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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, $\gamma.$

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  • $\begingroup$ 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
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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 $C^k$.

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  • $\begingroup$ 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

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