# Differentiation of a two dimensional inverse function

1. Aug 1, 2013

### Testguy

Hi

I have a question regarding differentiation of inverse functions that I am not capable of solving. I want to prove that

$\frac{\partial}{\partial y} h_y(h^{-1}_{y_0}(z_0))\bigg|_{y=y_0} = - \frac{\partial}{\partial y} h_{y_0}(h^{-1}_{y}(z_0))\bigg|_{y=y_0},$

where

$h_y(x)$ is considered as a function of x with a secondary variable y attached.
$h^{-1}_y(z)$ is the inverse function of $h$ written as a function of z, of course also depending of $y$, precisely given as the to solution to $z=h_y(x)$.

I have tested the relation with a wide range of easy-to-check h-functions and it holds in all cases I have checked. By using the chain rule I could rewrite the right-hand side to a quantity easier to handle, but as I am not able to re-write the left-hand side in any way this does not really help me.

As the derivative is with respect to the secondary y-variable and not the variable that that the inverse is taken with respect to I cannot apply the rule for differentiating inverse functions either.

Does anyone have any clue how I might prove that this holds, or have a counter-example showing that it does not hold? Any help or pointing to references are highly appreciated.

2. Aug 1, 2013

### pasmith

Is $h_y$ a bijection from the reals to themselves? If so, one can show the following:

Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable bijection, and for $y \neq 0$ let $h_y : \mathbb{R} \to \mathbb{R} : x \mapsto f(yx)$. This is a bijection since multiplication by $y \neq 0$ is a bijection, and $h_y^{-1} : \mathbb{R} \to \mathbb{R} : x \mapsto \frac{1}{y}f^{-1}(x)$.

Fix $x \in \mathbb{R}$ and $y_0 \in \mathbb{R} \setminus \{0\}$ and define $g : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ such that
$$g(y) = (h_y \circ h_{y_0}^{-1})(x) = f\left(\frac{y}{y_0}f^{-1}(x)\right).$$
Then
$$g'(y) = \frac{f^{-1}(x)}{y_0} f'\left(\frac{y}{y_0}f^{-1}(x)\right).$$
Define also $G : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ such that
$$G(y) = (h_{y_0} \circ h_y^{-1})(x) = f\left(\frac{y_0}{y}f^{-1}(x)\right).$$
Then
$$G'(y) = -\frac{y_0f^{-1}(x)}{y^2} f'\left(\frac{y_0}{y}f^{-1}(x)\right)$$
and we find that $g'(y_0) = -G'(y_0)$.

Thus if there is a counterexample then $h_y(x)$ is not of the form $f(yx)$ for any differentiable bijection $f$.

3. Aug 2, 2013

### Testguy

Thank you for your response, pasmith.
$h_y$ is indeed a bijection. (Further it generally maps $\Re$ to a bounded interval, say [0,1], if that is of any help.) I like your technique for showing this in your particular case.
If I am not mistaken one may use the same technique to show the same when $h_y(x) = f(d_1(x)d_2(y))$ and
$h_y(x) = f(d_1(x)+d_2(y))$. The algebra adds up, but I might need some additional restrictions on $d_1$ and $d_2$, at least $d_2$ must be invertible.

I stated the problem as a two-dimensional problem where $y \in \Re$, hoping that I would be able to generalize any proof to higher dimensional $y$, which is my case. So I am actually looking for a proof where $y \in \Re^m$. In the above special cases this should however not cause any problems.

I am however still interested in a proof (or counter-example) in the general setting, as there might be some cases that are not handled by the above. I was kind of hoping that there was some kind of theorem stating something like this. Does anyone know?

Any further help or pointing in a promising direction is appreciated. Thanks

4. Aug 2, 2013

### pasmith

I believe I have a proof, at least when $h_{y}(x)$ and $h_y^{-1}(x)$ are sufficiently differentiable with respect to both $x$ and $y \in \mathbb{R}$:

When $y = y_0$, $(h_{y_0} \circ h_y^{-1})$ and $(h_y \circ h_{y_0}^{-1})$ are both the identity function. Further each is the inverse of the other for any $y$.

Thus we should have
$$h_{y_0} \circ h_y^{-1} : x \mapsto x + (y - y_0)k(x) + \epsilon_1(x,y)\\ h_{y} \circ h_{y_0}^{-1} : x \mapsto x + (y - y_0)K(x) + \epsilon_2(x,y)$$
for some functions $k$ and $K$ (which are the partial derivatives with respect to $y$ at $y_0$) and for all $x$,
$$\lim_{y \to y_0} \epsilon_1(x,y) = \lim_{y \to y_0} \epsilon_2(x,y) = \lim_{y \to y_0} \frac{\epsilon_1(x,y)}{y - y_0} = \lim_{y \to y_0} \frac{\epsilon_2(x,y)}{y - y_0} = 0.$$
But we should then have
$$x = [x + (y - y_0)k(x) + \epsilon_1(x,y)] + (y - y_0)K(x + (y - y_0)k(x) +\epsilon_1(x,y)) + \epsilon_2(x + (y - y_0)k(x) +\epsilon_1(x,y),y)$$
so that
$$0 = (y - y_0)(k(x) + K[x + (y-y_0)k(x) + \epsilon_1(x,y)]) + \epsilon_1(x,y) + \epsilon_2(x + (y - y_0)k(x) +\epsilon_1(x,y),y)$$
and if $K$ and $\epsilon_2$ are continuous then the result follows on dividing by $y - y_0$ and taking the limit $y \to y_0$.

Last edited: Aug 2, 2013
5. Aug 2, 2013

### Testguy

Thank you once again for your help, this all seem correct to me. Making sure that $\epsilon_2$ could probably be problematic I guess? The extension to $y \in \mathbb{R}^k$ is probably a bit more messy as well. Thank you very much for the solution anyway.

I have in the meanwhile been looking on the more direct approach, but I am not sure if I am allowed to the operations where I switch limit operator.

I want to show that $\frac{\partial}{\partial y} (h_z \circ h_y^{-1})(x)\bigg|_{y=z} = -\frac{\partial}{\partial z} (h_z \circ h_y^{-1})(x) \bigg|_{z=y}$, which should be equivalent to what I want to show.

Let now $\Delta=y-z$ and $\delta=z-y=-\Delta$.
Thus we have that

$\frac{\partial}{\partial y} (h_z \circ h_y^{-1})(x)\bigg|_{y=z} = \lim_{\Delta \rightarrow 0} \frac{(h_z \circ h_{z+\Delta}^{-1})(x) - (h_z \circ h_{z}^{-1})(x)}{\Delta} = \lim_{y-z \rightarrow 0} \frac{(h_z \circ h_{y}^{-1})(x) - x }{y-z}$
$= \lim_{-\delta \rightarrow 0} \frac{(h_z \circ h_{y}^{-1})(x) - x}{-\delta} = -\lim_{\delta \rightarrow 0} \frac{(h_z \circ h_{y}^{-1})(x)- (h_y \circ h_{y}^{-1})(x)}{\delta}$
$= -\frac{\partial}{\partial z} (h_z \circ h_y^{-1})(x)\bigg|_{z=y}$, and we have shown what we wanted to show.

Does this seem right? I am a little troubled by changes in the limit operator by first letting $y$ be a constant and $z$ vary and then sort of change their role. Can anyone confirm that what I am doing is perfectly allowed? I guess I have to assume that the limit from above and below coincide, but that is the case for the derivative anyway.

6. Aug 2, 2013

### pasmith

The easiest approach is to define $f_x(y,z) = (h_z \circ h_y^{-1})(x)$ and note that $f_x(y,y) \equiv x$. Then if $f_x$ is differentiable the chain rule gives
$$\frac{\partial}{\partial y}f_x(y,g(y)) = \left.\frac{\partial f_x}{\partial y}\right|_{z = g(y)} + g'(y) \left.\frac{\partial f_x}{\partial z}\right|_{z= g(y)}$$

Now let $g(y) = y$, and the result follows.

(Actually it follows immediately from $f_x(y,y) = x$ that
$$\left( \left.\frac{\partial f_x}{\partial y}\right|_{y=z}, \left.\frac{\partial f_x}{\partial z}\right|_{y=z}\right) \cdot (1,1) = 0$$
and hence
$$\left.\frac{\partial f_x}{\partial y}\right|_{y=z} + \left.\frac{\partial f_x}{\partial z}\right|_{y=z} = 0$$
as required.)

Last edited: Aug 2, 2013