The real and the imaginary part of a vector

Question

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.

Answer 1

$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\C}{\mathbb C}$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 $X$. $\quad\Box$