# Proving that inverse of a smooth function is smooth

Suppose I have a smooth function $$g: \mathbb{R}^n \to \mathbb{R}^t$$ and write the variables as $$(x,y)$$ where $$x \in \mathbb{R}^t$$. Suppose the Jacobian matrix of $$g(\cdot, y)$$ is invertible at $$y = 0$$ for all $$y \in B_1(0)$$. Then by the inverse function theorem and a bit of work it follows that $$g(\cdot, y)$$ is invertible on $$B_{\epsilon}(0)$$ for all $$y \in B_{\epsilon}(0)$$ for some $$\epsilon > 0$$. Let $$G(u, y) = g(\cdot, y)^{-1}(u)$$

It follows from the inverse function theorem that for a fixed $$y$$, $$G(u, y)$$ is smooth in $$u$$. I am sure that $$G$$ is a smooth function in $$(u,y)$$ also, where $$G: B_{\epsilon'}(g(0,0) \times \{ 0 \}) \to \mathbb{R}^t$$ and $$\epsilon' > 0$$ is sufficiently small. We suppose such $$\epsilon'$$ can be found such that $$G$$ is well-defined on this open set.

I was not seeing how to prove $$G$$ is smooth... Any comments, explanation is appreciated!

• What's the domain of $G$? You have an open set for each fixed $y$ but I don't know if you can make that uniform on $y$. Commented Apr 19, 2021 at 8:57
• @Quimey that is a good point. thank you. I have fixed the problem now. Commented Apr 19, 2021 at 13:46
• Can't you use inverse function theorem with $(x, y) \mapsto (x, g(x, y))$ or something like that? Commented Apr 19, 2021 at 14:41
• @OliverDiaz Thank you for your comment. It seems that explanation is too advanced for me that I am having difficulties understanding fully. Would it be possible to make it more accessible? Is there a good reference? Commented Apr 23, 2021 at 20:12
• @JohnnyT. Writing a proof would be too long here. A nice way to approach this problems is through the the uniform contraction theorem. Here is a link to a note that I hope is useful to you. It does talk about the regularity of the solutions one get. The implicit function theorem (and so, the inverse function theorem) are based on finding a fixed point of a uniform contraction. Commented Apr 23, 2021 at 21:17

In your setup $$u=g(x,y)$$ is a $$C^k$$ map ($$k\geq 1$$) from $${\Bbb R}^t\times {\Bbb R}^{n-t}$$ with $$g(0,0)=0$$ and the assumption that $$A=\partial_x g_{|(0,0)} \in GL_t({\Bbb R})$$ is an invertible matrix. Let $$B=\partial_y g_{|(0,0)} \in M_{t, n-t}({\Bbb R})$$ be the corresponding partial derivative w.r.t. $$y$$. Suppose for a moment that $$g$$ was linear. Then the implicit function theorem amounts to solving the equation $$u = Ax + By$$ for $$x$$, which is easy enough since $$A$$ was invertible: $$x = A^{-1} u - A^{-1} B y$$ The linear part corresponds to an order 1 Taylor expansion. Now, $$g$$ in general has some non-linear part as well so we should really solve $$u=Ax+By +\delta g(x,y)$$ for $$x$$. Trying to do as before and isolating $$x$$ from the linear term, we get: $$x = A^{-1} u - A^{-1} By - A^{-1} \delta g(x,y)$$ where unfortunately $$x$$ appears also on the RHS. Nevertheless, without getting discouraged, we try to use this equation to get a solution by bootstrapping. You start with a reasonable guess, e.g. the one we had above from the linear solution and you plug that into $$x$$ on the RHS to get a better guess. Iterating the procedure and taking limits you hopefully wind up with a fixed point $$x$$ which of course depends upon $$u$$ and $$y$$, so $$x=G(u,y)$$. For this to work you need reasonable conditions on the RHS when iterating. In particular, the function $$x\mapsto \Gamma(x) = A^{-1} u - A^{-1} By - A^{-1} \delta g(x,y)$$
should be a uniformly contracting map on a neighborhood of $$x=0$$. Magically, it turns out to be enough to assume that $$g$$ is $$C^1$$. Then the non-linear perturbation is "small" enough for the iteration scheme to work. A harder part of the proof is to show that the fixed point $$G(u,y)$$ is then also $$C^1$$ in $$u$$ and $$y$$. The proof typically takes a couple of pages in textbooks (one of my favorites is Serge Lang: Real and Functional Analysis, which is, however, a bit abstract). Now, once you accept that $$G$$ exists and is at least $$C^1$$ then formulae for derivatives and further regularity actually comes almost for free, again by a kind of bootstrapping argument:
In a neighborhood of the origin we have the following identity between $$C^1$$ functions: $$u = g(x,y) = g(G(u,y),y)$$ so taking derivatives with respect to $$u$$ you get $${\bf 1}_t =\partial_x g (G(u,y),y) \ \partial_u G(u,y)$$ from which $$\partial_u G(u,y) = (\partial_x g(G(u,y))^{-1}.$$ If $$g$$ is $$C^2$$, then the RHS is $$C^1$$ (being compositions of $$C^1$$ maps) so $$\partial_u G(u,y)$$ must be $$C^1$$ and in a similar way you get that so is $$\partial_y G(u,y)$$. But if the partial derivatives of $$G$$ are $$C^1$$ then $$G$$ itself must be $$C^2$$. You may iterate this argument we see that $$G$$ is as smooth as $$g$$.