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We admit the following definition of Fréchet space.

The definition of Fréchet space: Let $X$ be a linear space and let $\{p_i:X\to\mathbb R^+\}_{i\in \mathbb N}$ be a countable family of seminorms such that $p_i(x)\ne 0$ for some $i\in\mathbb N$ whenever $x\ne 0$. Then we can equip $X$ with a bounded metric $$d(x,y)=\sum_{i=1}^\infty \frac{1}{2^i} \frac{p_i(x-y)}{1+p_i(x-y)}.$$ Such a metric space $(X,d)$ is called a Fréchet space if it is complete.

The question is:

Let $(X,\{p_i\}_{i\in\mathbb N})$ be a Fréchet space, and $A$ be a closed linear subspace of $X$. We want to show that $X/A$ is still a Fréchet space (in the sense of the above definition).

In Wikipedia, it says that a quotient of a Fréchet space by a closed subspace is a Fréchet space, so I believe that the answer is true.

My attempt

  1. First we can show that each seminorm $p$ on $X$ induces a seminorm $\tilde p$ on $X/A$ as follows:

The $\tilde q$ is given by $$ \tilde{p}:X/A\to\mathbb R,\quad [x]=x+A\mapsto \inf_{a\in A}p(x+a). $$ In the next we show that $\tilde q$ is a seminorm. Fix some $\varepsilon>0$. For $x,y\in X$, we pick $a,b\in A$ such that $$ p(x+a)-\varepsilon\le\tilde{p}([x])\quad\text{and}\quad p(y+b)-\varepsilon\le\tilde{p}([y]). $$ Therefore, $$ \tilde{p}([x])+\tilde{p}([y])\ge p(x+a)+p(y+b)-2\varepsilon \ge p(x+y+a+b)-2\varepsilon \ge \tilde{p}([x+y])-2\varepsilon. $$ By the arbitrariness of $\varepsilon$, we get the subadditivity of $\tilde{p}$. Since $A$ is a linear subspace, $$ \tilde{p}(c[x])=\inf_{a\in A}p(a+cx)=|c| \inf_{a\in A} p(a+x)=|c| \tilde{p}([x]). $$ So we get the absolute homogeneity.

  1. Therefore, each $p_i$ induces a seminorm $\tilde{p_i}$ on $X/A$. Then we want to show that $(X/A,\{\tilde{p_i}\}_{i\in\mathbb N})$ forms a Fréchet space.

But I got stuck on showing that $\tilde p_i([x])\ne 0$ for some $i\in\mathbb N$ whenever $[x]\ne 0$. I tried as follows.

Suppose $[x]\in A$ and $[x]\ne 0$. Then $x\in X\setminus A$. Since $X$ is complete and $A$ is closed, we can pick $a\in A$ with $d(x,a)=d(x,A)$. Since $x\in X\setminus A$, we know $d(x,a)>0$, i.e. $$ \inf_{a\in A} \sum_{i=1}^\infty \frac{1}{2^i}\frac{p_i(x-a)}{1+p_i(x-a)}>0. $$ What I need to show is that $\tilde p_i([x])=\inf_{a\in A} p_i(x-a)> 0$ for some $i$. I got stuck on this.

Any hints will be highly appreciated!

Chiquita
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1 Answers1

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In general, without further hypotheses on the seminorms, it is not true that $\tilde p_i([x])=0$ for every $i$ implies $[x]=0$. For example, take $X=\mathbb R^{\mathbb N}$ the space of all real sequences, $p_i(x):=|x_i|$, and $A=\mathbb R\cdot v$, where $v=(1,1,1,\dots)$. The subspace $A$ is closed, but the $\tilde p_i$ defined on $X/A$ are all identically zero.

To fix the above issue, it is sufficient to consider a family of seminorms on $X$ such that $p_i\leq p_j$ whenever $i<j$. You can always assume this condition in the definition of Fréchet space without loss of generality (if it’s not satisfied, consider the new family of seminorms $(q_i)_i$, with $q_i=\sum_{i=1}^ip_i$…; a space with such a family of seminorms is usually called graded), and it is often useful to consider for technical reasons. If this condition is true, then it is indeed true that the $\tilde p_i$ separate points: if $\tilde p_i([x])=0$ for all $i$, then there exists $a_n\in A$ such that $p_n(x-a_n)\leq 2^{-n}$. Since $p_i\leq p_n$ eventually, $p_i(x-a_n)\to 0$ for any fixed $i$, which implies that $a_n$ converges to $x$, thus $x\in A$ if $A$ is closed.

Lorenzo Pompili
  • 5,569