Roots of a perturbed equation

Question

I'm looking to show that the equation $$\displaystyle \psi(\delta) := e^{\alpha g(\delta)} - \delta$$ has a real root for $\alpha$ sufficiently small that converges to $\delta = 1$ as $\alpha \rightarrow 0$. Firstly, is this true assuming $g$ is nice? And if it is, does anyone have any idea how to prove it?

Answer 1

Introducing the function $f(\alpha,\delta) = e^{\alpha g(\delta)}-\delta$, you can invoke the implicit function theorem to give you the required result, as long as $f$ is continuously differentiable in a neighbourhood of $(0,1)$, which means that $g$ has to be continuously differentiable as well (around $\delta = 1$, at least). So, a continuously differentiable $g$ is sufficient.

On the other hand, $g$ does not have to be very 'nice' for this result to hold. Take for example a discontinuous but bounded $g$, such as \begin{equation} g(\delta) = \left\{ \begin{array} 11 & \delta > 1 \\ -1 & \delta < 1 \end{array}\right. , \end{equation} then it still holds that the solution of $f(\alpha,\delta) = 0$ converges to $\delta=1$ as $\alpha \to 0$. Even worse, take $g(\delta) = \frac{1}{\delta-1}$, which is discontinuous and unbounded as $\delta \to 1$. Still, the solution of $f(\alpha,\delta) = 0$ converges to $\delta=1$ as $\alpha \to 0$.

The reason for this is that the equation $f(\alpha,\delta) = 0$ is equivalent to \begin{equation} \alpha g(\delta) = \log \delta. \end{equation} So, as long as the graph of $g(\delta)$ exists for some interval left or right of $\delta = 1$, we can choose $\alpha$ sufficiently small to let that graph intersect the graph of $\log \delta$, where it's clear that that intersection will converge to $\delta=1$ as $\alpha \to 0$. So, with an exception of pathological examples, you can be very liberal in how 'nice' you want your function $g$ to be.
[Note that it could happen that only the left or only the right limit of $\alpha \to 0$ exists; take for example $g(\delta) = \pm \sqrt{\delta-1}$].

Answer 2

Yes, this is true.

$$\Psi(\delta)=e^{a \cdot g(\delta)}-\delta$$

We want to solve for its zero, so we have,

$$\delta=e^{a \cdot g(\delta)}$$

If we expand in powers of a, we get,

$$\delta=1+a \cdot g(\delta)+O(a^2)$$

You said that $g$ is nice, so we can simply eliminate $a$ by setting it to $0$.