(1+z\right)/2}\left(x-xz\right)^{-\left(1-z\right)/2}
\left(xz\right)^{-z}\right)\right\}.
\end{aligned}$$ For $z=\frac{I_{\delta}\left(x\right)}{x}$, $\hat{I}\left(x,z\right)=1$ implies $$\left(x^{x\left(1-1/\delta\right)}\left(1-x\right)^{\left(1-x\right)
\left(1-1/\delta\right)}\right)^{\frac{1}{x}}=\left(1-x-xz\right)^{\frac{1}{2x}
-\left(1+z\right)/2}\left(x-xz\right)^{\left(1-z\right)/2}\left(xz\right)^{z}$$ and hence $$\begin{aligned}
&\frac{\partial}{\partial x}\hat{I}\left(x,z\right)\\
&=\hat{I}\left(x,z\right)\left\{ \log\left(\left(\frac{x}{1-x}\right)^{\left(1-1/\delta\right)}
\left(x^{x\left(1-1/\delta\right)}\left(1-x\right)^{\left(1-x\right)
\left(1-1/\delta\right)}\right)^{-\frac{1}{x}}\left(1-x-xz\right)^{\frac{1}{2x}}\right)\right\}.
\label{eq:derivIhat}
\end{aligned}$$ The term inside the logarithm in (\[eq:derivIhat\]) simplifies to $\left(1-x\right)^{-\frac{1}{x}
\left(1-1/\delta\right)}\left(1-x-xz\right)^{\frac{1}{2x}}$, which satisfies $\left(1-x\right)^{-\frac{1}{x}\left(1-1/\delta\right)}\left(1-x-xz\right)^{\frac{1}{2x}}\geq1$ whenever $1-x-\left(1-x\right)^{2\left(1-1/\delta\right)} \geq xz=I_{\delta}\left(x\right)$. By Lemma \[lem:Ideltabound\], this is true for all $x\in\left(0,\frac{1}{2}\right]$, and so follows.

We can now 