Show that there is an innerproduct on $X/Y$ such that $\langle\langle{x+Y,x+Y}\rangle\rangle={\lvert\lvert\lvert{x+Y}\rvert\rvert\rvert}^2$ Let $X$ be an inner product space. Let $Y$ be a closed subset of $X$. Show that there is an innerproduct on $X/Y$ such that $\langle\langle{x+Y,x+Y}\rangle\rangle={\lvert\lvert\lvert{x+Y}\rvert\rvert\rvert}^2$
To begin with, let $H$ be the completion of $X$. Let $G$ be the closure of $Y$ in H. Then $H=G\oplus{G^{\perp}.}$Then for $x_1,x_2 \in H/G$, let's define $\langle\langle{x_1+G,x_2+G}\rangle\rangle=\langle{z_1,z_2}\rangle$, where $x_1=y_1+z_1,x_2=y_2+z_2$.
I am unable to relate this to $X/Y$ somehow although I see that $G\cap X=Y$.
 A: The norm on $X/Y$ is $\|x\|_{X/Y}=\inf_{y\in Y}\|x-y\|_{X}$; this is a valid norm because $Y$ is closed. If $x_1,x_2 \in X$ and $y_1,y_2 \in Y$, then the parallelogram law gives
$$
          \|(x_1-y_1)+(x_2-y_2)\|^{2}+\|(x_1-y_1)-(x_2-y_2)\|^{2}\\
       = 2\|x_1-y_1\|^{2}+2\|x_2-y_2\|^{2}.
$$
Therefore,
$$
            \|x_1+x_2\|_{X/Y}^2+\|x_1-x_2\|_{X/Y}^2 \le 2\|x_1-y_1\|^2+2\|x_2-y_2\|^{2}
$$
Taking inf over $y_1,y_2$ gives
$$
     \|x_1+x_2\|_{X/Y}^2+\|x_1-x_2\|_{X/Y}^2 \le 2\|x_1\|_{X/Y}^2+\|x_2\|_{X/Y}^2.
$$
Because you start with equality, you can also write
$$
     2\|x_1\|_{X/Y}^{2}+2\|x_2\|_{X/Y}^{2} \le \|x_1+x_2-y_1-y_2\|^2+\|x_1-x_2-y_1+y_2\|^2
$$
If $y_3,y_4 \in Y$, then $y_1+y_2=y_3$ and $y_1-y_2=y_4$ is solvable, which gives
$$
   2\|x_1\|_{X/Y}^{2}+2\|x_2\|_{X/Y}^{2} \le \|x_1+x_2-y_3\|^2+\|x_1-x_2-y_4\|^2
$$
Taking inf over $y_3, y_4 \in Y$, which are arbitrary elements of $Y$, gives
$$
   2\|x_1\|_{X/Y}^{2}+2\|x_2\|_{X/Y}^{2} \le \|x_1+x_2\|_{X/Y}^2+\|x_1-x_2\|_{X/Y}^2
$$
Therefore, the norm $\|\cdot\|_{X/Y}$ satisfies the parallelogram law, and that means that the following generates an inner product with associated norm $\|\cdot\|_{X/Y}$:
$$
             (x_1,x_2)_{X/Y}=\frac{1}{4}\sum_{n=0}^{3}i^{n}\|x_1+i^n x_2\|_{X/Y}^{2}
$$
