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The method of continuity for a uniformly estimated affine family of bounded operators

Statement

Assume Countable Choice. Let $X, Y$ be Banach spaces over the same field, let $L_{0}, L_{1} \in B (X, Y)$ and put $L_{t} : = (1 - t) L_{0} + t L_{1}$ for $t \in [0, 1]$ . Assume (i) $L_{0}$ is bijective; (ii) there is $0 \leq C < \infty$ with the uniform a priori estimate $∥ x ∥_{X} \leq C ∥ L_{t} x ∥_{Y}$ for every $t \in [0, 1]$ and every $x \in X$ . Then $L_{t}$ is bijective for every $t \in [0, 1]$ , and $∥ L_{t}^{- 1} ∥_{Y \to X} \leq C$ for every $t$ . No compactness or reflexivity hypothesis is used: the uniform estimate alone makes the bijectivity set closed, and the Neumann series makes it open.

Facts & Assumptions

Given: ${A C}_{ω}$ , Banach spaces $X, Y$ over the same field, operators $L_{0}, L_{1} \in B (X, Y)$ , the affine family $L_{t} = (1 - t) L_{0} + t L_{1}$ , and the hypotheses (i) $L_{0}$ bijective, (ii) $0 \leq C < \infty$ and $∥ x ∥ \leq C ∥ L_{t} x ∥$ for all $t \in [0, 1]$ , all $x \in X$ .

[A1]

The only choice assumption is Countable Choice ${A C}_{ω}$ , used through the sequential completeness conventions of the Banach spaces. No full Axiom of Choice is used. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

A Banach space is a normed space whose norm metric is complete, so every Cauchy sequence converges; limits in a metric space are unique. (Banach space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ )

[F2]

$B (X, Y)$ consists of the bounded linear maps $X \to Y$ , with pointwise operations, and $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ ; the operator norm is subadditive and homogeneous, so $∥ (1 - t) L_{0} + t L_{1} ∥ \leq (1 - t) ∥ L_{0} ∥ + t ∥ L_{1} ∥$ for $t \in [0, 1]$ , and for $T \in B (X, Y)$ , $S \in B (Y, Z)$ one has $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ . (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces, Composition satisfies |ST|\le|S|,|T|)

[F3]

If $X$ is a Banach space and $R \in B (X)$ with $∥ R ∥ < 1$ , then $I + R$ is invertible with inverse $\sum_{n \geq 0} (- R)^{n}$ and $∥ (I + R)^{- 1} ∥ \leq (1 - ∥ R ∥)^{- 1}$ ; if $A \in B (X, Y)$ is invertible and $E \in B (X, Y)$ satisfies $∥ A^{- 1} E ∥ < 1$ , then $A + E$ is invertible. (Neumann series and small perturbations of bounded inverses)

Proof

technique · direct

1.1givenF2algebra

Injectivity and uniform lower bound. Fix $t \in [0, 1]$ . If $L_{t} x = 0$ then [ii] gives $∥ x ∥ \leq C ∥ L_{t} x ∥ = 0$ , so $x = 0$ : every $L_{t}$ is injective. Moreover [ii] says exactly that $∥ L_{t}^{- 1} y ∥ \leq C ∥ y ∥$ for every $y$ in the range $L_{t} (X)$ , so whenever $L_{t}$ is surjective its inverse is bounded with norm at most $C$ .

2.1step 1.1F1F2algebra

The bijectivity set is closed in $[0, 1]$ . Let $t_{j} \to t$ in $[0, 1]$ with every $L_{t_{j}}$ bijective, let $f \in Y$ and put $u_{j} : = L_{t_{j}}^{- 1} f$ . Then $∥ u_{j} ∥ \leq C ∥ f ∥$ by step 1.1, and for all $j, k$ the identity $L_{t_{k}} (u_{j} - u_{k}) = L_{t_{k}} u_{j} - f = (L_{t_{k}} - L_{t_{j}}) u_{j} = (t_{k} - t_{j}) (L_{1} - L_{0}) u_{j}$ together with [ii] and [F2] gives $∥ u_{j} - u_{k} ∥ \leq C ∥ L_{t_{k}} (u_{j} - u_{k}) ∥ \leq C^{2} ∣ t_{k} - t_{j} ∣ ∥ L_{1} - L_{0} ∥ ∥ f ∥$ , so $(u_{j})$ is Cauchy in $X$ ; by [F1] it converges to some $u \in X$ . Since $∥ L_{t} - L_{t_{j}} ∥ \leq ∣ t - t_{j} ∣ ∥ L_{1} - L_{0} ∥$ by [F2] and the sequence $(∥ u_{j} ∥)$ is bounded by $C ∥ f ∥$ , $∥ L_{t} u - f ∥ \leq ∥ (L_{t} - L_{t_{j}}) u ∥ + ∥ L_{t_{j}} (u - u_{j}) ∥ + ∥ L_{t_{j}} u_{j} - f ∥ \leq ∣ t - t_{j} ∣ ∥ L_{1} - L_{0} ∥ ∥ u ∥ + ((1 - t_{j}) ∥ L_{0} ∥ + t_{j} ∥ L_{1} ∥) ∥ u - u_{j} ∥$ , and both terms tend to $0$ ; hence $L_{t} u = f$ . So $L_{t}$ is surjective, injective by step 1.1, and therefore bijective with $∥ L_{t}^{- 1} ∥ \leq C$ by step 1.1. This shows that a limit of bijective parameters is bijective, that is, the bijectivity set $I : = {t \in [0, 1] : L_{t} bijective}$ is closed in $[0, 1]$ .

2.2step 1.1F2F3casesalgebra

The bijectivity set is open in $[0, 1]$ . Let $t \in I$ . If $C = 0$ , the estimate implies $X = {0}$ , and bijectivity of $L_{0}$ implies $Y = {0}$ , so every $L_{s}$ is the unique bijection and $I = [0, 1]$ . Assume $C > 0$ . If $L_{1} = L_{0}$ then $L_{s} = L_{t}$ for every $s$ and the claim is trivial, so assume $∥ L_{1} - L_{0} ∥ > 0$ and let $s \in [0, 1]$ satisfy $∣ s - t ∣ < 1 / (C ∥ L_{1} - L_{0} ∥)$ . Write $L_{s} = L_{t} + (s - t) (L_{1} - L_{0}) = L_{t} (I + L_{t}^{- 1} (s - t) (L_{1} - L_{0}))$ , where $L_{t}^{- 1} \in B (Y, X)$ has norm at most $C$ by step 1.1. Since $∥ L_{t}^{- 1} (s - t) (L_{1} - L_{0}) ∥ \leq C ∣ s - t ∣ ∥ L_{1} - L_{0} ∥ < 1$ , the Neumann series [F3] makes $I + L_{t}^{- 1} (s - t) (L_{1} - L_{0})$ invertible on $X$ with inverse in $B (X)$ ; composing with the bijection $L_{t}$ shows that $L_{s}$ is bijective, with inverse $(I + L_{t}^{- 1} (s - t) (L_{1} - L_{0}))^{- 1} L_{t}^{- 1} \in B (Y, X)$ and norm at most $C (1 - C ∣ s - t ∣ ∥ L_{1} - L_{0} ∥)^{- 1}$ . Hence $I$ is open in $[0, 1]$ .

3.1step 1.1step 2.1step 2.2A1given∎

Conclusion. $I$ is nonempty because $0 \in I$ by (i), and it is open and closed in $[0, 1]$ by steps 2.1 and 2.2. Suppose $I \neq [0, 1]$ , and let $t : = \sup {y \in [0, 1] : [0, y] \subseteq I}$ , a set that contains $0$ and is nonempty. For $y < t$ one has $[0, y] \subseteq I$ , and closedness of $I$ gives $t \in I$ (if $t = 0$ , use $0 \in I$ ). If $t = 1$ this already gives $I = [0, 1]$ , a contradiction; so $t < 1$ ; openness of $I$ then gives $0 < δ \leq 1 - t$ with $(t - δ, t + δ) \cap [0, 1] \subseteq I$ , so $[0, t + δ / 2] \subseteq I$ , contradicting the definition of $t$ . Hence $I = [0, 1]$ : every $L_{t}$ is bijective, and $∥ L_{t}^{- 1} ∥ \leq C$ for every $t$ by step 1.1. No compactness, reflexivity or separability of $X$ or $Y$ was used anywhere; the only completeness used is that of $X$ in step 2.1 and the only choice principle is the sequential convention of [A1].

Remarks

If the uniform estimate [ii] holds only for $t$ in a subset $A \subseteq [0, 1]$ , the argument shows that the bijectivity set is relatively open and relatively closed in $A$ ; the interval $[0, 1]$ is used only to run the endpoint propagation in step 3.1.
The uniform lower bound controls the inverses and the Cauchy sequence in the closedness proof; both openness and closedness also use that $t \mapsto L_{t}$ is affine and hence Lipschitz with constant $∥ L_{1} - L_{0} ∥$ .

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Sources

Armin Schikorra, Partial Differential Equations I & II (version October 1, 2025; complete 281-page graduate lecture notes) (standard reference, not scraped)
Leon Simon, Lectures on Partial Differential Equations (Stanford University; complete 118-page author notes, Chapter 12 Schauder Theory) (standard reference, not scraped)
John Villavert, Elementary Theory and Methods for Elliptic Partial Differential Equations (2017; complete 220-page lecture notes) (standard reference, not scraped)