Theorem: Principle of Small Perturbations

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\GF$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a basic sequence with basis constant $K_b$.
Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $X$ with:
:$\ds 2 K_b \sum_{n \mathop = 1}^\infty \frac {\norm {x_n - y_n} } {\norm {x_n} } = \theta < 1$
Then $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ are congruent.
Further, $\sequence {y_n}_{n \mathop \in \N}$ is a basic sequence with basis constant $K_b' \le K_b \paren {1 + \theta} \paren {1 - \theta}^{-1}$.

Proof:
Let $Z = \sqbrk {x_n}_{n \mathop \in \N}$ be the closed linear span of $\sequence {x_n}_{n \mathop \in \N}$.
Let $\struct {Z^\ast, \norm {\, \cdot \,}_{Z^\ast} }$ be the normed dual space of $Z$.
Let $\map \BB Z$ be the space of bounded linear transformations on $Z$.
Let $\norm {\, \cdot \,}_{\map \BB Z}$ be the norm on $\map \BB Z$. 
Let $\sequence {x_n^\ast}_{n \mathop \in \N}$ be the coordinate functionals associated with $\sequence {x_n}_{n \mathop \in \N}$.
Let $\sequence {S_n}_{n \mathop \in \N}$ be the projections associated with $\sequence {e_n}_{n \mathop \in \N}$.
We have:
:$\map {x_n^\ast} x x_n = \map {S_n} x - \map {S_{n - 1} } x$ for each $x \in Z$ and $n \in \N$.
Hence we have, for $x \in Z$:
{{begin-eqn}}
{{eqn	| l = \cmod {\map {x_n^\ast} x} \norm {x_n}_X
		| r = \norm {\map {x_n^\ast} x x_n}_X 
		| c = {{NormAxiomVector|2}}
}}
{{eqn	| r = \norm {\map {S_n} x - \map {S_{n - 1} } x}_X 
}}
{{eqn	| o = \le
		| r = \norm {S_n - S_{n - 1} }_{\map \BB Z} \norm x_X 
		| c = Fundamental Property of Norm on Bounded Linear Transformation
}}
{{eqn	| o = \le
		| r = \paren {\norm {S_n}_{\map \BB Z} + \norm {S_{n - 1} }_{\map \BB Z} } \norm x_X 
		| c = {{NormAxiomVector|3}}
}}
{{eqn	| o = \le
		| r = 2 K_b \norm x_X 
		| c = {{Defof|Basis Constant}}
}}
{{end-eqn}}
We also have:
:$\map {x_1^\ast} x x_1 = \map {S_1} x$
and hence:
:$\cmod {\map {x_1^\ast} x} \norm {x_1}_X \le K_b \norm x_X \le 2 K_b \norm x_X$
Hence we have:
:$\norm {x_n^\ast}_{Z^\ast} \le \dfrac {2 K_b} {\norm {x_n}_X}$
by the definition of the norm on $Z^\ast$.
From Hahn-Banach Theorem: Complex Vector Space: Corollary, there exists an extension $\widehat {x_n^\ast}$ on $x_n^\ast$ to the whole of $X$ such that:
:$\norm {\widehat {x_n^\ast} }_{X^\ast} \le \dfrac {2 K_b} {\norm {x_n}_X}$
For each $x \in X$, we have:
{{begin-eqn}}
{{eqn	| l = \sum_{n \mathop = 1}^\infty \norm {\map {\widehat {x_n^\ast} } x \paren {y_n - x_n} }_X
		| r = \sum_{n \mathop = 1}^\infty \cmod {\map {\widehat {x_n^\ast} } x} \norm {y_n - x_n}_X
}}
{{eqn	| o = \le
		| r = \norm x \sum_{n \mathop = 1}^\infty \paren {\frac {2 K_b} {\norm {x_n}_X} } \norm {y_n - x_n}_X
		| c = Fundamental Property of Norm on Bounded Linear Functional
}}
{{eqn	| r = \theta \norm x
		| c = by hypothesis
}}
{{eqn	| o = <
		| r = \infty 
}}
{{end-eqn}}
Since $\struct {X, \norm {\, \cdot \,}_X}$ is a Banach space, we have that:
:$\ds \sum_{n \mathop = 1}^\infty \map {\widehat {x_n^\ast} } x \paren {y_n - x_n}$ converges in $\struct {X, \norm {\, \cdot \,}_X}$ for each $x \in X$
from Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach.
Define $T : X \to X$ by:
:$\ds T x = x + \sum_{n \mathop = 1}^\infty \map {\widehat {x_n^\ast} } x \paren {y_n - x_n}$ for each $x \in X$.
Note that since $\map {x_n^\ast} {x_j} = \delta_{n j}$ for each $n, j \in \N$ we have:
{{begin-eqn}}
{{eqn	| l = T x_j
		| r = x_j + \sum_{n \mathop = 1}^\infty \map {\widehat {x_n^\ast} } {x_j} \paren {y_n - x_n}
}}
{{eqn	| r = x_j + \sum_{n \mathop = 1}^\infty \map {x_n^\ast} {x_j} \paren {y_n - x_n}
		| c = $\widehat {x_n^\ast}$ extends $x_n^\ast$
}}
{{eqn	| r = x_j + \sum_{n \mathop = 1}^\infty \delta_{n j} \paren {y_n - x_n}
}}
{{eqn	| r = x_j + \paren {y_j - x_j}
}}
{{eqn	| r = y_j 
}}
{{end-eqn}}
We have further already established that:
:$\ds \sum_{n \mathop = 1}^\infty \norm {\map {\widehat {x_n^\ast} } x \paren {y_n - x_n} }_X \le \theta \norm x$
hence:
{{begin-eqn}}
{{eqn	| l = \norm {T x}_X
		| r = \norm x_X + \norm {\sum_{n \mathop = 1}^\infty \map {\widehat {x_n^\ast} } x \paren {y_n - x_n} }_X
}}
{{eqn	| o = \le
		| r = \norm x_X + \sum_{n \mathop = 1}^\infty \norm {\map {\widehat {x_n^\ast} } x \paren {y_n - x_n} }_X
		| c = {{NormAxiomVector|3}}, Modulus of Limit: Normed Vector Space
}}
{{eqn	| o = \le
		| r = \paren {1 + \theta} \norm x_X 
}}
{{end-eqn}}
Let $\map \BB X$ be the space of bounded linear transformations on $X$.
Let $\norm {\, \cdot \,}_{\map \BB X}$ be the norm on $\map \BB X$. 
Since:
:$\norm {T x}_X \le \paren {1 + \theta} \norm x_X$ for each $x \in X$
we have:
:$\norm T_{\map \BB X} \le 1 + \theta$
Let $I_X : X \to X$ be the identity mapping.
We have:
:$\ds \norm {T x - x} = \norm {\sum_{n \mathop = 1}^\infty \map {\widehat {x_n^\ast} } x \paren {y_n - x_n} }_X \le \theta$
Hence:
:$\norm {T - I}_{\map \BB X} \le \theta$
From Invertibility of Identity Minus Operator, $T$ is an invertible bounded linear transformation with:
:$\norm {T^{-1} }_{\map \BB X} \le \paren {1 - \theta}^{-1}$
Hence $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ are congruent.
Hence, from Sequence in Banach Space Congruent to Basic Sequence is Basic, $\sequence {y_n}_{n \mathop \in \N}$ is a basic sequence with basis constant $K_b' \le K_b \norm T_{\map \BB X} \norm {T^{-1} }_{\map \BB X} \le K_b \paren {1 + \theta} \paren {1 - \theta}^{-1}$.
{{qed}}