Prove that id the functions $f_1,..,f_n$ in the statement of the implicit function theorem are assumed to be k times continuously differentiable (i.e., all partial derivatives of order k exist and are continuous), then the same is true of the component function $\gamma_1,.., \gamma_n$ of $\gamma.$

I know this theorem is true, but I don't get how you can relate to the componet function, $\gamma.$

  • $\begingroup$ One has to assume $0=f(x,y)$ has local solutions $y=γ(x)$? You ever heard of "implicit differentiation"? $\endgroup$ – LutzL Mar 14 '16 at 20:55
  • $\begingroup$ I have, but how does it connect to $\gamma$? $\endgroup$ – Holly Mar 14 '16 at 20:57
  • $\begingroup$ It allows you to compute the derivatives of $γ(x)$. $0=f_x+f_yγ'$, $0=f_{xx}+2f_{xy}γ'+f_{yy}[γ',γ']+f_yγ''$ etc. has always as highest derivative term $f_yγ^{(k)}$ which allows to express this highest derivative in lower derivatives of $γ$ and partial derivatives of $f$. $\endgroup$ – LutzL Mar 14 '16 at 21:02
  • $\begingroup$ Could you show me the whole proof? I'm a physics major and found this topic to be interesting. $\endgroup$ – Holly Mar 14 '16 at 21:08

It follows from the formula for the (Fréchet) derivative of $\gamma$. Suppose $x^*,y^*,z^*$ are given points such that $g(x^*,y^*) = z^*$, where $g$ is $C^k$. If $\gamma$ is $C^1$ and solves (as the IFT gives you) $$g(x,\gamma(x)) = z^*$$ Then by the Chain Rule, $$d\gamma(x) = -[\partial_yg(x,\gamma(x))]^{-1}\partial_xg(x,\gamma(x))$$

Looking at this equation, we see that everything on the right hand side is indeed $C^1$; hence $d\gamma$ is $C^1$, so $\gamma$ is $C^2$. But by the same argument, we get $d\gamma$ is actually $C^2$, so in fact $\gamma$ is $C^3$. And so it continues, until you run out of continuous derivatives of $g$.

In coordinate form, we get by Chain rule on $z^* = g(x,\gamma(x))$

$$ \mathbf{0} = J_xg(x,\gamma(x)) + J_yg(x,\gamma(x))∇ \gamma(x)$$ where $\mathbf{0}$ is a zero matrix of appropriate size, $J_yg(a,b)$ is the Jacobian matrix of $g$ at $(a,b)$, considered as a function of $y$ only, and similarly with $J_xg(a,b)$. This gives us $$∇ \gamma(x) = \begin{pmatrix}\frac{\partial\gamma}{\partial x_1}(x) \\ \vdots \\ \frac{\partial\gamma}{\partial x_n}(x) \end{pmatrix} = -[J_yg(x,\gamma(x))]^{-1}J_xg(x,\gamma(x))$$

Now apply the same argument as before: everything on the right hand side (partials of g, $\gamma$) and all matrix operations (inversion) are $C^1$; hence the left hand side is $C^1$. But then this means that $\gamma$ is $C^2$. Repeating the argument inductively, we see that $\gamma$ is $C^k$.

  • $\begingroup$ Could you explain it without the usage of Fréchet derivative? I haven't learned that yet. $\endgroup$ – Holly Mar 14 '16 at 22:56
  • $\begingroup$ @Julie OK, can you give me the formula for $\partial \gamma_i /\partial x_j$ that your version of the IFT gives you? $\endgroup$ – Calvin Khor Mar 14 '16 at 23:13
  • $\begingroup$ docdro.id/RUbL3cE, it's the 3rd section, this is from roselincht $\endgroup$ – Holly Mar 14 '16 at 23:33
  • $\begingroup$ @Julie, I have added to the answer. Does it help? $\endgroup$ – Calvin Khor Mar 14 '16 at 23:52
  • $\begingroup$ Yes, a lot, but could you explain what argument as before? I'm trying to understand it, but it's not clicking yet. Thank you! $\endgroup$ – Holly Mar 15 '16 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.