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

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.

• Are you sure that your $E \times E$ is a complex vector space. Do you have $c_1c_2 \cdot (x,y) = c_1 \cdot (c_2 \cdot (x,y))$ ? I don't think your construction thus far is quite right. You can make $E \times E$ a complex vector space in much the same way as Gauss made $\mathbb{R} \times \mathbb{R}$ into $\mathbb{C}$... – James S. Cook Oct 28 '17 at 15:17
• @JamesS.Cook You are right, but the way as Gauss made $\mathbb{R}\times\mathbb{R}$ into $\mathbb{C}$ does not work in this context to define a norm in this complex structure. – Diego Fonseca Oct 28 '17 at 15:48
• @DanielFischer In the case of $\mathbb{R}\times \mathbb{R}$ we kanow that $\mathbb{R}$ have two perations, $+$ and $\cdot$, the operation $\cdot$ is important in the case $\mathbb{R}\times\mathbb{R}$, but in my case, $E$ have one operation $+$. – Diego Fonseca Oct 28 '17 at 15:59

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,

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

• I know I've seen specializations which do not require the integration, it must have been for a specific example. – James S. Cook Oct 28 '17 at 16:07
• This form of defining a $\mathbb{C}$-norm is unintuitive, is there any book where it is proved that this formulation is really a norm? since I have tried to prove it but since we do not know the form of the norm in $E$ then it is difficult to calculate the integral. – Diego Fonseca Oct 28 '17 at 16:22
• You don't need to calculate the integral. If you have a concrete normed space given, that is often easy enough, but in the abstract situation, you can only show that the integral has the required properties - and you don't need to do more. First note that $\varphi \mapsto \lVert e^{i\varphi}(x,y)\rVert_r$ is continuous for every fixed $(x,y) \in E\times E$. So the integral makes sense. Clearly $\lVert (x,y)\rVert_c \geqslant 0$ for all $x,y$, and for $(x,y) \neq (0,0)$ we have $\lVert (x,y)\rVert_c > 0$ since the integrand is strictly positive everywhere. – Daniel Fischer Oct 28 '17 at 17:43
• The triangle inequality for $\lVert\,\cdot\,\rVert_r$ implies that we have the triangle inequality pointwise for all $\varphi$, so we have the triangle inequality for $\lVert\,\cdot\,\rVert_c$. Finally, write $z = \lvert z\rvert e^{i\psi}$ and use the homogeneity of $\lVert\,\cdot\,\rVert_r$ to get $$\lVert z (x,y)\rVert_c = \frac{1}{2\pi} \int_0^{2\pi} \lVert \lvert z\rvert e^{i(\psi + \varphi)}(x,y)\rVert_r\,d\varphi = \lvert z\rvert \frac{1}{2\pi}\int_0^{2\pi} \lVert e^{i(\psi + \varphi)}(x,y)\rVert_r\,d\varphi = \lvert z\rvert\, \lVert (x,y)\rVert_c.$$ – Daniel Fischer Oct 28 '17 at 17:44

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.

• You are replicating $\mathbb{R}\times\mathbb{R}$, but in this case we kanow that $\mathbb{R}$ have two perations, $+$ and $\cdot$, the operation $\cdot$ is important to show that defined norm is a norm, but in my case, $E$ have one operation $+$. – Diego Fonseca Oct 28 '17 at 16:04
• If $E$ is a normed linear space then you are given a real scalar multiplication. The trick is to use the real multiplication to construct the corresponding complex scalar multiplication on $E \times E$. What is harder is to construct the norm. (but, Daniel gave a method) – James S. Cook Oct 28 '17 at 16:07