Canonical Banach complexification of a real Banach space

For every $(x,y)$ the set $\{\mathrm{\parallel }\cos \theta x-\sin \theta y\mathrm{\parallel }:\theta \in \mathbb{R}\}$ is nonempty and bounded above by $\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }$, because $\mathrm{\parallel }\cos \theta x-\sin \theta y\mathrm{\parallel }\le \mathrm{\mid }\cos \theta \mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\mid }\sin \theta \mathrm{\mid }\mathrm{\parallel }y\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }$; hence $\rho (x,y)$ is a well-defined real number with $0\le \rho (x,y)\le \mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }$, and choosing $\theta =0$ and $\theta =\frac{\pi }{2}$ gives $\mathrm{\parallel }x\mathrm{\parallel }\le \rho (x,y)$ and $\mathrm{\parallel }y\mathrm{\parallel }\le \rho (x,y)$.

$j(x)=(x,0)$ is real-linear, and $\rho (x,0)={\sup }_{\theta }\mathrm{\parallel }\cos \theta x\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ because the value at $\theta =0$ is $\mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\mid }\cos \theta \mathrm{\mid }\le 1$ bounds every other value by $\mathrm{\parallel }x\mathrm{\parallel }$; hence $j$ is an isometric real-linear embedding.

For a real-linear $T:X\to X$ the map $T_{\mathbb{C}}(x,y):=(Tx,Ty)$ is complex-linear: $T_{\mathbb{C}}((a+bi)(x,y))=T_{\mathbb{C}}(ax-by,bx+ay)=(aTx-bTy,bTx+aTy)=(a+bi)T_{\mathbb{C}}(x,y)$ by real-linearity of $T$.

In the situation of claim 3, every $z\in Z$ is $j_Z(x)+i{j}_Z(y)$ for unique $x,y\in X$ by hypothesis; hence $\mathrm{\Phi }(x,y):=j_Z(x)+i{j}_Z(y)$ is a well-defined bijection $X_{\mathbb{C}}\to Z$, and it is complex-linear because $j_Z$ is real-linear and $i^2=-1$.

The conjugation inverts the two components: $\sigma (j_Z(x)+i{j}_Z(y))=j_Z(x)-i{j}_Z(y)$, because $\sigma$ is real-linear, fixes $j_Z(X)$ pointwise and satisfies $\sigma (iw)=-i\sigma (w)$; consequently the formulas $j_Z(x)=\frac{1}{2}(z+\sigma (z))$ and $j_Z(y)=\frac{1}{2i}(z-\sigma (z))$ hold for $z=j_Z(x)+i{j}_Z(y)$.

$\rho (x,y)=0$ forces $x=y=0$ by the bound $\max (\mathrm{\parallel }x\mathrm{\parallel },\mathrm{\parallel }y\mathrm{\parallel })\le \rho (x,y)$ of step 1.1 and definiteness of the norm, so $\rho$ is definite.

$\rho$ satisfies the triangle inequality: for all $(x,y),(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$, and every $\theta$, $\mathrm{\parallel }\cos \theta (x+x^{\mathrm{\prime }})-\sin \theta (y+y^{\mathrm{\prime }})\mathrm{\parallel }\le \mathrm{\parallel }\cos \theta x-\sin \theta y\mathrm{\parallel }+\mathrm{\parallel }\cos \theta x^{\mathrm{\prime }}-\sin \theta y^{\mathrm{\prime }}\mathrm{\parallel }\le \rho (x,y)+\rho (x^{\mathrm{\prime }},y^{\mathrm{\prime }})$, so the supremum over $\theta$ gives $\rho (x+x^{\mathrm{\prime }},y+y^{\mathrm{\prime }})\le \rho (x,y)+\rho (x^{\mathrm{\prime }},y^{\mathrm{\prime }})$.

$\rho$ is absolutely homogeneous for complex scalars: writing $\mu =a+bi$, for $\theta \in \mathbb{R}$ one has $\cos \theta (ax-by)-\sin \theta (bx+ay)=(a\cos \theta -b\sin \theta )x-(b\cos \theta +a\sin \theta )y$; if $\mu \mathrm{\ne }0$ and $\mu =r(\cos \phi +i\sin \phi )$ by [L3], then $a=r\cos \phi$ and $b=r\sin \phi$, so [L4] turns the two coefficients into $r\cos (\phi +\theta )$ and $r\sin (\phi +\theta )$; the norm of the resulting vector is $r\mathrm{\parallel }\cos \psi x-\sin \psi y\mathrm{\parallel }$ with $\psi :=\phi +\theta$, and taking suprema over $\theta$ (equivalently over $\psi$) gives $\rho (\mu (x,y))=r\rho (x,y)=\mathrm{\mid }\mu \mathrm{\mid }\rho (x,y)$, while $\mu =0$ gives the zero vector.

If in addition $T$ is bounded, then $\rho (T_{\mathbb{C}}(x,y))={\sup }_{\theta }\mathrm{\parallel }T(\cos \theta x-\sin \theta y)\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }{\sup }_{\theta }\mathrm{\parallel }\cos \theta x-\sin \theta y\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }\rho (x,y)$ by [L5], and the reverse inequality follows by evaluating at $y=0$, where the supremum is $\mathrm{\parallel }Tx\mathrm{\parallel }$: the operator norm of $T_{\mathbb{C}}$ is exactly $\mathrm{\parallel }T\mathrm{\parallel }$. If $X=\{0\}$, both unit-ball suprema are zero by [L5], so this conclusion still holds.

The intertwining is a definitional identity: $T_Z:=\mathrm{\Phi }{T}_{\mathbb{C}}{\mathrm{\Phi }}^{-1}$ satisfies $T_Z(j_Z(x)+i{j}_Z(y))=\mathrm{\Phi }(T_{\mathbb{C}}(x,y))=j_Z(Tx)+i{j}_Z(Ty)$, so $\mathrm{\Phi }{T}_{\mathbb{C}}=T_Z\mathrm{\Phi }$ holds by construction.

If $Z$ carries the rotation-supremum norm relative to $j_Z$, then $\mathrm{\parallel }\mathrm{\Phi }(x,y){\mathrm{\parallel }}_Z={\sup }_{\theta }\mathrm{\parallel }\cos \theta j_Z(x)-\sin \theta j_Z(y){\mathrm{\parallel }}_Z={\sup }_{\theta }\mathrm{\parallel }{j}_Z(\cos \theta x-\sin \theta y){\mathrm{\parallel }}_Z={\sup }_{\theta }\mathrm{\parallel }\cos \theta x-\sin \theta y\mathrm{\parallel }=\rho (x,y)$, using real-linearity and isometry of $j_Z$; so $\mathrm{\Phi }$ is an isometry in that case. Conversely, if $\mathrm{\Phi }$ is isometric, its defining formula gives exactly this norm equality for every $(x,y)$, which is the stated rotation-supremum condition.

For $z=j_Z(x)+i{j}_Z(y)$ the component estimates $\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }{j}_Z(x)\mathrm{\parallel }\le \frac{1}{2}(\mathrm{\parallel }z\mathrm{\parallel }+\mathrm{\parallel }\sigma (z)\mathrm{\parallel })=\mathrm{\parallel }z\mathrm{\parallel }$ and $\mathrm{\parallel }y\mathrm{\parallel }\le \mathrm{\parallel }z\mathrm{\parallel }$ hold, because $\mathrm{\parallel }\sigma (z)\mathrm{\parallel }=\mathrm{\parallel }z\mathrm{\parallel }$ by hypothesis and $j_Z$ is isometric.

By steps 1.1, 2.2, 2.3 and 2.4, the function $\rho$ satisfies definiteness, the triangle inequality and absolute homogeneity, so it is a norm on the complex vector space $X_{\mathbb{C}}$ with scalar multiplication $(a+bi)(x,y)=(ax-by,bx+ay)$, which is associative and distributive because $\mathbb{C}$ is a field.

For $z=\mathrm{\Phi }(x,y)$ one has $\rho (x,y)\le \mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }\le 2\mathrm{\parallel }z\mathrm{\parallel }$ by [step 3.1] and $\mathrm{\parallel }\mathrm{\Phi }(x,y){\mathrm{\parallel }}_Z=\mathrm{\parallel }{j}_Z(x)+i{j}_Z(y)\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }\le 2\rho (x,y)$ by [step 1.1]; hence ${\mathrm{\Phi }}^{-1}$ and $\mathrm{\Phi }$ are bounded with norms at most $2$. Thus $\mathrm{\parallel }{T}_{Z}z\mathrm{\parallel }\le 4\mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }z\mathrm{\parallel }$, using its defining composition and step 2.5.

The norm $\rho$ is equivalent to the product maximum norm $\mathrm{\parallel }(x,y){\mathrm{\parallel }}_{\mathrm{\infty }}:=\max (\mathrm{\parallel }x\mathrm{\parallel },\mathrm{\parallel }y\mathrm{\parallel })$: indeed $\mathrm{\parallel }(x,y){\mathrm{\parallel }}_{\mathrm{\infty }}\le \rho (x,y)\le 2\mathrm{\parallel }(x,y){\mathrm{\parallel }}_{\mathrm{\infty }}$ by [step 1.1]. Consequently a $\rho$-Cauchy sequence in $X_{\mathbb{C}}$ is Cauchy for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$, hence its two coordinate sequences are Cauchy in $X$ and converge by completeness of $X$, and the coordinatewise limit is the $\rho$-limit by the same two-sided estimate; so $X_{\mathbb{C}}$ is complete for $\rho$ and is a complex Banach space.

Claims 1, 2 and 3 are established: [step 3.2] and [step 4.2] give the complex Banach space, [step 1.2] and [step 2.5] give the isometric embedding and the same-norm extension, and [step 4.1], [step 2.6] and [step 2.7] give the bounded canonical comparison, its intertwining property and its isometry in the equal-norm case.