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Differentiation of a two dimensional inverse function

  1. Aug 1, 2013 #1

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

    [itex]\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},[/itex]


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

    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. jcsd
  3. Aug 1, 2013 #2


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    Is [itex]h_y[/itex] a bijection from the reals to themselves? If so, one can show the following:

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

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

    Thus if there is a counterexample then [itex]h_y(x)[/itex] is not of the form [itex]f(yx)[/itex] for any differentiable bijection [itex]f[/itex].
  4. Aug 2, 2013 #3
    Thank you for your response, pasmith.
    [itex]h_y[/itex] is indeed a bijection. (Further it generally maps [itex]\Re[/itex] 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 [itex]h_y(x) = f(d_1(x)d_2(y))[/itex] and
    [itex]h_y(x) = f(d_1(x)+d_2(y))[/itex]. The algebra adds up, but I might need some additional restrictions on [itex]d_1[/itex] and [itex]d_2[/itex], at least [itex]d_2[/itex] must be invertible.

    I stated the problem as a two-dimensional problem where [itex]y \in \Re[/itex], hoping that I would be able to generalize any proof to higher dimensional [itex]y[/itex], which is my case. So I am actually looking for a proof where [itex]y \in \Re^m[/itex]. 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
  5. Aug 2, 2013 #4


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

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

    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 [itex]k[/itex] and [itex]K[/itex] (which are the partial derivatives with respect to [itex]y[/itex] at [itex]y_0[/itex]) and for all [itex]x[/itex],
    \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 [itex]K[/itex] and [itex]\epsilon_2[/itex] are continuous then the result follows on dividing by [itex]y - y_0[/itex] and taking the limit [itex]y \to y_0[/itex].
    Last edited: Aug 2, 2013
  6. Aug 2, 2013 #5
    Thank you once again for your help, this all seem correct to me. Making sure that [itex]\epsilon_2[/itex] could probably be problematic I guess? The extension to [itex]y \in \mathbb{R}^k[/itex] 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 [itex]\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}[/itex], which should be equivalent to what I want to show.

    Let now [itex]\Delta=y-z[/itex] and [itex]\delta=z-y=-\Delta[/itex].
    Thus we have that

    [itex]\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} [/itex]
    = \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}
    [/itex], 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 [itex]y[/itex] be a constant and [itex]z[/itex] 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.
  7. Aug 2, 2013 #6


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    The easiest approach is to define [itex]f_x(y,z) = (h_z \circ h_y^{-1})(x)[/itex] and note that [itex]f_x(y,y) \equiv x[/itex]. Then if [itex]f_x[/itex] 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)}[/tex]

    Now let [itex]g(y) = y[/itex], and the result follows.

    (Actually it follows immediately from [itex]f_x(y,y) = x[/itex] 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
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