Differentiation of a two dimensional inverse function

[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.

 

[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)<br /> = f\left(\frac{y}{y_0}f^{-1}(x)\right).
Then
<br /> g&#039;(y) = \frac{f^{-1}(x)}{y_0} f&#039;\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)<br /> = f\left(\frac{y_0}{y}f^{-1}(x)\right).
Then
<br /> G&#039;(y) = -\frac{y_0f^{-1}(x)}{y^2} f&#039;\left(\frac{y_0}{y}f^{-1}(x)\right)<br />
and we find that g&#039;(y_0) = -G&#039;(y_0).

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

 

[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

 

[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
<br /> h_{y_0} \circ h_y^{-1} : x \mapsto x + (y - y_0)k(x) + \epsilon_1(x,y)\\<br /> h_{y} \circ h_{y_0}^{-1} : x \mapsto x + (y - y_0)K(x) + \epsilon_2(x,y)<br />
for some functions k and K (which are the partial derivatives with respect to y at y_0) and for all x,
<br /> \lim_{y \to y_0} \epsilon_1(x,y) = <br /> \lim_{y \to y_0} \epsilon_2(x,y) = <br /> \lim_{y \to y_0} \frac{\epsilon_1(x,y)}{y - y_0} = <br /> \lim_{y \to y_0} \frac{\epsilon_2(x,y)}{y - y_0} = 0.<br />
But we should then have
<br /> 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)<br />
so that
<br /> 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)<br />
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.

 

[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} <br /> = -\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} <br /> = \lim_{\Delta \rightarrow 0} \frac{(h_z \circ h_{z+\Delta}^{-1})(x) - (h_z \circ h_{z}^{-1})(x)}{\Delta} <br /> = \lim_{y-z \rightarrow 0} \frac{(h_z \circ h_{y}^{-1})(x) - x }{y-z}
<br /> = \lim_{-\delta \rightarrow 0} \frac{(h_z \circ h_{y}^{-1})(x) - x}{-\delta}<br /> = -\lim_{\delta \rightarrow 0} \frac{(h_z \circ h_{y}^{-1})(x)- (h_y \circ h_{y}^{-1})(x)}{\delta} <br />
<br /> = -\frac{\partial}{\partial z} (h_z \circ h_y^{-1})(x)\bigg|_{z=y} <br />, 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.

 

[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
<br /> \frac{\partial}{\partial y}f_x(y,g(y)) = \left.\frac{\partial f_x}{\partial y}\right|_{z = g(y)}<br /> + g&#039;(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
<br /> \left( \left.\frac{\partial f_x}{\partial y}\right|_{y=z}, <br /> \left.\frac{\partial f_x}{\partial z}\right|_{y=z}\right) \cdot (1,1) = 0<br />
and hence
<br /> \left.\frac{\partial f_x}{\partial y}\right|_{y=z} + <br /> \left.\frac{\partial f_x}{\partial z}\right|_{y=z} = 0<br />
as required.)