This small step comes from my reading on saddle point approximation: suppose $$ w=\text{sign}(s)\sqrt{2(s K'(s)-K(s))}\tag{*} $$ where $K$ is a convex and infinitely differentiable function on some open interval $I$ containing $0$. Moreover, $K(0)=0$. I would like to show that $w$ is a continuously differentiable function of $s$ on an interval around $0$.
Things I tried:
- (*) implies: $\frac{w^2}{2}=sK'(s)-K(s)$ so that by "differentiating" both sides with respect to $s$, we get $w\frac{dw}{ds}=sK''(s)$ and so for $s\neq 0$, $\frac{dw}{ds}=\frac{s}{w}K''(s)$.
- $\lim_{s\to 0}(\frac{s}{w}K''(s))=\lim_{s\to 0}\frac{s}{w}\lim_{s\to 0}K''(s)=\frac{1}{\sqrt{K''(0)}}K''(0)=\sqrt{K''(0)}.$ "Therefore", $w'(0)=\sqrt{K''(0)}$.
- Consider $f(s,w)=-\frac{w^2}{2}+sK'(s)-K(s)$ hoping I could use the Implicit Function Theorem but $\frac{\partial f}{\partial w}$ is $0$ at $w=0$.
I put the words "differentiating" and "therefore" in quotations because they aren't justified until I can prove that $w$ is indeed a continuously differentiable function of $s$.
For $s>0$ or $s<0$, we have $w=+\sqrt{2(sK'(s)-K(s))}$ or $w=-\sqrt{2(sK'(s)-K(s))}$, respectively. In either case, $w$ is differentiable with $\frac{dw}{ds}=\frac{sK''(s)}{w}$.
Now, consider the derivative of $w$ at $s=0$. The right limit of the difference quotient is $$ \lim_{s\downarrow 0}\frac{\sqrt{2(sK'(s)-K(s))}-\sqrt{2(0-0)}}{s-0}=\lim_{s\downarrow 0}\frac{\sqrt{2(sK'(s)-K(s))}}{s}. $$ By Taylor's Theorem, $$ sK'(s)=\sum_{n=1}^{\infty}\frac{K^{(n)}(0)}{(n-1)!}s^n,\quad K(s)=K(0)+\sum_{n=1}^\infty\frac{K^{(n)}(0)}{n!}s^n=\sum_{n=1}^\infty\frac{K^{(n)}(0)}{n!}s^n. $$ Thus, $$ sK'(s)-K(s)=\sum_{n=1}^\infty\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)K^{(n)}(0)s^n=\sum_{n=2}^\infty\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)K^{(n)}(0)s^n. $$ It follows that $$ \lim_{s\downarrow 0}\frac{\sqrt{2(sK'(s)-K(s))}}{s}=\lim_{s\downarrow 0}\sqrt{2\left(\frac{1}{2}K''(0)+\sum_{n=3}^\infty\left(\frac{1}{(n-1)!}-\frac{1}{n!}\right)K^{(n)}(0)s^{n-2}\right)}. $$ Moving the limit inside the square root, we recognize the RHS above is just $\sqrt{K''(0)}$.
The left limit of the difference quotient at $s=0$ can be dealt with similarly and also yields $\sqrt{K''(0)}$. Thus, $w$ is differentiable at $0$ with the derivative equals $\sqrt{K''(0)}$.
Finally, to show that $w'$ is continuous, we need to show $\lim_{s\to 0}\frac{w}{s}=\sqrt{K''(0)}$. But this is done already because $\frac{w}{s}$ is just the difference quotient already considered above.