Quotient topology defined by seminorms

Question

Please Prove this:

If $p$ is a seminorm on $X$, where $M$ is a linear manifold in $X$ and ${p}_M:X/M \rightarrow[0,\infty)$ is defined by $p_M(x+M)=\inf\{p(x+y): y \in M\}$ then $p_M$ is a seminorm on $X/M$.

If $X$ is a locally convex space and $A$ is the family of all continuous seminorms on $X$ , then the family $B=\{p_M: p \in A\}$ defines the quotient topology on $X/M$.

Answer 1

The mapping $p_M:X/M\to[0,+\infty)$ is, indeed, a seminorm. Let us use the notation $[x]=x+M\in X/M$

First, for all $[x]\in X/M$, $$p_M([x])= p_M(x+M) = \inf\{p(x+y); y\in M\}\geq 0.$$

Second, \begin{align} p_M(\alpha [x]) &= p_M(\alpha x + M) = \inf \{p(\alpha x + y); y \in M\}\\ &=\inf \{p(\alpha (x + \tfrac{y}{\alpha})); y \in M\}\\ &=\inf \{|\alpha| p(x + z); z \in M\}\\ &=|\alpha|p_M([x]). \end{align}

Third, for $[x],[x']\in X/M$, \begin{align} p_M([x]+[x]') &= p_M(x+x'+M)\\ &=\inf \{p(x+x'+y); y \in M\}\\ &=\inf \{p(x+x'+z+z'); z+z'\in M\}\\ &\leq \inf \{p(x+x'+z+z'); z,z'\in M\}\\ &\leq \inf \{p(x+z)+p(x'+z'); z,z'\in M\}\\ &= p([x]) + p([x']). \end{align}

Note that we can also prove that $p_M$ is a norm if $M$ is taken to be a closed subspace.

For the second question, we claim that the sets $V_\epsilon = \{[x]\in X/M; p_M([x]) < \epsilon\}$ with $\epsilon>0$ produce the topology in $X/M$.

We will in fact show that the topology $\tau'=\{p_M^{-1}(a,b), a, b\in \mathbb{R}, 0 \leq a < b\}$ is equal to the topology $\tau$ of $X/M$.

I will assume that $X$ is a locally convex space whose topology is exactly that generated by the seminorms of $A$. It suffices to show that $p_M$ are continuous. We have

\begin{align} p_M([x]) = \inf \{p(x+y);y\in M\} \overset{y=0}{\leq} p(x), \end{align}

therefore, we can see that $p_M$ is continuous at 0, thus continuous (because $p_M$ is a sublinear functional).

This proves that $\tau' \subseteq \tau$. Indeed, by definition $p_M^{-1}((a,b))$ is an open set in $X/M$ because of the continuity of $p_M$.

It is known that seminorms, which are not everywhere equal to $0$ $^\ast$, are open mappings (proof), i.e., they map open sets in their domain space to open sets in their range. Any $W\in \tau$ is mapped to an open set $p_M(W)$ in $\mathbb{R}_+$, meaning $\tau \subseteq \tau'$ which proves the assertion.

$^\ast$ We assumed that there exists at least one $p\in A$ which is not equal to $0$ over $X$.