Inverse of analytic functions #

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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).

Main statements #

• p.left_inv i: the formal left inverse of the formal multilinear series p, for i : E ≃L[𝕜] F which coincides with p₁.
• p.right_inv i: the formal right inverse of the formal multilinear series p, for i : E ≃L[𝕜] F which coincides with p₁.
• p.left_inv_comp says that p.left_inv i is indeed a left inverse to p when p₁ = i.
• p.right_inv_comp says that p.right_inv i is indeed a right inverse to p when p₁ = i.
• p.left_inv_eq_right_inv: the two inverses coincide.
• p.radius_right_inv_pos_of_radius_pos: if a power series has a positive radius of convergence, then so does its inverse.

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 $$ q_n = - (p_1)^{-n} \sum_{k=0}^{n-1} \sum_{i_1 + \dotsc + i_k = n} q_k (p_{i_1}, \dotsc, p_{i_k}). $$ Assume for simplicity that we are in dimension 1 and p₁ = 1. In the formula for qₙ, 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 qₙ 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 $$ q_n = - (p_1)^{-1} \sum_{k=2}^n \sum_{i_1 + \dotsc + i_k = n} p_k (q_{i_1}, \dotsc, q_{i_k}). $$ 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 Qₙ = ‖qₙ‖. Bounding ‖pₖ‖ with C r^k gives an inequality $$ Q_n ≤ C' \sum_{k=2}^n r^k \sum_{i_1 + \dotsc + i_k = n} Q_{i_1} \dotsm Q_{i_k}. $$

This formula is not enough to prove by naive induction on n a bound of the form Qₙ ≤ D R^n. However, assuming that the inequality above were an equality, one could get a formula for the generating series of the Qₙ:

$$ \begin{align} Q(z) & := \sum Q_n z^n = Q_1 z + C' \sum_{2 \leq k \leq n} \sum_{i_1 + \dotsc + i_k = n} (r z^{i_1} Q_{i_1}) \dotsm (r z^{i_k} Q_{i_k}) \\ & = Q_1 z + C' \sum_{k = 2}^\infty (\sum_{i_1 \geq 1} r z^{i_1} Q_{i_1}) \dotsm (\sum_{i_k \geq 1} r z^{i_k} Q_{i_k}) \\ & = Q_1 z + C' \sum_{k = 2}^\infty (r Q(z))^k = Q_1 z + C' (r Q(z))^2 / (1 - r Q(z)). \end{align} $$

One can solve this formula explicitly. The solution is analytic in a neighborhood of 0 in ℂ, hence its coefficients grow at most geometrically (by a contour integral argument), and therefore the original Qₙ, 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 [Pos17]. Let $S_n = \sum_{k=1}^n Q_k a^k$ (where a is a positive real parameter to be chosen suitably small). The above computation but with finite sums shows that

$$ S_n \leq Q_1 a + C' \sum_{k=2}^n (r S_{n-1})^k. $$

In particular, $S_n \leq Q_1 a + C' (r S_{n-1})^2 / (1- r S_{n-1})$. Assume that $S_{n-1} \leq K a$, where K > Q₁ is fixed and a is small enough so that r K a ≤ 1/2 (to control the denominator). Then this equation gives a bound $S_n \leq Q_1 a + 2 C' r^2 K^2 a^2$. 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 Sₙ ≤ K a for all n, which gives in particular the fact that aⁿ Qₙ remains bounded.