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Consider $X$ a normed space with norm $\|\cdot\|$ and $M$ a closed subspace of $X$. In the quotient space $X/M$ we define the quotient norm: $$|||\hat{x}||| = \inf_{y\in M} \|\hat{x}+y\|, \quad \hat{x}\in X/M.$$

I'm trying to prove the well-definiton of this norm, that is, given $\hat{x_1}$, $\hat{x_2}\in X/M$ such that $x_1-x_2\in M$ then it must ve verified that $|||\hat{x_1}|||=|||\hat{x_2}|||$.

By means of triangle inequality property of the norm $\|\cdot\|$ I managed to show that $$|||\hat{x_1}|||-|||\hat{x_2}||| \leq \|x_1-x_2\|,$$ but that doesn't help to conclude what I want. I would really appreciate if someone can please help me with this.

Zim
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Juan David Samboní
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2 Answers2

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Suppose $\hat{x}_1-\hat{x}_2=m\in M$. Since $M$ is a subspace, taking the $\inf$ over arbitrary $z\in M$ yields the same result as taking the $\inf$ over $m-y$ for arbitrary $y\in M$ (since $m-y$ is still in $M$ and any value at $z$ is achieved when $y$ is $m-z$). Using this change of variables ($z$ for $m-y$),

$$|||\hat{x}_1|||=\inf_{z\in M}\|\hat{x}_1-z\|=\inf_{y\in M}\|\hat{x}_1-m+y\|=\inf_{y\in M}\|\hat{x}_2+y\|=|||\hat{x}_2|||$$

Zim
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Here is a proof for general linear spaces that admit a translation invariant metric. The case of a norm will follow in part from this.

Suppose that $d$ is translation invariant metric on linear topological space $X$ compatible with the linear topology $\tau$. Let $M$ be a closed linear subspace in $X$, $\pi$ the quotient map, and $\tau_M$ the topology on $X/M$ induced by $\pi$. Define $$ \rho(\pi(x),\pi(y)):=\inf\{d(x-y,z): x\in M\}=d(x-y,M) $$ Notice that $d(x-y,M)=0$ iff $x-y\in \overline{M}=M$. Hence $$ \rho(\pi(x),\pi(y))=\rho(\pi(x)-\pi(y),\pi(0)), $$ and $\rho(\pi(x),\pi(y))=0$ iff $\pi(x)=\pi(y)$. Since $$ d(x-y,z)\leq d(0,z+y-x)=d(-z,y-x)=d(y-x,-z), $$ it follows that $\rho(\pi(x),\pi(y))=\rho(\pi(y),\pi(x))$. From $$ d(x-y,z)\leq d(x-y,u-y+z')+d(u-y+z',z)=d(x-u,z')+d(u-y,z-z'), $$ we conclude that $\rho(\pi(x),\pi(y))\leq \rho(\pi(x),\pi(u))+\rho(\pi(u),\pi(y))$. This shows that $\rho$ is a translation invariant metric on $X/M$. Since $d(x,0)=d(x+z,z)$ for all $z$, $$ \pi\big(\{x:d(x,0)<r\}\big)=\{\pi(x):\rho(\pi(x),\pi(0))<r\} $$ All this shows that if $d$ is a translation invariant metric that generates the topology $\tau$ on $X$ then, $\rho$ is a translation invariant metric on $X/M$ that generates $\tau_M$.

Back to norms:

If $d$ corresponds to a norm, then $$ \|\pi(x)\|_M:=\inf\|x-z,z\in M\|=d(x,M) $$ defines a norm on $X/M$. It suffices to show that $\|\pi(\alpha x)\|_M=\alpha\|\pi(x)\|_M$. If $\alpha=0$ there is nothing to prove. If $\alpha\neq0$ then, from $\|\alpha x-z\|=|\alpha|\|x-\alpha^{-1}z\|$ and $\alpha^{-1}M=M$ we conclude that $\|\pi(\alpha x)\|_M=\alpha\|\pi(x)\|_M$. Thus, $\|\;\|_M$ is a norm on $X/M$.

Complete metrics (Banach space in case of complete norm):

Suppose that $d$ is a complete translation invariant metric generating $\tau$. and let $\{\pi(x_n):n\in\mathbb{N}\}$ be a Cauchy sequence in $(X/M,\rho)$. Without loss of generality we may assume that $\rho(\pi(x_{n+1}-x_n),\pi(0))<2^{-n}$. Set $z_1=0$ and choose $z_2\in M$ such that $$ d(x_2+z_2-(x_1+z_1),0)<\frac12. $$ Proceeding by induction, we obtain a sequence $\{z_n:n\in\mathbb{N}\}\subset M$ such that $$ \rho(\pi(x_{n+1}),\pi(x_n)) \leq d(x_{n+1}+z_{n+1},x_n+z_n)<2^{-n} $$ Since $X$ is complete, there is $x^*\in X$ such that $d(x_n+z_n,x^*)\rightarrow0$. The continuity of $\pi$ implies that $\lim_n\pi(x_n+z_n)=\lim_n\pi(x_n)=\pi(x^*)$. This shows that $\rho$ is complete.

Mittens
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