$\varphi(x)= \min\{\||x-u\||^2 ; u \in U \}$ is a quadratic form

Question

Let $V$ be a finitely generated Euclidean vector space, $U \subset V$ a $\mathbb {R}$ -subvector space. The mapping $\varphi: V \rightarrow \mathbb{R}$ is through

$\varphi(x)= \min\{\||x-u\||^2 ; u \in U \}$

defined. Now I need to show that $\varphi$ is a quadratic form on $V$.

So show following properties:

1. $q(a x)=a^{2} q(x) \quad \forall a \in K, \forall x \in V $

2. $q(x+y)+q(x-y)=2 q(x)+2 q(y) \quad \forall x, y \in V$

Here is what I got for 1.:

$\min\{\||ax-u\||^2 : u \in U\} = ... = \min\{-2a \langle x, u \rangle + a^2 \langle x,x \rangle + \langle u,u \rangle \}$

In the end I got this for 2.:

$\min\{2(\langle x, y\rangle- \langle x, u\rangle - \langle y, u\rangle)\} + \min \{2(- \langle x, y\rangle - \langle x, u\rangle + \langle y, u \rangle)\} = 2 \min\{-2 \langle x, u\rangle) + 2 \min\{-2 \langle y, u \rangle\}$

After several attempts of calculations, I'm done with my ideas to both of them. Maybe someone can give me the name of the problem so I can look it up online.

Thanks very much!

Answer 1

Per my comment, define the projection operator $\pi_U$ as the projection of any vector $x$ onto the subspace $U$. (In details, you may pick an arbitrary orthonormal basis $u_1,\cdots,u_k$ of $U$ and you can concretely define $$\pi_U:V\ni x\mapsto \sum_{i=1}^k\langle x, u_i\rangle u_i$$ But this is perhaps not necessary for your current purpose). Now prove that $$\phi(x)=\langle x - \pi_U(x), x - \pi_U(x)\rangle \tag{1}$$ by, for example, writing out the complete orthonormal basis as $u_1,\cdots,u_k,u_{k+1},\cdots,u_n$ where the first $k$ vectors span $U$, and then rewriting $\|x-u\|^2$ under this set of coordinates as $$\langle x, x \rangle + \sum_{i=1}^k \langle u, u_i\rangle^2-2\sum_{i=1}^n\langle x, u_i \rangle\langle u, u_i \rangle = \text{constant} + \sum_{i=1}^k \langle u, u_i\rangle^2-2\sum_{i=1}^k\langle x, u_i \rangle\langle u, u_i \rangle$$ In the above, treating $x$ as fixed and optimising over each $\{\langle u, u_i\rangle\}_{i=1}^n$ gives the optimal solution $\langle u, u_i\rangle=\langle x, u_i\rangle$, and therefore $u=\pi_U(x)$ indeed minimises $\|x-u\|^2$, which proves the equation $(1)$.

Note that the existence and the minimising property of the projection operator is guaranteed in any Hilbert space, so the actual proof can be much more general than my current version.