Theorem: Sufficient Condition for Perturbation of Basic Sequence to be Basic Sequence

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual of $\struct {X, \norm {\, \cdot \,}_X}$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a basic sequence such that there exists $f \in X^\ast$ with:
:$\map f {x_n} = 0$ for each $n \in \N$.
Let $\sqbrk {x_n}_{n \mathop \in \N}$ be the closed linear span of $\sequence {x_n}_{n \mathop \in \N}$.
Let $u \not \in \sqbrk {x_n}_{n \mathop \in \N}$.
Then $\sequence {x_n + u}_{n \mathop \in \N}$ is a basic sequence.

Proof:
Let $I_X : X \to X$ be the identity mapping.
From Existence of Distance Functional, there exists $\widetilde g \in X^\ast$ such that:
:$\norm {\widetilde g}_{X^\ast} = 1$
:$\ds \map {\widetilde g} u = \map {\operatorname {dist} } {u, \sqbrk {x_n}_{n \mathop \in \N} } := d$
:$\map {\widetilde g} x = 0$ for all $x \in \sqbrk {x_n}_{n \mathop \in \N}$.
By rescaling:
:$\ds g = -\frac {\map f u} d \widetilde g$
we obtain $g \in X^\ast$ such that:
:$\map g u = -\map f u$
and:
:$\map g x = 0$ for all $x \in \sqbrk {x_n}_{n \mathop \in \N}$.
Define $x^\ast : X \to \GF$ by:
:$\map {x^\ast} x = \map f x + \map g x$ for each $x \in X$.
From Normed Dual Space is Banach Space, $x^\ast \in X^\ast$. 
Further, we have $\map {x^\ast} {x_n} = 1$ for each $n \in \N$ and $\map {x^\ast} u = 0$.
Define $T : X \to X$ by:
:$T x = \map {x^\ast} x u$ for all $x \in X$.
We have that:
{{begin-eqn}}
{{eqn	| l = \norm {T x}_X
		| r = \norm {\map {x^\ast} x u}_X 
}}
{{eqn	| o = \le
		| r = \cmod {\map {x^\ast} x} \norm u_X
		| c = {{NormAxiomVector|1}}
}}
{{eqn	| o = \le 
		| r = \paren {\norm {x^\ast}_{X^\ast} \norm u_X} \norm x_X
		| c = Fundamental Property of Norm on Bounded Linear Functional
}}
{{end-eqn}}
Hence $T$ is a bounded linear transformation.
Hence $I_X + T$ and $I_X - T$ are bounded linear transformations from Space of Bounded Linear Transformations forms Vector Space.
We can then calculate, for each $x \in X$:
{{begin-eqn}}
{{eqn	| l = \paren {I_X + T} \paren {I_X - T} x
		| r = \map {\paren {I_X + T} } {x - \map {x^\ast} x u}
}}
{{eqn	| r = \paren {x - \map {x^\ast} x u} + \map {x^\ast} {x - \map {x^\ast} x u} u
}}
{{eqn	| r = \paren {x - \map {x^\ast} x u} + \map {x^\ast} x u - \map {x^\ast} x \map {x^\ast} u u
		| c = $x^\ast$ is linear
}}
{{eqn	| r = x - \map {x^\ast} x u + \map {x^\ast} x u
		| c = $\map {x^\ast} u = 0$
}}
{{eqn	| r = x
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn	| l = \paren {I_X - T} \paren {I_X + T} x
		| r = \map {\paren {I_X - T} } {x + \map {x^\ast} x u}
}}
{{eqn	| r = \paren {x + \map {x^\ast} x u} - \map {x^\ast} {x + \map {x^\ast} x u} u
}}
{{eqn	| r = \paren {x + \map {x^\ast} x u} - \map {x^\ast} x u - \map {x^\ast} x \map {x^\ast} x \map {x^\ast} u u
		| c = $x^\ast$ is linear
}}
{{eqn	| r = \paren {x + \map {x^\ast} x u} - \map {x^\ast} x u 
		| c = $\map {x^\ast} u = 0$
}}
{{eqn	| r = x
}}
{{end-eqn}}
Hence $I_X + T$ is an invertible bounded linear transformation with inverse $I_X - T$.
We have:
:$\map {\paren {I_X + T} } {x_n} = x_n + \map f {x_n} u = x_n + u$ for each $n \in \N$.
Hence $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {x_n + u}_{n \mathop \in \N}$ are congruent. 
Recall that $\sequence {x_n}_{n \mathop \in \N}$ is basic.
Hence from Sequence in Banach Space Congruent to Basic Sequence is Basic, $\sequence {x_n + u}_{n \mathop \in \N}$ is basic.
{{qed}}