# Defining an inner product from a norm which satisfies parallelogram law

Suppose we define inner product on complex inner-product space as the following :

$$\langle u,v\rangle =\frac{\|u+v\|^2 - \|u-v\|^2 + \|u+iv\|^2i - \|u-iv\|^2i}{4}$$

Given that the norm satisfies parallelogram inequality, I want to check the linearity condition of this inner product (I have checked the other conditions). How to apply parallelogram law to show that $\langle u+w,v\rangle = \langle v\rangle + \langle w,v\rangle$

$\newcommand{\norm}[1]{\left\| #1 \right\|} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}}$ It so happens that I wrote up a proof of this pretty recently. Here it is, mostly unaltered, except I changed the variables in question to $u$ and $v$. I am using $(u,v)_\C$ here to denote your $\langle u,v \rangle$; I first solve the real case and then the complex case follows directly.
First consider $(u_1 + u_2, v)_\R = \frac{\norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2}{4}$. Applying the parallelogram law to each factor, we write \begin{align*} \norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2 &= \left( 2\norm{u_1 + v}^2 + 2\norm{u_2}^2 - \norm{(u_1 + v) - u_2}^2 \right) \\ &\quad - \left( 2\norm{u_1}^2 + 2\norm{u_2 - v}^2 - \norm{u_1 + (u_2 - v)} \right) \\ &= 2\norm{u_2}^2 - 2\norm{u_1}^2 + 2\norm{u_1 + v}^2 - 2\norm{u_2 - v}^2 \end{align*} By symmetry (switching $u_1$ and $u_2$) we have the two equations \begin{align*} \norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2 &= 2\norm{u_2}^2 - 2\norm{u_1}^2 + 2\norm{u_1 + v}^2 - 2\norm{u_2 - v}^2 \\ \norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2 &= 2\norm{u_1}^2 - 2\norm{u_2}^2 + 2\norm{u_2 + v}^2 - 2\norm{u_1 - v}^2 \end{align*} Averaging (arithmetically) the two equations we vet \begin{align*} \norm{u_1 + u_2 + v}^2 - \norm{u_1 + u_2 - v}^2 &= \norm{u_1 + v}^2 - \norm{u_1 - v}^2 + \norm{u_2 + v}^2 - \norm{u_2 - v}^2 \end{align*} which when divided by four is $$(u_1 + u_2,v)_\R = (u_1,v)_\R + (u_2,v)_\R.$$
• How can we prove that $\langle \lambda x, y \rangle = \lambda \langle x, y \rangle$ ? Commented Nov 15, 2019 at 19:49