Theorem: Sobczyk's Theorem

Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a separable Banach space over $\GF$.
Let $E$ be a closed vector subspace of $X$. 
Let $\struct {\map {c_0} \GF, \norm {\, \cdot \,}_\infty}$ be the space of zero-limit sequences.
Let $\struct {\map \BB {E, c_0}, \norm {\, \cdot \,}_{\map \BB {E, c_0} } }$ and $\struct {\map \BB {X, c_0}, \norm {\, \cdot \,}_{\map \BB {X, c_0} } }$ be the bounded linear transformations $E \to \map {c_0} \GF$ and $X \to \map {c_0} \GF$ respectively.
Let $T : E \to \map {c_0} \GF$ be a bounded linear transformation.
Then there exists a bounded linear transformation $\widetilde T : X \to \map {c_0} \GF$ that extends $T$ and satisfies:
:$\norm {\widetilde T}_{\map \BB {X, c_0} } \le 2 \norm T_{\map \BB {E, c_0} }$
That is, $\map {c_0} \GF$ is separably injective.

Proof:
Note that if $T = {\mathbf 0}_{\map \BB {E, c_0} }$, then we can simply set:
:$\widetilde T x = {\mathbf 0}_{c_0}$ for each $x \in X$. 
Then $\widetilde T$ extends $T$ and has:
:$0 = \norm {\widetilde T}_{\map \BB {X, c_0} } \le 2 \norm T_{\map \BB {E, c_0} } = 0$ 
We now initially assume that $\norm T_{\map \BB {E, c_0} } = 1$.  
Let $\struct {E^\ast, \norm {\, \cdot \,}_{E^\ast} }$ be the normed dual space of $E$.
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $X$.
From Characterization of Bounded Linear Transformation from Normed Vector Space to Space of Bounded Sequences, there exists a sequence $\sequence {f_n^\ast}_{n \mathop \in \N}$ in $E^\ast$ such that:
:$T x = \sequence {\map {f_n^\ast} x}_{n \mathop \in \N}$ for each $x \in E$
and:
:$\ds \norm T_{\map \BB {E, c_0} } = \sup_{n \mathop \in \N} \norm {f_n^\ast}_{E^\ast} = 1$ 
Hence $\norm {f_n^\ast}_{E^\ast} \le 1$ for each $n \in \N$.
Further, since $\sequence {\map {f_n} x}_{n \mathop \in \N} \in \map {c_0} \GF$ for each $x \in E$, we have that:
:$\map {f_n} x \to 0$ for each $x \in E$. 
By the Hahn-Banach Theorem, for each $n \in \N$ there exists an extension $\phi_n^\ast \in X^\ast$ of $f_n^\ast$ such that:
:$\norm {\phi_n^\ast}_{X^\ast} \le 1$
Note that there is no reason to think that:
:$\map {\phi_n^\ast} x \to 0$ for each $x \in X$
and indeed we do not prove this.
Let $w^\ast$ denote the weak-$\ast$ topology.
Let $B_{X^\ast}^-$ be the closed unit ball of $X^\ast$. 
We have $\phi_n^\ast \in B_{X^\ast}^-$ for each $n \in \N$. 
From Closed Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable, $\struct {2 B_{X^\ast}^-, w^\ast}$ is metrizable with metric $d$ satisfying:
:$\map d {f, g} = \map d {f - g, {\mathbf 0}_X}$ for each $f, g \in 2 B_{X^\ast}^-$ with $f - g \in 2 B_{X^\ast}^-$
In particular, from {{NormAxiomVector|3}}, we have:
:$\map d {f, g} = \map d {f - g, {\mathbf 0}_X}$ for each $f, g \in B_{X^\ast}^-$.
Let $E^\bot \subseteq X^\ast$ be the annihilator of $E$.
From Annihilator of Subspace of Banach Space is Weak-* Closed, $E^\bot$ is closed in $\struct {X^\ast, w^\ast}$.
From the Banach-Alaoglu Theorem, $\struct {B_{X^\ast}^-, w^\ast}$ is compact. 
From Closed Subspace of Compact Space is Compact, we have that $\struct {E^\bot \cap B_{X^\ast}, w^\ast}$ is compact. 
We show that:
:$\map d {\phi_n^\ast, E^\bot \cap B_{X^\ast}^- } \to 0$ as $n \to \infty$
{{AimForCont}} otherwise.
Then there exists $\epsilon > 0$ and a subsequence $\sequence {\phi_{n_k}^\ast}_{k \mathop \in \N}$ such that:
:$\map d {\phi_{n_k}^\ast, E^\bot \cap B_{X^\ast}^- } \ge \epsilon$ for each $k \in \N$.
From Compactness and Sequential Compactness are Equivalent in Metric Spaces, we have that:
:$\struct {B_{X^\ast}^-, w^\ast}$ is sequentially compact in itself.
Hence there exists a further subsequence $\sequence {\phi_{n_{k_j} }^\ast}_{j \mathop \in \N}$ and $\phi^\ast \in X^\ast$ such that:
:$\sequence {\phi_{n_{k_j} } }_{j \mathop \in \N}$ converges weak-$\ast$ to $\phi^\ast$. 
For each $x \in E$, we have:
:$\map {f_{n_{k_j} }^\ast} x = \map {\phi_{n_{k_j} } } x \to \map {\phi^\ast} x$ as $j \to \infty$
On the other hand, we have established that since $\sequence {\map {f_n^\ast} x}_{n \mathop \in \N} \in \map {c_0} \GF$, we have:
:$\map {f_n^\ast} x \to 0$ as $n \to \infty$ for each $x \in E$.
From Limit of Subsequence equals Limit of Sequence, we therefore have:
:$\map {\phi^\ast} x = 0$ for each $x \in E$.
That is::$\phi^\ast \in E^\bot$
From Closed Unit Ball in Normed Dual Space is Weak-* Closed, $B_{X^\ast}$ is closed in $\struct {X^\ast, w^\ast}$. 
Hence, since $\phi_n^\ast \in B_{X^\ast}$ for each $n \in \N$, we have $\phi^\ast \in B_{X^\ast}$.
Hence:
:$\phi^\ast \in E^\bot \cap B_{X^\ast}$
It follows that:
:$\map d {\phi^\ast_{n_{k_j} }, \phi^\ast} \ge \epsilon$ for each $j \in \N$.
This contradicts that:
:$\sequence {\phi_{n_{k_j} } }_{j \mathop \in \N}$ converges weak-$\ast$ to $\phi^\ast$.
We conclude that:
:$\map d {\phi_n^\ast, E^\bot \cap B_{X^\ast}^- } \to 0$ as $n \to \infty$.
From Distance from Point to Compact Set in Metric Space is Attained, there exists $v_n^\ast \in E^\bot \cap B_{X^\ast}^-$ such that:
:$\map d {\phi_n^\ast, E^\bot \cap B_{X^\ast}^-} = \map d {\phi_n^\ast, v_n^\ast}$
Hence:
:$\map d {\phi_n^\ast - v_n^\ast, {\mathbf 0}_{X^\ast} } = \map d {\phi_n^\ast, x_n^\ast} \to 0$ 
We therefore have that:
:$\sequence {\phi_n^\ast - v_n^\ast}_{n \mathop \in \N}$ converges weak-$\ast$ to ${\mathbf 0}_{X^\ast}$.
That is:
:$\sequence {\map {\phi_n^\ast} x - \map {v_n^\ast} x}_{n \mathop \in \N} \in \map {c_0} \GF$
Further, we have:
:$\norm {\phi_n^\ast - v_n^\ast}_{X^\ast} \le \norm {\phi_n^\ast}_{X^\ast} + \norm {v_n^\ast}_{X^\ast} \le 2$
Hence by Characterization of Bounded Linear Transformation from Normed Vector Space to Space of Bounded Sequences, we can define $\widetilde T : X \to \map {c_0} \GF$ by:
:$\widetilde T x = \sequence {\map {\phi_n^\ast} x - \map {v_n^\ast} x}_{n \mathop \in \N}$ for each $x \in X$
with:
:$\ds \norm {\widetilde T}_{\map \BB {X, c_0} } = \sup_{n \mathop \in \N} \norm {\phi_n^\ast - v_n^\ast}_{X^\ast} \le 2$ 
Since $v_n^\ast \in E^\bot$, we have $\map {v_n^\ast} x = 0$:
:$\widetilde T x = \sequence {\map {\phi_n^\ast} x}_{n \mathop \in \N} = T x$
Hence $\widetilde T$ is our desired map.
Now suppose that $\norm T_{\map \BB {E, c_0} } > 0$. 
Then, by what we have just proven, there exists a bounded linear transformation $T' : X \to \map {c_0} \GF$ such that:
:$\ds T' x = \frac {T x} {\norm T_{\map \BB {E, c_0} } }$ for each $x \in E$
and:
:$\ds \norm {T'}_{\map \BB {x, c_0} } \le 2 \norm {\frac T {\norm T_{\map \BB {E, c_0} } } }_{\map \BB {e, c_0} }$
Hence taking:
:$\widetilde T = \norm T_{\map \BB {E, c_0} } T'$
we have that $\widetilde T$ extends $T$ and satisfies:
:$\norm {\widetilde T}_{\map \BB {X, c_0} } \le 2 \norm T_{\map \BB {E, c_0} }$
{{qed}}
{{Namedfor|Andrew Sobczyk|cat = Sobczyk}}