I am trying to prove the Inverse Function Theorem from the Implicit Function Theorem for Banach spaces. My attempt so far is as follows:

Let $f:\mathbf{X}\to \mathbf{Y}$ be a $\mathcal{C}^k$ function between Banach spaces, and let $x^*\in\mathbf{X}$ and $y^*:=f(x^*)$ be such that the Fréchet derivative $\mathrm{d}\,f(x^*):\mathbf{X}\to\mathbf{Y}$ is bounded with bounded inverse.

Consider $F:\mathbf{X}\times\mathbf{Y}\to\mathbf{Y}$ given by $F(x,y):=f(x)-y$. Then the partial Fréchet derivative $\partial_xF(x^*,y^*)\equiv \mathrm{d}\,f(x^*)$ is bounded with a bounded inverse, so $F$ satisfies the hypotheses of the Implicit Function Theorem. Hence, there are open sets $U\subseteq\mathbf{X}$ and $V\subseteq\mathbf{Y}$ containing $x^*$ and $y^*$, respectively, and a $\mathcal{C}^k$ function $g:V \to U$ such that $F(g(y), y) = 0$ for all $y\in V$, i.e. $f(g(y)) = y$.

The above shows that $g$ is a right inverse of $f$, but I want to be able to conclude, further, that $g$ is a left inverse of $f$ on $U$, i.e. $g(f(x)) = x$ for all $x \in U$. I know this to be equivalent to showing that the restriction $f|_U$ is injective (possibly by choosing a smaller neighbourhood of $x^*$ in $U$), but I am not sure how to do this.

I would greatly appreciate any hints!

  • $\begingroup$ Doesn't the classical formulation of the implicit function theorem say $f:X\oplus Y \to Z$ etc $\endgroup$ – JJR May 16 '17 at 11:11
  • $\begingroup$ @JJR Are you referring to my writing '$X \times Y$' instead of '$X \oplus Y$? The statement of the theorem as I know it has $F$ defined on an open set $W \subseteq X \times Y$ (I have taken $W = X \times Y$ for simplicity). $\endgroup$ – ryan221b May 16 '17 at 13:10
  • $\begingroup$ Oh sry my mistake I thought you were proving implicit function theorem from inverse function theorem not the other way around $\endgroup$ – JJR May 16 '17 at 13:15

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