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Using the implicit function theorem one can prove the following:

Let $X,Y$ be Banach-spaces, $U\subset X$ open, $f\colon U\to \mathbf{R}$, $g\colon U\to Y$ continuously differentiable function. If $f|_{g^{-1}(0)}$ has a local extremum at $x$, and $g'(x)\in L(X;Y)$ has a right inverse in $L(Y;X)$, then there is a unique $\lambda\in Y^*$, such that $f'(x)=\lambda\circ g'(x)$.

One can also give some second order necessary and sufficient conditions. This generalizes the case where $Y$ is finite dimensional, and $g'(x)$ is surjective. However I have seen some texts that claimed that only surjectivity is sufficient even in the infinite dimensional case.

My question is does that hold true? If it does what is the method of the proof, because I cannot figure how the implicit function theorem could be used. If the proof is complicated I would appreciate even just some good references.

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up vote 2 down vote accepted

Yes, surjectivity of dg(x) is sufficient.

The proof is not very complicated, but takes some work. The reference cited by Planetmath is: Eberhard Zeidler. Applied functional analysis: main principles and their applications. Springer-Verlag, 1995. Take a look at page 268 and following.

The key ingredient is a weaker formulation of the implicit function theorem: Suppose $F: X \times Y \to Z$ is $C^1$ in a neighbourhood of $0$, with $F(0,0) = 0$ and $D_Y F(0,0) : Y \to Z$ is surjective. Then,

(1) For each $r > 0$, there exists $\rho > 0$ so for every $|| u || < \rho$, there exists $v = v(u)$ with $F(u, v(u) ) = 0$ and $|| v || < r$.

(2) There exists $d > 0$ so that $|| v(u) || \le d || D_Y F(0,0) v(u) ||$.

The proof of this is a bit more delicate than the standard implicit function theorem, but you use the closed range theorem to construct a surrogate for the inverse, and then use this to construct your iteration.

You then prove the Lagrange multiplier theorem from this, essentially following the standard proof (presumably the one you had in mind above).

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