2.2 Radii Polynomial Approach on Banach Spaces
This section extends the radii polynomial method to infinite-dimensional Banach spaces, allowing for maps between potentially different spaces \(E\) and \(F\).
2.2.1 Banach Space Setup
We work with two Banach spaces \(E\) and \(F\) over \(\mathbb {R}\). For each space \(X \in \{ E, F\} \):
(1) NormedAddCommGroup X: \(X\) has a norm satisfying definiteness, symmetry, triangle inequality
(2) NormedSpace \(\mathbb {R}\) X: The norm is compatible with scalar multiplication
(3) CompleteSpace X: Every Cauchy sequence converges (crucial for fixed point theorems)
This framework supports:
(4) Fréchet derivatives (via the norm structure)
(5) Fixed point theorems (via completeness)
(6) Mean Value Theorem (via the metric structure)
(7) Linear operator theory (via the vector space structure)
2.2.2 Neumann Series and Operator Invertibility
The Neumann series provides a constructive way to show operators close to the identity are invertible.
Let \(E\) be a Banach space and \(B: E \rightarrow E\) a continuous linear operator. If \(\| I_E - B\| {\lt} 1\), then \(B\) is invertible (a unit in the multiplicative sense).
Write \(B = I_E - (I_E - B)\). Since \(\| I_E - B \| {\lt} 1\), the Neumann series \(\sum _{n=0}^{\infty } (I_E - B)^n\) converges absolutely to \((I_E - (I_E - B))^{-1} = B^{-1}\). This is a direct application of Mathlib’s isUnit_one_sub_of_norm_lt_one.
If \(\| I_E - B\| {\lt} 1\) for \(B: E \rightarrow E\), then there exists \(B^{-1}: E \rightarrow E\) such that
If \(\| I_E - B\| {\lt} 1\), then there exists \(B^{-1}\) such that
2.2.3 Newton-Like Operators for E to F Maps
For a function \(f: E \rightarrow F\) and an approximate inverse \(A: F \rightarrow E\), the Newton-like map is
Note that \(T: E \rightarrow E\) even though \(f\) maps between different spaces.
Let \(f: E \rightarrow F\) and \(A: F \rightarrow E\) be injective. Then for the Newton-like operator \(T(x) = x - A(f(x))\):
This holds even when \(E \neq F\); injectivity of \(A\) is sufficient.
2.2.4 Radii Polynomial Definitions
For constants \(Y_0, Z_0, Z_1 \geq 0\) and function \(Z_2: (0,\infty ) \rightarrow [0,\infty )\), the general radii polynomial is
The combined bound is
The general radii polynomial can be rewritten as \(p(r) = (Z(r) - 1)r + Y_0\).
If \(Y_0 \geq 0\), \(r_0 {\gt} 0\), and \(p(r_0) {\lt} 0\) for the general radii polynomial, then \(Z(r_0) {\lt} 1\).
For \(Y_0 \geq 0\) and \(Z: (0,\infty ) \rightarrow [0,\infty )\), the simple radii polynomial is
If \(Y_0 \geq 0\), \(r_0 {\gt} 0\), and \(p(r_0) {\lt} 0\) for the simple radii polynomial, then \(Z(r_0) {\lt} 1\).
2.2.5 Operator Bounds
If \(\| A(f(\bar{x}))\| \leq Y_0\), then for \(T(x) = x - A(f(x))\): \(\| T(\bar{x}) - \bar{x}\| \leq Y_0\).
\(\| T(\bar{x}) - \bar{x}\| = \| (\bar{x} - A(f(\bar{x}))) - \bar{x}\| = \| -A(f(\bar{x}))\| = \| A(f(\bar{x}))\| \leq Y_0\).
For \(T(x) = x - A(f(x))\) where \(f: E \rightarrow F\) is differentiable: \(\text{DT}(x) = I_E - A \circ \text{Df}(x)\).
By the chain rule: \(\text{D}[x \mapsto A(f(x))] = A \circ \text{Df}(x)\). Since \(\text{D}[\text{id}] = I_E\), we have \(\text{DT}(x) = I_E - A \circ \text{Df}(x)\).
Suppose for all \(c \in \overline{B}_r(\bar{x})\):
Then \(\| \text{DT}(c)\| \leq Z_0 + Z_1 + Z_2(r) \cdot r = Z(r)\).
Decompose using \(A^\dagger \):
By triangle inequality: \(\| \text{DT}(c)\| \leq Z_0 + Z_1 + Z_2(r) \cdot r\).
When \(A^\dagger = \text{Df}(\bar{x})\) (so \(Z_1 = 0\)), for all \(c \in \overline{B}_r(\bar{x})\): if \(\| I_E - A \circ \text{Df}(\bar{x})\| \leq Z_0\) and \(\| A \circ (\text{Df}(c) - \text{Df}(\bar{x}))\| \leq Z_2(r) \cdot r\), then \(\| \text{DT}(c)\| \leq Z_0 + Z_2(r) \cdot r\).
2.2.6 Helper Lemmas
In a complete space \(E\), closed balls \(\overline{B}_r(x)\) are complete.
In normed spaces, extended distance is always finite: \(d_{\text{ext}}(x,y) \neq \top \).
If \(A: F \rightarrow E\) is injective and \(\| I_E - A \circ B\| {\lt} 1\) for \(B: E \rightarrow F\), then \(B\) is invertible with inverse \(B^{-1} = (A \circ B)^{-1} \circ A\).
By Lemma 2.2.3, \(A \circ B\) is invertible with inverse \((A \circ B)^{-1}\). Let \(B^{-1} = (A \circ B)^{-1} \circ A\).
Left inverse: \(B(B^{-1}(x)) = B((A \circ B)^{-1}(A(x)))\). Apply \(A\): \(A(B(B^{-1}(x))) = (A \circ B)((A \circ B)^{-1}(A(x))) = A(x)\). By injectivity of \(A\): \(B(B^{-1}(x)) = x\).
Right inverse: \(B^{-1}(B(x)) = (A \circ B)^{-1}(A(B(x))) = (A \circ B)^{-1}((A \circ B)(x)) = x\).
Given \(T: E \rightarrow E\) differentiable with: (a) \(\| T(\bar{x}) - \bar{x}\| \leq Y_0\); (b) \(\| \text{DT}(c)\| \leq Z(r_0)\) for all \(c \in \overline{B}_{r_0}(\bar{x})\); (c) \(Z(r_0) \geq 0\); (d) \(p(r_0) = (Z(r_0) - 1)r_0 + Y_0 {\lt} 0\). Then \(T: \overline{B}_{r_0}(\bar{x}) \rightarrow \overline{B}_{r_0}(\bar{x})\).
From \(p(r_0) {\lt} 0\): \(Z(r_0) \cdot r_0 + Y_0 {\lt} r_0\). The segment \([\bar{x}, x]\) lies in \(\overline{B}_{r_0}(\bar{x})\) by convexity. By Mean Value Theorem: \(\| T(x) - T(\bar{x})\| \leq Z(r_0) \cdot \| x - \bar{x}\| \leq Z(r_0) \cdot r_0\). By triangle inequality:
2.2.7 Main Theorems
Let \(T: E \rightarrow E\) be Fréchet differentiable and \(\bar{x} \in E\). Suppose:
Define \(p(r) := (Z(r) - 1)r + Y_0\).
If there exists \(r_0 {\gt} 0\) such that \(p(r_0) {\lt} 0\), then there exists a unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) such that \(T(\tilde{x}) = \tilde{x}\).
Step 1: From \(p(r_0) {\lt} 0\) and Lemma 2.2.11: \(Z(r_0) {\lt} 1\). Also \(Z(r_0) \geq 0\) since \(\| \text{DT}(\bar{x})\| \geq 0\).
Step 2: By Lemma 2.2.19, \(T: \overline{B}_{r_0}(\bar{x}) \rightarrow \overline{B}_{r_0}(\bar{x})\).
Step 3: \(T\) restricted to \(\overline{B}_{r_0}(\bar{x})\) is a contraction with constant \(Z(r_0) {\lt} 1\). For \(x, y \in \overline{B}_{r_0}(\bar{x})\), by Mean Value Theorem: \(\| T(x) - T(y)\| \leq Z(r_0) \| x - y\| \).
Step 4: The closed ball is complete by Lemma 2.2.16.
Step 5: Apply Theorem 1.1.3 to get unique fixed point.
Let \(E\) and \(F\) be Banach spaces and \(f: E \rightarrow F\) be Fréchet differentiable. Suppose \(\bar{x} \in E\), \(A^\dagger : E \rightarrow F\), and \(A: F \rightarrow E\) with \(A\) injective. Assume:
Define \(p(r) := Z_2(r)r^2 - (1 - Z_0 - Z_1)r + Y_0\).
If there exists \(r_0 {\gt} 0\) such that \(p(r_0) {\lt} 0\), then there exists a unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) with \(f(\tilde{x}) = 0\).
Step 1: Let \(T(x) = x - A(f(x))\). Then \(T: E \rightarrow E\) is differentiable.
Step 2: By Lemma 2.2.12: \(\| T(\bar{x}) - \bar{x}\| \leq Y_0\). By Lemma 2.2.14: \(\| \text{DT}(c)\| \leq Z(r_0)\) for \(c \in \overline{B}_{r_0}(\bar{x})\). By Lemma 2.2.8: \(p(r_0) = (Z(r_0) - 1)r_0 + Y_0 {\lt} 0\).
Step 3: Apply Theorem 2.2.20: unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) with \(T(\tilde{x}) = \tilde{x}\).
Step 4: By Proposition 2.2.5 with injectivity of \(A\): \(f(\tilde{x}) = 0\).
Given \(f: E \rightarrow F\) Fréchet differentiable and injective \(A: F \rightarrow E\) satisfying \(\| A(f(\bar{x}))\| \leq Y_0\), \(\| I_E - A \circ \text{Df}(\bar{x})\| \leq Z_0\), and \(\| A \circ [\text{Df}(c) - \text{Df}(\bar{x})]\| \leq Z_2(r) \cdot r\) for \(c \in \overline{B}_r(\bar{x})\). If \(p(r_0) = Z_2(r_0)r_0^2 - (1-Z_0)r_0 + Y_0 {\lt} 0\), then there exists unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) with \(f(\tilde{x}) = 0\).
Consider \(f: E \rightarrow E\) Fréchet differentiable, \(\bar{x} \in E\), and \(A: E \rightarrow E\) injective. Assume \(\| A(f(\bar{x}))\| \leq Y_0\), \(\| I_E - A \circ \text{Df}(\bar{x})\| \leq Z_0\), and \(\| A \circ [\text{Df}(c) - \text{Df}(\bar{x})]\| \leq Z_2(r) \cdot r\) for \(c \in \overline{B}_r(\bar{x})\). Define \(p(r) := Z_2(r)r^2 - (1-Z_0)r + Y_0\).
If \(p(r_0) {\lt} 0\), then there exists unique \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) with \(f(\tilde{x}) = 0\) and \(\text{Df}(\tilde{x})\) invertible.
Steps 1–4: As in Theorem 2.2.21, get fixed point \(\tilde{x}\) with \(f(\tilde{x}) = 0\).
Step 5 (Invertibility): Since \(\tilde{x} \in \overline{B}_{r_0}(\bar{x})\) and \(Z(r_0) {\lt} 1\) (by Lemma 1.4.5), \(\| I_E - A \circ \text{Df}(\tilde{x})\| \leq Z(r_0) {\lt} 1\). Apply Lemma 2.2.18: \(\text{Df}(\tilde{x})\) is invertible.