Module docstring
{"# Inverse of analytic functions
We construct the left and right inverse of a formal multilinear series with invertible linear term, we prove that they coincide and study their properties (notably convergence). We deduce that the inverse of an analytic partial homeomorphism is analytic.
Main statements
p.leftInv i x: the formal left inverse of the formal multilinear seriesp, with constant coefficientx, fori : E βL[π] Fwhich coincides withpβ.p.rightInv i x: the formal right inverse of the formal multilinear seriesp, with constant coefficientx, fori : E βL[π] Fwhich coincides withpβ.p.leftInv_compsays thatp.leftInv i xis indeed a left inverse topwhenpβ = i.p.rightInv_compsays thatp.rightInv i xis indeed a right inverse topwhenpβ = i.p.leftInv_eq_rightInv: the two inverses coincide.p.radius_rightInv_pos_of_radius_pos: if a power series has a positive radius of convergence, then so does its inverse.PartialHomeomorph.hasFPowerSeriesAt_symmshows that, if a partial homeomorph has a power seriespat a point, with invertible linear part, then the inverse also has a power series at the image point, given byp.leftInv. ","### The left inverse of a formal multilinear series ","### The right inverse of a formal multilinear series ","### Coincidence of the left and the right inverse ","### Convergence of the inverse of a power series
Assume that p is a convergent multilinear series, and let q be its (left or right) inverse.
Using the left-inverse formula gives
$$
qn = - (p1)^{-n} \sum{k=0}^{n-1} \sum{i1 + \dotsc + ik = n} qk (p{i1}, \dotsc, p{i_k}).
$$
Assume for simplicity that we are in dimension 1 and pβ = 1. In the formula for qβ, the term
q_{n-1} appears with a multiplicity of n-1 (choosing the index i_j for which i_j = 2 while
all the other indices are equal to 1), which indicates that qβ might grow like n!. This is
bad for summability properties.
It turns out that the right-inverse formula is better behaved, and should instead be used for this
kind of estimate. It reads
$$
qn = - (p1)^{-1} \sum{k=2}^n \sum{i1 + \dotsc + ik = n} pk (q{i1}, \dotsc, q{i_k}).
$$
Here, q_{n-1} can only appear in the term with k = 2, and it only appears twice, so there is
hope this formula can lead to an at most geometric behavior.
Let Qβ = βqββ. Bounding βpββ with C r^k gives an inequality
$$
Qn β€ C' \sum{k=2}^n r^k \sum{i1 + \dotsc + ik = n} Q{i1} \dotsm Q{i_k}.
$$
This formula is not enough to prove by naive induction on n a bound of the form Qβ β€ D R^n.
However, assuming that the inequality above were an equality, one could get a formula for the
generating series of the Qβ:
$$ \begin{align} Q(z) & := \sum Qn z^n = Q1 z + C' \sum{2 \leq k \leq n} \sum{i1 + \dotsc + ik = n} (r z^{i1} Q{i1}) \dotsm (r z^{ik} Q{ik}) \\ & = Q1 z + C' \sum{k = 2}^\infty (\sum{i1 \geq 1} r z^{i1} Q{i1}) \dotsm (\sum{ik \geq 1} r z^{ik} Q{ik}) \\ & = Q1 z + C' \sum{k = 2}^\infty (r Q(z))^k = Q_1 z + C' (r Q(z))^2 / (1 - r Q(z)). \end{align} $$
One can solve this formula explicitly. The solution is analytic in a neighborhood of 0 in β,
hence its coefficients grow at most geometrically (by a contour integral argument), and therefore
the original Qβ, which are bounded by these ones, are also at most geometric.
This classical argument is not really satisfactory, as it requires an a priori bound on a complex analytic function. Another option would be to compute explicitly its terms (with binomial coefficients) to obtain an explicit geometric bound, but this would be very painful.
Instead, we will use the above intuition, but in a slightly different form, with finite sums and an
induction. I learnt this trick in [poeschel2017siegelsternberg]. Let
$Sn = \sum{k=1}^n Q_k a^k$ (where a is a positive real parameter to be chosen suitably small).
The above computation but with finite sums shows that
$$ Sn \leq Q1 a + C' \sum{k=2}^n (r S{n-1})^k. $$
In particular, $Sn \leq Q1 a + C' (r S{n-1})^2 / (1- r S{n-1})$.
Assume that $S{n-1} \leq K a$, where K > Qβ is fixed and a is small enough so that
r K a β€ 1/2 (to control the denominator). Then this equation gives a bound
$Sn \leq Q_1 a + 2 C' r^2 K^2 a^2$.
If a is small enough, this is bounded by K a as the second term is quadratic in a, and
therefore negligible.
By induction, we deduce Sβ β€ K a for all n, which gives in particular the fact that aβΏ Qβ
remains bounded.
","### The inverse of an analytic partial homeomorphism is analytic
"}