If over the complex numbers it is Banach, then so it is over the reals.

Question

My book has the following exercise:

Let $X$ be a complex Banach space. Let $X_\mathbb R=X$ as a set, but consider $X_\mathbb R$ as a vector space over the real field. That is, vector addition in $X_\mathbb R$ is defined just as in $X$, but scalar multiplication in $X_\mathbb R$ is restricted to multiplication by real scalars. Let $\left\|\cdot\right\|_\mathbb R=\left\|\cdot\right\|$, and show that $\left(X_\mathbb R,\left\|\cdot\right\|_\mathbb R\right)$ is a real Banach space.

I feel like I am missing something important: does this not follow immediately? That is, if $\left\{x_n\right\}\subseteq X_\mathbb R=X$ is Cauchy, then $x_n\to x\in X=X_\mathbb R$. Hence, $X_\mathbb R$ is complete. I did not use the fact that $X$ is complex or that $X_\mathbb R$ is real.

Answer 1

Yes, it is direct. But checking Cauchy implies convergent is not the only thing that needs checking. One also needs to check that $\|\cdot\|_{\mathbb{R}}$ is a norm. It is still easy, but well, it needs to be done.

1. For $r\in\mathbb{R}$, $\|rx\|_{\mathbb{R}}=\|rx\|=|r|_{\mathbb{C}}\|x\|=|r|\|x\|_{\mathbb{R}}$. This one uses that $|r|$ (complex absolute value) for $r\in\mathbb{R}$ is the same as the absolute value of the reals.
2. $\|x+y\|_{\mathbb{R}}=\|x+y\|\leq \|x\|+\|y\|=\|x\|_{\mathbb{R}}+\|y\|_{\mathbb{R}}$
3. $\|x\|_{\mathbb{R}}=\|x\|>0$ for $x\neq0$.