# Is the solution to this functional equation differentiable?

Let $$R(s)$$ be the inverse of a differentiable CDF $$F:[0,1] \rightarrow [0,1]$$. Therefore $$R()$$ is an increasing function, $$R:[0,1]\rightarrow [0,1].$$

[Aside: I earlier mistakenly thought if $$F$$ is differentiable $$R$$ must be differentiable. But this need not be the case, as pointed out by copper.hat in the comments using the counterexample $$F(x)=x^2$$, so that $$R(x)=\sqrt{x}$$ which is not differentiable at $$x=0$$.]

Let $$\alpha \in [0,1)$$. We are looking for a solution $$\beta(\alpha)$$ to the following functional equation:

$$\int\limits^{\beta(\alpha)}_{\alpha} R(s)ds=\left(\frac{\beta(\alpha)-\alpha}{1-\alpha}\right) \int\limits^{1}_{\beta(\alpha)} R(s)ds$$

such that $$\beta(\alpha)>\alpha$$. (i.e. we are not interested in the trivial solution $$\beta(\alpha)=\alpha$$.)

My question is: Is $$\beta(\alpha)$$ always differentiable? If yes, how can we show it? If not, what additional conditions do we need on $$F$$ so it is?

Define $$\phi(\cdot)$$ as: $$\phi(x)=\int\limits^x_0R(s)ds$$ for all $$x\in[0,1]$$.

Note that a unique solution $$\beta(\alpha) \in (\alpha,1)$$ always exists. This can be seen from writing the previous equation as:

$$\frac{\phi(\beta)-\phi(\alpha)}{\phi(1)-\phi(\beta)} = \frac{\beta - \alpha}{1-\alpha}$$

and noting that the LHS is a convex increasing function of $$\beta$$ for a given $$\alpha$$ and then applying standard intermediate value theorem arguments to the LHS and RHS which are both continuous functions of $$\beta$$ for a given $$\alpha$$.

(Edit) My attempt: Define $$W:[0,1] \rightarrow [0,1]$$, $$W(x)=\phi(1)-\phi(x)$$.

The original equation can be written as:

$$\int\limits_{\alpha}^{\beta}\left((1-\alpha)R(s)-W(\beta)\right)ds=0$$

which gives us, $$\left((1-\alpha)R(\alpha)-W(\beta)\right)<0$$ and $$\left((1-\alpha)R(\beta)-W(\beta)\right)>0$$.

We can differentiate both sides of the original equation using fundamental theorem of calculus and get:

$$((1-\alpha)R(\beta)-(W(\beta)-(\beta-\alpha)R(\beta)))\beta'(\alpha)= ((1-\alpha)R(\alpha)-\left(\frac{1-\beta}{1-\alpha}\right)W(\beta))$$

The inequalities derived above show that $$((1-\alpha)R(\beta)-(W(\beta)-(\beta-\alpha)R(\beta)))>0$$, so $$\beta'(\alpha)$$ exists.

Is this a valid proof of the existence of $$\beta'(\alpha)$$ always? I'm confused because we (probably) need to know $$\beta'(\alpha)$$ exists before taking derivative of the original equation, and not the other way round? Any help is appreciated.

Write your equation above in the form $$G(\alpha,\beta)=0$$ and note that $$G$$ is $$C^1$$ in a neighborhood of your solution (indeed: $$\phi$$ is $$C^1$$ because $$R$$ is continuous as an inverse of an injective continuous function on an interval).
The implicit function theorem states that a sufficient (and I guess in your case also neccesary) condition for the differentiability of the solution $$\beta$$ at a point $$\alpha_0$$ is that $$\frac{dG}{d\beta}(\alpha_0,\beta(\alpha_0))\ne0$$.
Thus, the (simple) exercise is to find $$G$$, and differentiate it with respect to the variable $$\beta$$. The more difficult part is to check whether this derivative can be zero on the graph of your solution. If not, your solution is even continuously differentiable.
• When you say differentiate $G$ wrt the variable $\beta$, do you mean partial derivative treating $\alpha$ as constant? I have already tried to give an argument by totally differentiating $G$ wrt $\alpha$. Do you think that makes sense? Sorry I mistakenly posted this comment before it was complete. May 7 '21 at 6:28
• > When you say differentiate $G$ wrt the variable $\beta$, do you mean partial derivative treating $\alpha$ as constant? Yes, that's how the partial derivative is defined. I haven't checked your argument, but if it shows that this partial derivative is nonzero, it should be correct. May 7 '21 at 6:31
• I now understand - this condition boils down to exactly the condition I've suggested in my answer, which is: Calculate $\beta'(\alpha)$ as if it exists, and show that its denominator cannot be equal to zero for any $(\alpha, \beta(\alpha))$. One more question - Do we need continuity of $R$ (or, continuity of $F$) for $\beta'(\alpha)$ to exist? (denominator $\neq 0$ doesn't require it)Ideally I want to use $R$ as the generalized inverse of any CDF $F$ if possible. But I do need differentability of $\beta(\alpha)$. Of course, if not, I'll stick to the current formulation of "very nice" $F$. May 7 '21 at 6:52
• The classical implicit function theorem assumes that the function is $C^1$ but has a stronger conclusion than you want (namely that the inverse exists locally uniquely). In Theorem 8.8 of my book Topological Analysis, I have proved a “pointwise” version which requires only differentiability of $G$ but assumes existence, uniqueness and continuity of the implicit function. However, I am not sure whether this helps here, because if $R$ has a jump (because $F$ is constant on some interval), then $G$ is not even differentiable. May 8 '21 at 8:23