Inner product of a quotient space Let $V$ be an $K$-linear space with inner product, $S \subset V$ a linear subspace and $V/S$ the quotient space. I want to prove that the quadratic form $\psi: V/S \to K$ with $\psi([v])=min\{\|v'\|^2: [v]=[v']\}$ defines an inner product in the quotient.
Should I use the polarization formula $\varphi(v,w)=\frac{1}{4}(\psi(v+w)-\psi(v-w))$ to define the bilinear form, and then use that to prove the properties of the inner product? I'm really lost in that question. Thanks.
 A: As @user8675309 pointed out in their comment, the question features a mistake of category: after all, an inner product on the vector space $V/S$ is defined to be a map
\begin{align}
\varphi:V/S\times V/S&\to\mathbb K\\
(x,y)&\mapsto\varphi(x,y)
\end{align}
with certain properties. In particular it needs two arguments, while $\psi$ as defined in the question only takes one argument as input. Hence the question is not "Is $\psi$ an inner product?" -- because that would mix two different types of objects together -- but rather "Can $\psi$ somehow generate an inner product on $V/S$, and if so, how?". And for that second question your idea already goes into the right direction.
In terms of category, $\psi$ is closer to a norm on $V/S$ (one input, non-negative output) so

*

*relating $\psi$ to a norm

*showing that this norm satisfies the parallelogram law

*defining an inner product via this norm (polarization identity), as is already written in the original question

to me seems like the best course of action.
One readily verifies that given any $v\in V$
$$
\psi([v])=\min_{s\in S}\|v+s\|^2=\min_{s\in S}\langle v+s,v+s\rangle
$$
and thus $\psi(\alpha x)=|\alpha|^2\psi(x)$ for all $\alpha\in\mathbb K$, $x\in V/S$. But a norm needs to be homogeneous, that is, $\|\alpha x\|=|\alpha|\|x\|$ which suggests taking the square root of $\psi$. Hence the object of interest becomes
\begin{align}
\sqrt\psi:V/S&\to[0,\infty)\\
x&\mapsto\min_{y\in x}\|y\|\,.
\end{align}
Let us quickly show that $\sqrt\psi$ is indeed a norm: it is homogeneous (by construction), positive definite (if $\sqrt\psi(x)=0$, then $0\in x$ by definiteness of $\|\cdot\|$, so $x=[0]$), and satisfies the triangle inequality: for all $v,w\in V$
\begin{align}
\sqrt\psi([v]+[w])=\sqrt\psi([v+w])&=\min_{s\in S}\|v+w+s\|\\
&\leq \min_{s\in S}\Big(\Big\|v+\frac{s}2\Big\|+\Big\|w+\frac{s}2\Big\|\Big)\\
&\leq \Big(\min_{s\in S}\Big\|v+\frac{s}2\Big\|\Big)+\Big(\min_{s\in S}\Big\|w+\frac{s}2\Big\|\Big)\\
&=\big(\min_{\tilde s\in S}\|v+\tilde s\|\big)+\big(\min_{\tilde s\in S}\|w+\tilde s\|\big)=\sqrt\psi([v])+\sqrt\psi([w])\,.
\end{align}
Now because $\sqrt\psi$ is a norm on $V/S$, all that is left to show is that it satisfies the parallelogram law, i.e. for all $x,y\in V/S$
$$
2\psi(x)+2\psi(y)=\psi(x+y)+\psi(x-y)\tag{1}
$$
Then it would generate an inner product on $V/S$ via
\begin{align}
\varphi:V/S\times V/S&\to\mathbb K\\
(x,y)&\mapsto\frac{ (\sqrt\psi(x+y))^2+(\sqrt\psi(x-y))^2 }4=\frac{ \psi(x+y)+\psi(x-y)}4\tag{2}
\end{align}
if $\mathbb K=\mathbb R$ (and analogously if $\mathbb K=\mathbb C$).
In order to show that (1) holds, we have to "explicitly" calculate $\psi([v])$. For this we need the following auxiliary result which is another formulation of the Hilbert projection theorem:

Lemma. Given $v\in V$ and any orthonormal basis $(z_i)_{i=1}^m$ of the orthogonal complement $S^\perp$ of $S$ in $V$, one has
$$
\psi([v])=\sum_{i=1}^m|\langle z_i,v\rangle|^2\,.
$$
Proof: By the projection theorem the minimizer $s_v\in S$ of $\min_{s\in S}\|v-s\|$ is the unique element in $S$ such that $v-s\in S^\perp$. Hence $s_v=\pi_S(v)$ is the orthogonal projection of $v$ onto $S$, and using the Pythagorean theorem we compute
$$
\psi([v])=\|v-\pi_S(v)\|^2=\|\pi_{S^\perp}(v)\|^2=\Big\|\sum_{i=1}^m\langle z_i,v\rangle z_i\Big\|^2=\sum_{i=1}^m|\langle z_i,v\rangle|^2\,.\tag*{$\square$}
$$

With this we can verify (1) by means of a direct computation: given any orthonormal basis $(z_i)_{i=1}^m$ of $S^\perp$ one (in abuse of notation) finds
\begin{align}
\psi([v+w])+\psi([v-w])&=\sum_{i=1}^m|\langle z_i,v\pm w\rangle|^2\\
&=\sum_{i=1}^m\langle z_i,v\pm w\rangle\langle v\pm w,z_i\rangle\\
&= \sum_{i=1}^m\big(\langle z_i,v\rangle\langle v,z_i\rangle\underbrace{\pm \langle z_i,v\rangle\langle w,z_i\rangle}_{\text{cancel out}}\pm \underbrace{\langle z_i,w\rangle\langle v,z_i\rangle}_{\text{cancel out}}+ \langle z_i,w\rangle\langle w,z_i\rangle\big)\\
&=2\sum_{i=1}^m|\langle z_i,v\rangle|^2+2\sum_{i=1}^m|\langle z_i,w\rangle|^2=2\psi([v])+2\psi([w])\,.
\end{align}
And with this we are done -- as explained above, because $\sqrt\psi$ is a norm which satisfies the parallelogram law it generates an inner product on $V/S$ via (2).
