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A definition in my notes state:

If $Df(a)$ is invertible (as a matrix), then $f$ is invertible on an open set $U$ containing $a$.
So given that $f(x,y) = (a,b)$ and there exists a $C^1$ local inverse near (x,y) with derivative $Df^{-1}(a,b) = (Df(x,y))^{-1}$.

My query is on the $C^1$. What does this mean in this context or any context? A lot of questions in my homework also go along the lines of "Show that $f$ has a local $C^1$ inverse near ...".

Twenty-six colours
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2 Answers2

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A function of class $C^n$ is a function that is $n$ times continuously differentiable. The inverse function theorem states that the inverse function is continuous and differentiable, so you can safely take one time derivative (it does not imply second derivative also exists) so it is of class $C^1$.

user440429
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  • So does this imply that $f$ has continuous partial derivatives (i.e. is differentiable and has derivative equal to the jacobian matrix)? – Twenty-six colours Apr 27 '17 at 10:44
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    That's right. Derivative of $f$ with respect to different independent variables form elements of Jacobian matrix. – user440429 Apr 27 '17 at 10:57
  • Thanks. A slight unrelated question: If $f$ is proven to be $C^1$, can we show that it is $C^2$ (for some point)? So if $f$ is differentiable with derivative $Jf$ then can we show that $f$ is twice differentiable with $J(Jf)$? This seems weird because the Jacobian of $f$ already has four elements... – Twenty-six colours Apr 27 '17 at 11:24
  • It should be checked. It may also be of class $C^2$. The statement that $f$ is of class $C^1$ says nothing about higher order derivatives.
    • – user440429 Apr 27 '17 at 11:30
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  • Are elements of J constant? If so the answer of second question is yes. Otherwise, you have to take derivative of $J$ into account, too.
    • – user440429 Apr 27 '17 at 11:32
  • Might be out of my course content, but how would we find $J(Jf)$ even if the elements of $Jf$ are constants?
    What I understand from $Jf$ is that the first column of $Jf$ are all partials of 1st,2nd,3rd, ... etc components of $f$ with respect to $x_1$. Similarly for other columns, but if $Jf$ has say 2 columns, 2 rows, how would we take the jacobian of this?     – Twenty-six colours Apr 27 '17 at 11:43
  • Well. $J.f = g$ is again a column vector. Now you have a column vector $g$ and you can now take Jacobian as you did for $f$. The only difference is that the elements of $g$ are now new functions. – user440429 Apr 27 '17 at 12:17
  • Hmm, I think $Jf$ is a vector with more than one column isn't it?
    For example, if $f(x,y) = \begin{pmatrix} x^2+y^2 \ x^2 y \end{pmatrix}$,
    $Jf = \begin{pmatrix} 2x && 2y \ 2xy && x^2 \end{pmatrix}$. Correct me if I'm wrong.     – Twenty-six colours Apr 27 '17 at 12:24
  • It's not correct. $f$ is a $2\times 1$ matrix (a column), $J$ is a $2\times 2$ matrix and so their product becomes $g_{2\times 1} = J_{2\times 2} * f_{2\times 1}$. – user440429 Apr 27 '17 at 12:29
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    What you have evaluated is not $Jf$, it is exactly $J$. – user440429 Apr 27 '17 at 12:31
  • Sorry, my class have been using the notation that $Jf$ is the Jacobian matrix of $f$ at $(x,y)$. In this context, the derivative of f is just the jacobian matrix. – Twenty-six colours Apr 27 '17 at 12:34
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    It's not a common notation. I thought that you mean by $Jf$ the matrix multiplication of $J$ and $f$. So I have to correct my statement. The Jacobian is defined for a vector with respect to another one (it may be easily generalized for derivative of matrix with respect to vector that gives a 3Ds matrix). – user440429 Apr 27 '17 at 12:41