Theorem: Coordinate Functionals Associated with Schauder Basis of Banach Space are Bounded

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\GF$.
Let $E = \sequence {e_n}_{n \mathop \in \N}$ be a Schauder basis for $X$.
Let $\sequence {e_n^\ast}_{n \mathop \in \N}$ be the coordinate functionals associated with $\sequence {x_n}_{n \mathop \in \N}$. 
Then each linear functional $e_n^\ast$ is bounded.

Proof:
We define the projections associated with $\sequence {e_n}_{n \mathop \in \N}$ by:
:$\ds \map {S_n} x = \sum_{k \mathop = 1}^n \map {e_k^\ast} x e_k$
Define $\norm {\, \cdot \,}_E : X \to \R$ by:
:$\ds \norm x_E = \sup_{n \mathop \in \N} \norm {\map {S_n} x}$ for each $x \in X$. 
From Norm Induced by Projections Associated with Schauder Basis is Norm, $\norm {\, \cdot \,}_E$ is a norm.
Consider the identity mapping $I : \struct {X, \norm {\, \cdot \,}_E} \to \struct {X, \norm {\, \cdot \,} }$.
Let $\epsilon > 0$.
From Projections Associated with Schauder Basis Converge to Identity in Strong Operator Topology and Characterization of Convergence in Strong Operator Topology we have:
:$\map {S_n} x \to x$ for each $x \in X$.
Hence from Modulus of Limit: Normed Vector Space, we have:
:$\norm {\map {S_n} x} \to \norm x$
and there exists $N \in \N$ such that:
:$\norm x - \epsilon < \norm {\map {S_N} x}$
So we have:
:$\ds \norm x - \epsilon < \sup_{n \mathop \in \N} \norm {\map {S_n} x} = \norm x_E$ for all $\epsilon > 0$.
That is:
:$\norm x \le \norm x_E$
So $I$ is bounded.
With view to apply Complete Norms on Vector Space Equivalent iff Identity Mapping Bounded, we show that $\struct {X, \norm {\, \cdot \,}_E}$ is a Banach space.
That is, every Cauchy sequence in $\struct {X, \norm {\, \cdot \,}_E}$ convergent sequence.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {X, \norm {\, \cdot \,}_E}$.
Let $\epsilon > 0$. 
Since $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {X, \norm {\, \cdot \,}_E}$, there exists $N \in \N$ such that:
:$\ds \sup_{j \mathop \in \N} \norm {\map {S_j} {x_n} - \map {S_j} {x_m} } < \frac \epsilon 3$
for $n, m \ge N$.
Fix $n, m \ge N$. 
Recall that:
:$\map {S_j} {x_n} \to x_n$
and:
:$\map {S_j} {x_m} \to x_m$
as $j \to \infty$.
Hence we can take $N' \ge N$ such that:
:$\ds \norm {\map {S_j} {x_n} - x_n} < \frac \epsilon 3$
and:
:$\ds \norm {\map {S_j} {x_m} - x_m} < \frac \epsilon 3$
for $j \ge N'$.
By {{NormAxiomVector|3}}, we have:
:$\norm {x_n - x_m} \le \norm {x_n - \map {S_j} {x_n} } + \norm {\map {S_j} {x_n} - \map {S_j} {x_m} } + \norm {\map {S_j} {x_m} - x_m}$
Since:
:$\ds \sup_{j \mathop \in \N} \norm {\map {S_j} {x_n} - \map {S_j} {x_m} } < \frac \epsilon 3$
we in particular have:
:$\ds \norm {\map {S_j} {x_n} - \map {S_j} {x_m} } < \frac \epsilon 3$ for all $j \ge N'$. 
Hence we have $\norm {x_n - x_m} < \epsilon$ for $n, m \ge N$.
So $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {X, \norm {\, \cdot \,} }$. 
Since $\struct {X, \norm {\, \cdot \,} }$ is a Banach space, $\sequence {x_n}_{n \mathop \in \N}$ converges to a limit $x$ in $\struct {X, \norm {\, \cdot \,} }$.
We show that $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$ in $\struct {X, \norm {\, \cdot \,}_E}$
Since $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {X, \norm {\, \cdot \,}_E}$, for each $\epsilon > 0$ there exists $N \in \N$ such that:
:$\ds \sup_{k \mathop \in \N} \norm {\map {S_k} {x_n} - \map {S_k} {x_m} } < \epsilon$ for each $n, m \ge N$.
In particular, for each fixed $k \in \N$ we have:
:$\norm {\map {S_k} {x_n} - \map {S_k} {x_m} } < \epsilon$ for each $n, m \ge N$.
That is:
:$\sequence {S_k x_n}_{k \mathop \in \N}$ is a Cauchy sequence in $\struct {X, \norm {\, \cdot \,} }$ for each $j \in \N$. 
Since $\struct {X, \norm {\, \cdot \,} }$ is a Banach space, $\sequence {S_j x_n}_{n \mathop \in \N}$ converges in $\struct {X, \norm {\, \cdot \,} }$ to some $y_j \in X$. 
For each $k \in \N$, we have:
:$S_k x_n \in \map \cl {\map \span {\set {e_1, \ldots, e_k} } }$
From Finite Dimensional Subspace of Normed Vector Space is Closed and Set is Closed iff Equals Topological Closure, we have:
:$\map \cl {\map \span {\set {e_1, \ldots, e_k} } } = \map \span {\set {e_1, \ldots, e_k} }$
Hence from Subset of Metric Space contains Limits of Sequences iff Closed:
:$S_k x_n \in \map \span {\set {e_1, \ldots, e_k} }$ for each $k \in \N$
and:
:$y_k \in \map \span {\set {e_1, \ldots, e_k} }$
We now show that:
:$\map {e_k^\ast} {y_k} = \map {e_k^\ast} {y_j}$ for $j \ge k$.
Fix $j \ge k$.
We have, for all $n \in \N$:
{{begin-eqn}}
{{eqn	| l = \map {e_k^\ast} {S_j x_n}
		| r = \map {e_k^\ast} {\sum_{i \mathop = 1}^j \map {e_i^\ast} {x_n} e_n}
}}
{{eqn	| r = \map {e_k^\ast} {\map {e_k^\ast} {x_k} e_k + \sum_{1 \le i \le j, \, i \ne k} \map {e_i^\ast} {x_n} e_n}
}}
{{eqn	| r = \map {e_k^\ast} {x_n} 
		| c = {{Defof|Coordinate Functionals Associated with Schauder Basis}}
}}
{{eqn	| r = \map {e_k^\ast} {\sum_{i \mathop = 1}^k \map {e_i^\ast} {x_n} e_n}
		| c = {{Defof|Coordinate Functionals Associated with Schauder Basis}}
}}
{{eqn	| r = \map {e_k^\ast} {S_k x_n}
}}
{{end-eqn}}
From Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous:
:the restriction of $e_k^\ast$ to $\map \span {\set {e_1, \ldots, e_k} }$ is bounded.
Since $S_k x_n, S_j x_n \in \map \span {\set {e_1, \ldots, e_k} }$, we can take $n \to \infty$ in:
:$\map {e_k^\ast} {S_j x_n} = \map {e_k^\ast} {S_k x_n}$
to obtain:
:$\map {e_k^\ast} {y_j} = \map {e_k^\ast} {y_k}$
One may now reason that the $y_j$ is "refined", in the sense that once a coordinate is determined it will not be changed, for increasing $j \in \N$.
We further reason that we should have $y_j \to x$ in $\struct {X, \norm {\, \cdot \,} }$.
Since $y_j$ "refines" with respect to $\sequence {e_n}_{n \mathop \in \N}$ and is contained in $\span \set {e_1, e_2, \ldots, e_j}$, we will obtain that $y_j = \map {S_j} x$.
Hence we will have $\map {S_j} {x_n} \to \map {S_j} x$ as $n \to \infty$ for each $n \in \N$. 
From this we will be able to find that $x_n \to x$ in $\struct {X, \norm {\, \cdot \,}_E}$. 
We now make this sketch precise.
Let:
:$\alpha_k = \map {e_k^\ast} {y_k}$
We then have:
{{begin-eqn}}
{{eqn	| l = y_k
		| r = \sum_{j \mathop = 1}^k \map {e_j^\ast} {y_k} e_k
		| c = {{Defof|Coordinate Functionals Associated with Schauder Basis}}
}}
{{eqn	| r = \sum_{j \mathop = 1}^k \map {e_j^\ast} {y_j} e_j
		| c = since $\map {e_j^\ast} {y_j} = \map {e_j^\ast} {e_k}$ for $k \ge j$
}}
{{eqn	| r = \sum_{j \mathop = 1}^k \alpha_j e_j
}}
{{end-eqn}}
We aim to show that:
:$\ds x = \sum_{j \mathop = 1}^\infty \alpha_j e_j$
which is precisely the statement:
:$y_k \to x$ in $\struct {X, \norm {\, \cdot \,} }$. 
From the uniqueness of $\alpha_j$, we will then obtain:
:$\map {e_j^\ast} x = \alpha_j$
and hence that:
:$y_j = \map {S_j} x$ 
which is our current goal.
Let $\epsilon > 0$.
By {{NormAxiomVector|3}}, we write:
:$\norm {x - y_j} \le \norm {y_j - \map {S_j} {x_n} } + \norm {\map {S_j} {x_n} - x_n} + \norm {x_n - x}$
for all $n \in \N$.
We can write:
{{begin-eqn}}
{{eqn	| l = \norm {y_j - \map {S_j} {x_n} }
		| r = \lim_{m \mathop \to \infty} \norm {\map {S_j} {x_m} - \map {S_j} {x_n} }
		| c = Modulus of Limit: Normed Vector Space
}}
{{eqn	| o = \le
		| r = \limsup_{m \to \infty} \sup_{j \mathop \in \N} \norm {\map {S_j} {x_m} - \map {S_j} {x_n} }
		| c = Squeeze Theorem
}}
{{eqn	| r = \limsup_{m \to \infty} \norm {x_m - x_n}_E
}}
{{end-eqn}}
Since $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy in $\struct {X, \norm {\, \cdot \,}_E}$, for each $\delta > 0$ there exists $N \in \N$ such that for $m \ge n \ge N_1$, we have: 
:$\norm {x_n - x_m}_E < \dfrac \epsilon 3$.
That is, for $n \ge N_1$, we have:
:$\ds \limsup_{m \to \infty} \norm {x_m - x_n}_E \le \dfrac \epsilon 3$ 
giving:
:$\norm {y_j - \map {S_j} {x_n} } \le \dfrac \epsilon 3$
uniformly in $j \in \N$.
Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$, we can take $N_2 \in \N$ such that:
:$\norm {x_n - x} < \dfrac \epsilon 3$ for $n \ge N_2$. 
Now fix $n = \max \set {N_1, N_2}$. 
For this $n$, we can take $J \in \N$ such that:
:$\norm {\map {S_j} {x_n} - x_n} < \dfrac \epsilon 3$ for $j \ge J$
by Projections Associated with Schauder Basis Converge to Identity in Strong Operator Topology and Characterization of Convergence in Strong Operator Topology.
Putting everything together, we have:
:$\norm {y_j - x} < \epsilon$ for $j \ge J$. 
Since $\epsilon > 0$ was arbitrary, we have that $y_j \to x$ in $\struct {X, \norm {\, \cdot \,} }$. 
Hence, as previously discussed, we have $y_j = \map {S_j} x$ for each $j \in \N$.
Hence $\map {S_j} {x_n} \to \map {S_j} x$ for each $j \in \N$. 
We can write:
{{begin-eqn}}
{{eqn	| l = \norm {\map {S_j} {x_n} - \map {S_j} x}
		| r = \lim_{m \mathop \to \infty} \norm {\map {S_j} {x_n} - \map {S_j} {x_m} }
		| c = Modulus of Limit: Normed Vector Space
}}
{{eqn	| o = \le
		| r = \limsup_{m \mathop \to \infty} \sup_{j \mathop \in \N} \norm {\map {S_j} {x_n} - \map {S_j} x}
}}
{{eqn	| r = \limsup_{m \mathop \to \infty} \norm {x_n - x_m}_E 
}}
{{end-eqn}}
Taking the supremum over $j \in \N$:
:$\ds \sup_{j \mathop \in \N} \norm {\map {S_j} {x_n} - \map {S_j} x} \le \limsup_{m \mathop \to \infty} \norm {x_n - x_m}_E$
That is:
:$\ds \norm {x_n - x}_E \le \limsup_{m \mathop \to \infty} \norm {x_n - x_m}_E$
Again $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy in $\struct {X, \norm {\, \cdot \,}_E}$, so given $\epsilon > 0$ there exists $N \in \N$ such that $\norm {x_n - x_m}_E < \epsilon$ for $n, m \ge N$. Hence:
:$\ds \limsup_{m \mathop \to \infty} \norm {x_n - x_m}_E \le \epsilon$ for $n \ge N$.
So $\norm {x_n - x}_E \le \epsilon$ for $n \ge N$. 
Hence $x_n \to x$ in $\struct {X, \norm {\, \cdot \,}_E}$. 
Since $\sequence {x_n}_{n \mathop \in \N}$ was an arbitrary Cauchy sequence in $\struct {X, \norm {\, \cdot \,}_E}$, we have that $\struct {X, \norm {\, \cdot \,}_E}$ is a Banach space.
We have also established that the identity mapping $I : \struct {X, \norm {\, \cdot \,}_E} \to \struct {X, \norm {\, \cdot \,} }$ is a bounded surjection. 
Hence by the Complete Norms on Vector Space Equivalent iff Identity Mapping Bounded, the norms $\norm {\, \cdot \,}$ and $\norm {\, \cdot \,}_E$ are equivalent. 
Hence there exists $K > 0$ such that:
:$\norm x_E \le K \norm x$ for each $x \in X$.
That is:
:$\norm {\map {S_n} x} \le K \norm x$ for each $x \in X$ and $n \in \N$.
We can now write:
:$\map {e_n^\ast} x e_n = \map {S_n} x - \map {S_{n - 1} } x$
Hence by {{NormAxiomVector|3}}:
:$\norm {\map {e_n^\ast} x e_n} = \norm {\map {S_n} x - \map {S_{n - 1} } x} \le \norm {\map {S_n} x} + \norm {\map {S_{n - 1} } x} \le 2 K \norm x$
By {{NormAxiomVector|2}}, we have:
:$\norm {\map {e_n^\ast} x e_n} = \norm {e_n} \cmod {\map {e_n^\ast} x}$
By Schauder Basis is Linearly Independent, we have that $e_n \ne 0$ for each $n \in \N$, and hence:
:$\ds \cmod {\map {e_n^\ast} x} \le \frac {2 K} {\norm {e_n} } \norm x$
Hence $e_n^\ast$ is bounded.
{{qed}}