I recently came across a generalized Implicit Function Theorem (Jittorntrum (1978) and Kumagai (1980)) that relaxes the requirements of differentiability and non-singularity (essentially just requires the function to be continuous and locally injective/one-to-one). The following question came to mind: is there an an analogue of implicit differentiation for one-sided derivatives?

For concreteness, consider $F(x,y) = 0$ where $F:\mathbb{R}^2\to\mathbb{R}$ is a continuous function. Suppose that for some ($x_0,y_0)\in\mathbb{R}^2$, there exist $\varepsilon_x,\varepsilon_y>0$ s.t. for every fixed $y\in(y_0-\varepsilon_y,y_0+\varepsilon_y)$, $F(\cdot,y):(x_0-\varepsilon_x,x_0+\varepsilon_x)\to\mathbb{R}$ is locally one-to-one. (This supposition is just so that Theorem 2.1 from Jittorntrum (1978) holds).

Now, let $\partial_x^- F(x,y) = \lim_{h\nearrow 0} \frac{F(x+h,y)-F(x,y)}{h} $ denote the left-hand partial derivative w.r.t. $x$ and define $\partial_y^- F(x,y)$ similarly. Then, is the following true
$$ \partial_x^-F(x,y)\cdot\underset{=1}{\underbrace{\partial_x^- x}} + \partial_y^- F(x,y) \cdot \partial_x^- y = 0 $$ for all $(x,y) \in (x_0-\varepsilon_x,x_0+\varepsilon_x) \times (y_0-\varepsilon_y,y_0+\varepsilon_y) $?

Thanks so much for reading! Any insights for why the above is (or is not) true would be very much appreciated!

  • $\begingroup$ The issue with implicit differentiation is that you need to show that the implicit function exists, I believe there are implicit function theorems that require local Lipschitzness rather than differentiability in which case the above would hold (assuming that the one sided derivatives exist). See Clarke's generalised derivative. $\endgroup$
    – copper.hat
    Sep 10 '20 at 3:37

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