Theorem: Construction of Schauder Basis from Bounded Projections

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be an infinite-dimensional normed vector space over $\GF$.
Let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of bounded linear transformations such that:
:$(1) \quad$ $\dim S_n \sqbrk X = n$ for all $n \in \N$
:$(2) \quad$ $S_n S_m = S_m S_n = S_{\min \set {m, n} }$ for each $m, n \in \N$
:$(3) \quad$ $\map {S_n} x \to x$ for each $x \in X$. 
Let $e_1 \in S_1 \sqbrk X \setminus \set { {\mathbf 0}_X}$. 
For each $n \ge 2$, let $e_n \in \paren {S_n \sqbrk X \cap S_{n - 1}^{-1} \sqbrk {\set { {\mathbf 0}_X} } } \setminus \set { {\mathbf 0}_X}$.
Then $\sequence {e_n}_{n \mathop \in \N}$ is a Schauder basis for $X$. 
Further, $\sequence {S_n}_{n \mathop \in \N}$ are the projections associated with $\sequence {e_n}_{n \mathop \in \N}$.

Proof:
Let $\BB X$ be the space of bounded linear transformations on $X$.
Let $\norm {\, \cdot \,}_{\map \BB X}$ be the norm on $\map \BB X$. 
From Scalar Multiple of Schauder Basis is Schauder Basis, $\sequence {e_n}_{n \mathop \in \N}$ is a Schauder basis {{iff}} $\ds \sequence {\frac {e_n} {\norm {e_n} } }_{n \mathop \in \N}$ is a Schauder basis.
{{WLOG}} we can therefore assume that $\norm {e_n} = 1$ for each $n \in \N$. 
Our goal will be to use Condition for Sequence to be Schauder Basis in terms of Coordinate Functionals.
We must therefore find bounded linear functionals $\sequence {e_n^\ast}_{n \mathop \in \N}$ on $X$ such that:
:$(1') \quad$ $\map {e_n^\ast} {e_j} = \delta_{i j}$ for each $i, j \in \N$
:$(2') \quad$ $\ds x = \sum_{n \mathop = 1}^\infty \map {e_n^\ast} x e_n$ for each $x \in X$
where $\delta$ denotes the Kronecker $\delta$. 
Note that since:
:$\dim S_1 \sqbrk X = 1$
by property $(1)$, we have that:
:$S_1 \sqbrk X = \map \span {e_1}$
Hence we can define a function $e_1^\ast : X \to \C$ by:
:$\map {e_1^\ast} x e_1 = \map {S_1} x$ for each $x \in X$.
We first show that $e_1^\ast$ is linear.
Let $x, y \in X$ and $\lambda \in \GF$.
Then:
{{begin-eqn}}
{{eqn	| l = \map {e_1^\ast} {x + \lambda y} e_1
		| r = \map {S_1} {x + \lambda y}
}}
{{eqn	| r = \map {S_1} x + \lambda \map {S_1} y
}}
{{eqn	| r = \map {e_1^\ast} x e_1 + \lambda \map {e_1^\ast} y e_1
}}
{{eqn	| r = \paren {\map {e_1^\ast} x + \lambda \map {e_1^\ast} y} e_1
}}
{{end-eqn}}
Since $e_1 \ne {\mathbf 0}_X$, we have:
:$\map {e_1^\ast} {x + \lambda y} = \map {e_1^\ast} x + \lambda \map {e_1^\ast} y$ 
Hence $e_1^\ast$ is indeed a linear functional.
We now show that $e_1^\ast$ is bounded.
Since $\norm {e_1} = 1$, we have:
:$\cmod {\map {e_1^\ast} x} = \norm {\map {e_1^\ast} x e_1} = \norm {\map {S_1} x}$ 
From Fundamental Property of Norm on Bounded Linear Transformation we then have:
:$\cmod {\map {e_1^\ast} x} \le \norm {S_1}_{\map \BB X} \norm x$
Hence $e_1^\ast$ is bounded.
Finally, since $e_1 \in S_1 \sqbrk X$, there exists $u_1 \in X$ such that $\map {S_1} {u_1} = e_1$. 
Then we have $\map {S_1} {e_1} = \map {S_1^2} {u_1} = \map {S_1} {u_1} = e_1$ by property $(2)$. 
Now we have:
:$\dim S_2 \sqbrk X = 2$
by property $(1)$.
Recalling that:
:$\map {S_1} {e_1} = e_1$
we have:
:$\map {S_2} {e_1} = \map {S_1} {e_1} = e_1$
by property $(2)$. 
From Linear Functional on Vector Space is Zero or Surjective, we therefore have that $e_2$ and $e_1$ are linearly independent vectors in the two-dimensional space $S_2 \sqbrk X$.
Hence by Linearly Independent Set is Contained in some Basis: Finite Dimensional Case, $S_2 \sqbrk X = \map \span {\set {e_1, e_2} }$.
We can therefore write:
:$\map {S_2} x = \map {\alpha_1} x e_1 + \map {\alpha_2} x e_2$ for each $x \in X$
for some $\alpha_1, \alpha_2$ to be understood.
Applying $S_1$ to this equation, we have:
:$\map {S_1} x = \map {\alpha_1} x \map {S_1} {e_1} + \map {\alpha_2} x \map {S_1} {e_2}$ for each $x \in X$ 
by property $(2)$, since $\min \set {1, 2} = 1$.
Again by the choice of $e_2$, we have $\map {S_1} {e_2} = 0$. 
By the definition of $S_1$ we have $\map {S_1} {e_1} = e_1$. 
Hence we have $\map {S_1} x = \map {\alpha_1} x$ and we obtain:
:$\map {S_2} x - \map {S_1} x = \map {\alpha_2} x e_2$
At this point, we have seen that $\map {e_1^\ast} {e_i} = \delta_{1 i}$ for $i \in \set {1, 2}$. 
We define $e_2^\ast = \alpha_2$.
Finally, for $x, y \in X$ and $\lambda \in \GF$, we have:
{{begin-eqn}}
{{eqn	| l = \map {e_2^\ast} {x + \lambda y} e_2
		| r = \map {S_2} {x + \lambda y} - \map {S_1} {x + \lambda y}
}}
{{eqn	| r = \paren {\map {S_2} x + \lambda \map {S_2} y} - \paren {\map {S_1} x + \lambda \map {S_1} y}
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = \paren {\map {S_2} x - \map {S_1} x} + \lambda \paren {\map {S_2} y - \map {S_1} y}
}}
{{eqn	| r = \paren {\map {e_2^\ast} x + \lambda \map {e_2^\ast} y} e_2
}}
{{end-eqn}}
Since $e_2$, we have:
:$\map {e_2^\ast} {x + \lambda y} = \map {e_2^\ast} x + \lambda \map {e_2^\ast} y$
Hence $e_2^\ast$ is a linear functional.
Further, we have:
:$\map {e_2^\ast} {e_1} e_2 = \map {S_2} {e_1} - \map {S_1} {e_1} = {\mathbf 0}_X$
Since $e_2 \ne {\mathbf 0}_X$, we have $\map {e_2^\ast} {e_1} = 0$. 
Further, we have $\map {S_1} {e_2} = {\mathbf 0}_X$.
Since $e_2 \in S_2 \sqbrk X$, there exits $u_2 \in X$ such that $\map {S_2} {u_2} = e_2$.
From property $(2)$, we then have $\map {S_2} {e_2} = \map {S_2^2} {u_2} = \map {S_2} {u_2} = e_2$. 
Hence $\map {e_2^\ast} {e_2} e_2 = \map {S_2} x = e_2$. 
Hence $\map {e_2^\ast} {e_2} = 1$. 
We conclude that $\map {e_j^\ast} {e_i} = \delta_{i j}$ for $1 \le i, j \le 2$. 
Suppose now that we have defined bounded linear functionals $e_1^\ast, e_2^\ast, \ldots, e_{n - 1}^\ast$ such that:
:$\map {e_j^\ast} x e_j = \map {S_j} x - \map {S_{j - 1} } x$ for $2 \le j \le n - 1$
and:
:$\map {e_j^\ast} {e_i} = \delta_{i j}$ for $1 \le i, j \le n - 1$
for some $n - 1 \ge 2$. 
Suppose that in doing so we have proven that $S_j \sqbrk X = \map \span {\set {e_1, \ldots, e_j} }$ and $\map {S_j} {e_j} = e_j$ for $1 \le j \le n - 1$ and $e_{j + 1} \not \in \map \span {\set {e_1, \ldots, e_j} }$ for $1 \le j \le n - 2$. 
We quickly establish that the condition $e_{j + 1} \not \in \map \span {\set {e_1, \ldots, e_j} }$  for $1 \le j \le n - 2$ implies that $\set {e_1, \ldots, e_{n - 1} }$ is linearly independent.
Otherwise, there would exist $c_1, \ldots, c_{n - 1} \in \GF$ not all $0$ such that:
:$\ds \sum_{j \mathop = 1}^{n - 1} c_j e_j = {\mathbf 0}_X$
Fix $1 \le j^\ast \le n - 1$ greatest such that $c_{j^\ast} \ne 0$.
Then we have:
:$\ds e_{j^\ast} = -\frac 1 {c_{j^\ast} } \sum_{1 \le j < j^\ast} c_j e_j$ 
This contradicts that $e_{j^\ast} \not \in \map \span {\set {e_1, \ldots, e_{j^\ast - 1} } }$. 
Hence $\set {e_1, \ldots, e_{n - 1} }$ is linearly independent.
We firstly prove the facts in the previous line for $S_n$ and $e_n$. 
We have $\map {S_{n - 1} } {e_n} = {\mathbf 0}_X$, hence:
:$\map {e_{n - 1}^\ast} {e_n} e_{n - 1} = \map {S_{n - 1} } {e_n} - \map {S_{n - 2} } {e_n} = -\map {S_{n - 2} } {e_n}$
Since $\map {S_{n - 2} } {e_n} \in \map \span {\set {e_1, \ldots, e_{n - 2} } }$ and $e_{n - 1} \not \in \map \span {\set {e_1, \ldots, e_{n - 2} } }$, we have that $\map {e_{n - 1}^\ast} {e_n} = 0$ and $\map {S_{n - 2} } {e_n} = {\mathbf 0}_X$.
We then have:
:$\map {e_{n - 2}^\ast} {e_n} e_{n - 2} = \map {S_{n - 2} } {e_n} - \map {S_{n - 3} } {e_n} = -\map {S_{n - 3} } {e_n}$
and we can see again that $\map {S_{n - 3} } {e_n} = {\mathbf 0}_X$ and $\map {e_{n - 2}^\ast} {e_n} = 0$.
This way, we can show that $\map {S_j} {e_n} = {\mathbf 0}_X$ and $\map {e_j^\ast} {e_n} = 0$ for $1 \le j \le n - 1$.
Note again that $\map {S_{n - 1} } {e_n} = {\mathbf 0}_X$.
If it was the case that $e_n \in \map \span {\set {e_1, \ldots, e_{n - 1} } }$, then we would have $e_n \in S_{n - 1} \sqbrk X$.
Hence we could take $v_n \in X$ such that $e_n = \map {S_{n - 1} } {v_n}$. 
Then $\map {S_{n - 1} } {e_n} = \map {S_{n - 1}^2} {v_n} = \map {S_{n - 1} } {v_n} = e_n$. 
But $\map {S_{n - 1} } {e_n} = {\mathbf 0}_X \ne e_n$, so we must have $e_n \not \in \map \span {\set {e_1, \ldots, e_{n - 1} } }$.
We have hence established that $\set {e_1, \ldots, e_n}$ is linearly independent. 
From property $(2)$, we have:
:$\map {S_n} {e_j} = \map {S_n} {\map {S_j} {e_j} } = \map {S_j} {e_j} = e_j$
for $1 \le j \le n - 1$.
Hence $\set {e_1, \ldots, e_{n - 1}, e_n} \subseteq S_n \sqbrk X$ .
Since $e_n \not \in \map \span {\set {e_1, \ldots, e_{n - 1} } }$ and $\dim S_n \sqbrk X = n$ by property $(1)$, we must have:
:$S_n \sqbrk X = \map \span {\set {e_1, \ldots, e_n} }$ 
Finally, $e_n \in S_n \sqbrk X$ so there exists $u_n \in X$ with $\map {S_n} {u_n} = e_n$. 
Then we have $\map {S_n} {e_n} = \map {S_n^2} {u_n} = \map {S_n} {u_n} = e_n$.
With this all done, it remains to define $e_n^\ast$ and show that it satisfies:
:$\map {e_j^\ast} {e_i} = \delta_{i j}$ for $1 \le i, j \le n$
Recall that we have:
:$\map {e_j^\ast} x e_j = \map {S_j} x - \map {S_{j - 1} } x$ for $2 \le j \le n - 1$. 
Summing from $j = 2$ to $j = n - 1$ we obtain:
:$\ds \sum_{j \mathop = 2}^{n - 1} \map {e_j^\ast} x e_j = \map {S_{n - 1} } x - \map {S_1} x = \map {S_{n - 1} } x - \map {e_1^\ast} x e_1$
So that:
:$\ds \map {S_{n - 1} } x = \sum_{j \mathop = 1}^{n - 1} \map {e_j^\ast} x e_j$
Since $\map {S_n} x \in \map \span {\set {e_1, \ldots, e_n} }$, there exists $\map {\beta_1} x, \ldots, \map {\beta_n} x \in \GF$ such that:
:$\ds \map {S_n} x = \sum_{j \mathop = 1}^n \map {\beta_j} x e_j$ 
We have:
{{begin-eqn}}
{{eqn	| l = \map {S_{n - 1} } x
		| r = \map {S_{n - 1} } {\map {S_n} x}
		| c = property $(2)$
}}
{{eqn	| r = \map {S_{n - 1} } {\sum_{j \mathop = 1}^n \map {\beta_j} x e_j}
}}
{{eqn	| r = \sum_{j \mathop = 1}^{n - 1} \map {\beta_j} x \map {S_{n - 1} } {e_j} + \map {\beta_n} x \map {S_{n - 1} } {e_n}
}}
{{eqn	| r = \sum_{j \mathop = 1}^{n - 1} \map {\beta_j} x e_j 
}}
{{end-eqn}}
hence $\map {\beta_j} x = \map {e_j^\ast} x$ for $1 \le j \le n - 1$ by the linear independence of $\set {e_1, \ldots, e_{n - 1} }$.
Hence we have:
:$\map {\beta_n} x e_n = \map {S_n} x - \map {S_{n - 1} } x$
Let $e_n^\ast = \beta_n$. 
As before, $e_n^\ast$ is a linear functional.
Further, from Fundamental Property of Norm on Bounded Linear Transformation, we have:
:$\cmod {\map {e_n^\ast} x} \le \norm {S_n - S_{n - 1} }_{\map \BB X} \norm x$
for each $x \in X$, since $\norm {e_n} = 1$. 
That is, from {{NormAxiomVector|3}}, we have:
:$\cmod {\map {e_n^\ast} x} \le \paren {\norm {S_n}_{\map \BB X} + \norm {S_{n - 1} }_{\map \BB X} } \norm x$
Hence $e_n^\ast$ is bounded.
It remains to show that $\map {e_n^\ast} {e_j} = 0$ for $1 \le j \le n - 1$ and $\map {e_n^\ast} {e_n} = 1$.
We recall that $\map {S_n} {e_j} = e_j$ for each $1 \le j \le n$. 
Writing out:
:$\ds \map {S_n} {e_j} = \sum_{k \mathop = 1}^n \map {e_k^\ast} {e_j} e_k = e_j = \sum_{k \mathop = 1}^n \map {\delta_{k j} } e_j$
Then, from the linear independence of $\set {e_1, \ldots, e_n}$, we have that:
:$\map {e_k^\ast} {e_j} = \delta_{k j}$ for $1 \le k, j \le n$. 
This is what we wished to show. 
We now have defined bounded linear functionals $e_1^\ast, e_2^\ast, \ldots, e_{n}^\ast$ such that:
:$\map {e_j^\ast} x e_j = \map {S_j} x - \map {S_{j - 1} } x$ for $2 \le j \le n$
and:
:$\map {e_j^\ast} {e_i} = \delta_{i j}$ for $1 \le i, j \le n$
In doing so we have proven that $S_j \sqbrk X = \map \span {\set {e_1, \ldots, e_j} }$ and $\map {S_j} {e_j} = e_j$ for $1 \le j \le n$ and $e_{j + 1} \not \in \map \span {\set {e_1, \ldots, e_j} }$ for $1 \le j \le n - 1$. 
The construction may then continue.
Hence we have now obtained bounded linear functionals $\sequence {e_n^\ast}_{n \mathop \in \N}$ such that:
:$\map {e_j^\ast} {e_i} = \delta_{i j}$ for $i, j \in \N$
and:
:$\ds \map {S_n} x = \sum_{j \mathop = 1}^n \map {e_j^\ast} x e_j$ for each $x \in X$ and $n \in \N$. 
By property $(3)$, we have $\map {S_n} x \to x$. 
Hence we have:
:$\ds x = \sum_{j \mathop = 1}^\infty \map {e_j^\ast} x e_j$
By Condition for Sequence to be Schauder Basis in terms of Coordinate Functionals, we have that $\sequence {e_n}_{n \mathop \in \N}$ is a Schauder basis.
Further from:
:$\ds \map {S_n} x = \sum_{j \mathop = 1}^n \map {e_j^\ast} x e_j$ for each $x \in X$ and $n \in \N$
we have that $\sequence {S_n}_{n \mathop \in \N}$ are the projections associated with $\sequence {e_n}_{n \mathop \in \N}$.
{{qed}}