Is the space $\ell^2_1$ injective? ($\ell^2_1$ = 2 dimensional(complex) space with 1-norm) A Banach space $Z$ is said to be injective
 if  for for any  bounded linear map $\varphi: X
\rightarrow Z$ and for any  Banach space $Y$ containing $X$ as a
closed subspace, there exists a bounded linear extension
$\tilde{\varphi}: Y \rightarrow Z $ such that
 $ \left\|\tilde{ \varphi
}\right\| =  \left\| \varphi\right\|$.
In the real case...the answer is yes. But in the complex case, $\mathbb{C}^2$, answer is No. How to establish this? Some examples?
 A: The one-dimensional space $\mathbb R$ is injective, by the Hahn-Banach theorem. 
If $X$ and $Y$ are injective, then $X\oplus_{\infty} Y$ is injective (subscript means using the norm $\|(x,y)\|_\infty = \max (\|x\|,\|y\|)$ on the direct sum.) This is because any bounded linear map into $X\oplus_{\infty} Y$ can be extended component-wise, and the definition of the norm is such that the norm of operator into $X\oplus_{\infty} Y$ is the maximum of the norms of its components. 
Hence, $\ell_\infty^n$ is injective for every $n$ (so is $\ell_\infty$, but this is not needed here.) In particular, $\ell_\infty^2$ is injective. And $\ell_\infty^2$ is isometric to $\ell_1^2$: see here.
A: Here is a explicit example showing that the Banach space $X=(\mathbf{C}^2,||.||_1)$
is not injective. (There may be a much simpler one.)
Consider the isometric embedding of $X$ into the $C(\mathbf{T})$ (the space of
complex-valued continuous functions on the unit circle $\mathbf{T} \subset \mathbf{C}$) given by
$$ \phi (z_1,z_2) \mapsto (\lambda \mapsto z_1+z_2\lambda). $$
Assume by contradiction that there is a contraction $\psi : C(\mathbf{T}) \to X$ such that $\psi \circ \phi = \mathrm{id}$. By Riesz theorem, such a $\psi$ has the form $\psi(f) = (\int f \, \mathrm{d}\mu, \int f \, \mathrm{d}\nu)$ for complex Borel measures $\mu$, $\nu$. The condition $||\psi|| \leq 1$ is equivalent to $\|\mu + \alpha \nu\| \leq 1$ for every $\alpha \in \mathbf{T}$.
Evaluating the identity $\psi \circ \phi = \mathrm{id}$ on $(1,1)$ shows that $2 = \int_{\mathbf{T}} (1+\lambda) \, \mathrm{d}(\mu+\nu)(\lambda)$. Since $\|\mu+\nu\|_{TV}=1$, this implies that $\mu + \nu = \delta_1$ (there is equality in the inequality $\int f \, \mathrm{d} \mu \leq \|f\| \cdot \|\mu\|$ only when $\mu$ has support in $\{|f|=1\}$). A similar argument evaluating on $(1,-1)$ shows that $\mu-\nu = \delta_{-1}$, and we get quickly a contradiction.
