There exists on $E\times E$ a structure of complex normed vector space such that the inclusions are $\mathbb{R}$-linear isometries.

Question

Let $E$ be a real normed vector space. Show that there exists on $E\times E$ a structure of complex normed vector space such that the inclusions $x\mapsto (x,0)$ and $y\mapsto (0,y)$ are $\mathbb{R}$-linear isometries.

My attempt: The structure of complex normed vector space in $E\times E$ is defined by operations $$\begin{array}{rcl}(x,y)+(z,w)&=&(x+z,y+w)\\ \alpha (x,y)&=&(ax-by,ay+bx) \quad \mathrm{where}\: \alpha=a+ib\in\mathbb{C}. \end{array}$$ With this operations $E\times E$ is a $\mathbb{C}$-vector space.

The problem is define a norm in this vector space, I have not been able to define a norm in this space, which does not allow me to move forward in order to analyze the inclusions.

Answer 1

We can define an $\mathbb{R}$-norm on $E\times E$ by setting

$$\lVert (x,y)\rVert_r = \sqrt{\lVert x\rVert^2 + \lVert y\rVert^2}.$$

This will usually not be a $\mathbb{C}$-norm, however. But we can make one from it by averaging,

$$\lVert (x,y)\rVert_c = \frac{1}{2\pi} \int_0^{2\pi} \lVert e^{i\varphi}(x,y)\rVert_r\,d\varphi.$$

Answer 2

Let $E \times E = \{ (x,y) \ | \ x,y \in E \}$ as usual and define $$ (a+ib)(x,y) = (ax-by, bx+ay) $$ You show $c_1c_2(x,y) = c_1(c_2(x,y))$ with respect to the complex scalar multiplication given above. Essentially, $(x,y)$ functions as $x+iy$ in this construction, but, if in doubt it is probably best to suffer the $(x,y)$ notation until we really understand more deeply.

The norm can be constructed on this space. But, I leave that to Daniel Fishcher.