Norms Induce by Inner Products (Complex Case)

Question

I have just proved that if $||\cdot||$ satisfy $||u+v||^{2}+||u-v||^{2}=2||u||^{2}+2||v||^{2}$, then there exists an inner product such that $||u||^{2}= \langle u,u \rangle$ is given by $\langle u,v \rangle =\frac{1}{4}(||u+v||^{2}-||u-v||^{2})$. Now, I want to extend this to the complex numbers. My friend insists that one would have to repeat all the arguments necessary to prove the real case separately for the complex numbers, but is that really necessary? Isn't sufficient to simply define $\langle u,v \rangle:= \frac{1}{4} \sum_{k=0}^{3}i^{k}||u+i^{k}v||^{2}$ show that $\langle iu,v \rangle =i\langle u,v \rangle $, $\langle u,v \rangle=\overline{\langle v,u \rangle}$ and then simply refer to the proof for the real numbers (point out off course that the imaginary and real parts are separated)?

Answer 1

You don't have to repeat all arguments. You are probably aware that the real and complex inner product are related by $\DeclareMathOperator{\real}{Re} \DeclareMathOperator{ima}{Im}$

$$\langle u,v\rangle_{\mathbb{R}} = \operatorname{Re} \langle u,v\rangle_{\mathbb{C}}.$$

So if you have a complex inner product, you get a real inner product for free, and from a real inner product, you obtain a complex inner product via

$$\begin{align} \langle u,v\rangle_{\mathbb{C}} &= \operatorname{Re} \langle u,v\rangle_{\mathbb{C}} + i\operatorname{Im} \langle u,v\rangle_{\mathbb{C}}\\ &= \real \langle u,v\rangle_{\mathbb{C}} - i \real \left(i\langle u,v\rangle_{\mathbb{C}}\right)\\ &= \real \langle u,v\rangle_{\mathbb{C}} - i\real \left(\langle u,iv\rangle_{\mathbb{C}}\right)\\ &= \langle u,v\rangle_{\mathbb{R}} - i \langle u,iv\rangle_{\mathbb{R}}. \end{align}$$

By that construction, it is clear that $\langle\cdot,\cdot\rangle_\mathbb{C}$ is a real-bilinear form, and it only remains to show that it is complex sesquilinear, that is, $\langle iu,v\rangle_\mathbb{C} = - i\langle u,v\rangle_{\mathbb{C}}$ and $\langle u, iv\rangle_\mathbb{C} = i\langle u,v\rangle_\mathbb{C}$, and hermitian, $\langle v,u\rangle_\mathbb{C} = \overline{\langle u,v\rangle_\mathbb{C}}$.