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 $X/M$.


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 $X$.

  • $\begingroup$ Thanks for answering my question. But I am sorry I am not quite understand the answer of the second question. Why to show pM are continuous? Should we show that the quotient topology is equal to the topology defined by pM? Thank you! $\endgroup$ – Answer Lee Jun 21 '17 at 15:19
  • $\begingroup$ @AnswerLee I updated my answer. Let me know if it is clear now. $\endgroup$ – Pantelis Sopasakis Jun 23 '17 at 16:43
  • $\begingroup$ It is perfect right now. Thank you so much! $\endgroup$ – Answer Lee Jul 4 '17 at 18:47
  • $\begingroup$ You're welcome. You can also accept the answer if you like :) $\endgroup$ – Pantelis Sopasakis Jul 4 '17 at 22:53
  • $\begingroup$ Can you help me with my latest question about alaoglu's Theorem? It drives me crazy!! Thank you so much! Link: math.stackexchange.com/questions/2346510/… $\endgroup$ – Answer Lee Jul 5 '17 at 3:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.