In an infinite-dimensional Banah space $(X, \|\cdot\|)$ with a countable Schauder basis $\{x\_n\}$, define:


$$
F\_r: \operatorname{Span}(\{x\_n\}) \rightarrow \operatorname{Span}(\{x\_n\}), \hspace{0.3cm} \sum\_{i\leq N} \lambda\_i x\_i \mapsto \sum\_{i\leq N} \operatorname{Re}(\lambda\_i) x\_i
$$


and:


$$
F\_i: \operatorname{Span}(\{x\_n\}) \rightarrow \operatorname{Span}(\{x\_n\}), \hspace{0.3cm} \sum\_{i\leq N} \lambda\_i x\_i \mapsto \sum\_{i\leq N} \operatorname{Im}(\lambda\_i) x\_i
$$


1. Can $F\_r$ be always continuously extended to the entire $X$?


Next, define another norm $\|\cdot\|\_1$ in $\operatorname{Span}(\{x\_n\})$ as $\|x\|\_1 = \|F\_r(x)\| + \|F\_i(x)\|$ and let $X\_1$ be the completion of $\operatorname{Span}(\{x\_n\})$ with respect to $\|\cdot\|\_1$. Notice that in the case when $F\_r$ could be continuously extended, $F\_i$ could be too and we could find $M>0$ such that $\|x\|\_1 \leq M\|x\|$ for each $x\in X$. Then:


2. In the case when $F\_r$ (or $F\_i$) could be continuously extended, can $\|\cdot\|$ an $\|\cdot\|\_1$ be equivalent?


**Update**: Thanks for losifs' answers, if we define $\|\cdot\|\_r = \|F\_r(x)\|$, then in the completion of the complex span of $\{e\_n\}$ with respect to $\|\cdot\|\_r$, say $(\ell')^2$, we have $\ell^2\subseteq (\ell')^2$ but $\|\cdot\|\_r$ is not equivalent to $\|\cdot\|$.


Similar to how the idea of proving the real version of the Hahn-Banach Theorem is extended to the complex version one, when we deal with problems in $X^{\ast}$, we could always start from a real bounded linear functional. However, I am not sure if, when dealing with problems related to Banach space geometry, it can be assumed what holds in a real Banach space will also hold in a complex one.



The real and the imaginary part of a vector
$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\C}{\mathbb C}$No, in general the map $F\_r$
\begin{equation\*}
 \sum\_1^n w\_j b\_j\mapsto \sum\_1^n \Re(w\_j) b\_j \tag{10}\label{10}
\end{equation\*}
from the span of the $b\_j$'s into itself cannot be continuously extended to the entire $X$, where $(b\_j)$ is a Schauder basis of $X$, $n$ is any natural number, and the $w\_j$'s are any complex numbers.


Indeed, let $X$ be the complex space $\ell^2$ and let $(e\_k)$ be the standard basis of $X=\ell^2$. For any natural $n$, any $c\in\C^n=\C^{n\times1}$ with norm $|c|=1$, and any $\ep\in(0,1)$, let
\begin{equation\*}
 Z:=Z^{n,c,\ep}:=(\ep-1)I\_n+cc^\*=:(z\_{jk}=z\_{jk}^{n,c,\ep}\colon(j,k)\in[n]\times[n]),
\end{equation\*}
where $I\_n$ is the $n\times n$ identity matrix, $c^\*$ is the complex conjugate of the $n\times1$ row matrix $c^\top$, and $[n]:=\{1,\dots,n\}$. The eigenvectors of $Z$ are the nonzero multiples of $c$ (with eigenvalue $\ep$) and the nonzero vectors in $\C^n$ orthogonal to $c$ (with eigenvalue $\ep-1$). So, the matrix $Z$ is nonsingular.


For $q=0,1,\dots$, let $K^q:=\{2^q,\dots,2^{q+1}-1\}$ and then let
\begin{equation\*}
 b\_j:=\sum\_{k\in K^q}z\_{jk}^q e\_k \tag{20}\label{20}
\end{equation\*}
for $j\in K^q$, where $z\_{jk}^q:=z\_{jk}^{2^q,c^q,\ep^q}$, $\ep^q\in(0,1)$,
$c^q:=x^q+iy^q$, and $x^q$ and $y^q$ are any vectors in $\R^{K^q}$ such that $|x^q|=|y^q|=1/\sqrt2$ and $(y^q)^\top x^q=0$, so that $|c^q|=1$.
Since the matrix $Z^q:=Z^{2^q,c^q,\ep^q}$ is nonsingular for each $q$, we see that $(b\_j)$ is a Schauder basis of $X=\ell^2$.


Next, writing $c^q=(c^q\_k\colon k\in K\_q)$, by \eqref{20} we have
\begin{equation\*}
 \sum\_{k\in K\_q}c^q\_k b\_k=\sum\_{k\in K\_q}(Z^qc^q)\_k e\_k=\ep^q\sum\_{k\in K\_q}c^q\_k e\_k
 =\ep^q c^q, 
\end{equation\*}
so that for all $q$
\begin{equation\*}
 \Big|\sum\_{k\in K\_q}c^q\_k b\_k\Big|=\ep^q. \tag{30}\label{30} 
\end{equation\*}


On the other hand, for all $q$ we have  

\begin{equation\*}
 |\Im(Z^q x^q)|=|x^q|^2|y^q|=A:=\frac1{2\sqrt2}>0 
\end{equation\*}
and
\begin{equation\*}
 \sum\_{k\in K\_q}\Re(c^q\_k) b\_k
= \sum\_{k\in K\_q}x^q\_k b\_k
 =\sum\_{k\in K\_q}(Z^qx^q)\_k e\_k=Z^qx^q, 
\end{equation\*}
and hence
\begin{equation\*}
 \Big|\sum\_{k\in K\_q}\Re(c^q\_k) b\_k\Big|
\ge|\Im(Z^q x^q)|=A>0. \tag{40}\label{40}
\end{equation\*}
Letting now $\ep^q\downarrow0$ as $q\to\infty$, we see from \eqref{30} and \eqref{40} that $\big|\sum\_{k\in K\_q}c^q\_k b\_k\big|\to0$ while $\big|\sum\_{k\in K\_q}\Re(c^q\_k) b\_k\big|
\ge|\Im(Z^q x^q)|=A>0$. So, the map $F\_r$ cannot be continuously extended to the entire $X$. $\quad\Box$


