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