Theorem: Bessaga-Pełczyński Selection Principle

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\GF$.
Let $\sequence {e_n}_{n \mathop \in \N}$ be a Schauder basis for $\struct {X, \norm {\, \cdot \,} }$ with basis constant $K_b$. 
Let $\sequence {e_n^\ast}_{n \mathop \in \N}$ be the coordinate functionals associated with $\sequence {e_n}_{n \mathop \in \N}$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ such that:
:$\ds \inf_n \norm {x_n} > 0$
and:
:$\map {e_k^\ast} {x_n} \to 0$ as $n \to \infty$ for each $k \in \N$.
Then $\sequence {x_n}_{n \mathop \in \N}$ contains a subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ that is congruent to a block basic sequence $\sequence {y_k}_{k \mathop \in \N}$ of $\sequence {e_n}_{n \mathop \in \N}$.
Further, for each $\epsilon > 0$, this subsequence to have basis constant at most $K_b + \epsilon$.

Proof:
Let:
:$\ds \alpha = \inf_n \norm {x_n} > 0$
Let $\nu \in \openint 0 {1/4}$.
Take $n_1 = 1$ and $r_0 = 0$.
Let $\sequence {S_n}_{n \mathop \in \N}$ be the projections associated with $\sequence {e_n}_{n \mathop \in \N}$.
From Projections Associated with Schauder Basis Converge to Identity in Strong Operator Topology: Corollary, we have $\map {S_n} {x_{n_1} } \to x_{n_1}$.
Hence we can take $r_1 > r_0$ such that:
:$\ds \norm {x_{n_1} - \map {S_{r_1} } {x_{n_1} } } < \frac {\nu \alpha} {2 K_b}$
We have that:
:$\ds \map {S_{r_1} } x = \sum_{j \mathop = 1}^{r_1} \map {e_j^\ast} x e_j$ for each $x \in X$.
Since:
:$\map {e_j^\ast} {x_n} \to 0$ as $n \to \infty$ for each $1 \le j \le r_1$
we have that:
:$\map {S_{r_1} } {x_n} \to {\mathbf 0}_X$ as $n \to \infty$.
from Linear Combination of Convergent Sequences in Topological Vector Space is Convergent and Multiple of Vector in Topological Vector Space Converges.
Hence we can take $n_2 > n_1$ such that:
:$\ds \norm {S_{r_1} x_{n_2} } < \frac {\nu^2 \alpha} {2 K_b}$
Again, we have $S_r x_{n_2} \to x_{n_2}$ as $r \to \infty$, hence we can take $r_2 > r_1$ such that:
:$\ds \norm {x_{n_2} - S_{r_2} x_{n_2} } < \frac {\nu^2 \alpha} {2 K_b}$
Continuing this way we obtain strictly increasing sequences $\sequence {n_k}_{k \mathop \in \N}$ and $\sequence {r_k}_{k \mathop \in \N}$ such that:
:$\ds \norm {S_{r_{k - 1} } x_{n_k} } < \frac {\nu^k \alpha} {2 K_b}$
and:
:$\ds \norm {x_{n_k} - S_{r_k} x_{n_k} } < \frac {\nu^k \alpha} {2 K_b}$
For each $k \in \N$, let:
:$y_k = \map {S_{r_k} } {x_{n_k} } - \map {S_{r_{k - 1} } } {x_{n_k} }$
Hence we have:
:$\ds y_k = \sum_{j \mathop = r_{k - 1} + 1}^{r_k} \map {e_j^\ast} {x_{n_k} } e_j$
Hence $\sequence {y_k}_{k \mathop \in \N}$ is a block basic sequence of $\sequence {e_j}_{j \mathop \in \N}$.
From Block Basic Sequence is Basic Sequence, $\sequence {y_k}_{k \mathop \in \N}$ is a basic sequence.
For each $k \in \N$, we have:
{{begin-eqn}}
{{eqn	| l = \norm {y_k - x_{n_k} } 
		| o = \le
		| r = \norm {S_{r_k} x_{n_k} - S_{r_{k - 1} } x_{n_k} - x_{n_k} }
}}
{{eqn	| o = \le
		| r = \norm {S_{r_{k - 1} } x_{n_k} } + \norm {x_{n_k} - S_{r_k} x_{n_k} }
		| c = {{NormAxiomVector|3}}
}}
{{eqn	| o = <
		| r = 2 \times \frac {\nu^k} {2 K_b}
}}
{{eqn	| r = \frac {\nu^k} {K_b}
}}
{{end-eqn}}
By the  Reverse Triangle Inequality: Normed Vector Space, we have:
:$\norm {y_k} \ge \norm {x_{n_k} } - \norm {y_k - x_{n_k}  }$
We have:
:$\norm {x_{n_k} } \ge \alpha$
and:
:$\ds \norm {y_k - x_{n_k}  } < \frac {\nu^k \alpha} {K_b}$
Since $\nu \in \openint 0 1$, we have $\nu^k \le \nu$, and hence:
:$\ds \norm {y_k - x_{n_k} } < \frac {\nu \alpha} {K_b}$
So:
:$\ds \norm {y_k} \ge \alpha - \frac {\nu \alpha} {K_b} = \paren {1 - \frac {\nu} {K_b} } \alpha$
From Basis Constant is Greater Than or Equal to One, we have:
:$\dfrac {\nu} {K_b} \le \nu$
so:
:$1 - \dfrac \nu {K_b} \ge 1 - \nu$ 
and hence:
:$\ds \paren {1 - \frac {\nu} {K_b} } \alpha \ge \paren {1 - \nu} \alpha$
We now have:
:$\ds \frac {\norm {y_k - x_{n_k} } } {\norm {y_k} } < \frac {\frac {\nu^k \alpha} {K_b} } {\paren {1 - \nu} \alpha} = \frac {\nu^k} {1 - \nu} \times \paren {K_b}^{-1}$
From Sum of Infinite Geometric Sequence: Corollary 1, we have:
:$\ds \sum_{k \mathop = 1}^\infty \nu^k = \frac \nu {1 - \nu}$
We then have:
:$\ds 2 K_b \sum_{k \mathop = 1}^\infty \frac {\norm {y_k - x_{n_k} } } {\norm {y_k} } < 2 K_b \sum_{k \mathop = 1}^\infty \frac {\nu^k} {1 - \nu} \times \paren {K_b}^{-1} = 2 \nu \paren {1 - \nu}^{-2}$
For brevity write $c_\nu = \nu \paren {1 - \nu}^{-2}$.
By the Principle of Small Perturbations, $\sequence {x_{n_k} }_{k \mathop \in \N}$ is also a basic sequence with basis constant $\widetilde {K_b} \le K_b \paren {1 + c_\nu} \paren {1 - c_\nu}^{-1}$
As $\nu \to 0^+$, we have $\nu \paren {1 - \nu}^{-2} \to 0$ and so:
:$\paren {1 + c_\nu} \paren {1 - c_\nu}^{-1} \to 1$ as $\nu \to 0^+$.
Let $\epsilon > 0$.
We can take $N > 4$ such that $\paren {1 + c_\nu} \paren {1 - c_\nu}^{-1} < \dfrac {K_b + \epsilon} {K_b}$ with $\nu = N^{-1}$.
Constructing the subsequence corresponding to this $\nu$, we have $\widetilde {K_b} < K_b + \epsilon$.
{{qed}}
{{Namedfor|Czeslaw Bessaga|name2 = Aleksander Pełczyński|cat = Bessaga|cat2 = Pełczyński}}