Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $V$ be a vector space with a seminorm $\|\cdot\|_s$. Then apparently we can turn $\|\cdot\|_s$ into a norm $\|\cdot\|$ on $V/W$ by defining $\|v + W\| = \inf_{w \in W} \|v + w\|_s$ where $W$ is any closed subspace of $V$.

It's clear to me that if $V_0$ denotes the kernel of the seminorm $\|\cdot\|_s$ then $\|\cdot\|_s$ turns into a norm on $V/V_0$. What is not so intuitive to me is what happens if $W$ is disjoint from $V_0$. I think the fact that the norm $\|\cdot\|$ defined above then is a norm means that for every $v_0 \in V_0$ there is a sequence in $W$ converging to it. This holds because $0 \in W$ hence there is a sequence $w_n$ converging to $0$ and if $v_0 \in V_0$ then $v_0$ is in every neighbourhood of $0$, hence $w_n$ also converges to $v_0$. Is this correct?

If yes, would you show me some concrete examples illustrating this to help me develop some intuition? Thanks.

share|cite|improve this question
You cannot «turn $\lVert\cdot\rVert_s$ into a norm» on the same vector space. It would be best to make your first paragraph be more precise. – Mariano Suárez-Alvarez Jul 9 '12 at 7:00

In general, we can turn $\|\cdot \|_s$ into a norm on $V/(W+ V_0)$ by the method described. To see this, note that if a norm is defined in this way on some quotient space $V'$ of $V$ we automatically get a surjection $V/W\to V'$ while in order for the norm to be a norm at all it the kernel must be $0$, so we get a surjection $V/V_0\to V'$. Putting these together gives us a surjection $V/(W+V_0)\to V'$, so all that remains is to verify $\|\cdot\|$ is a norm on $V/(W+V_0)$, which follows from composing the obvious maps $V\to V/V_0\to V/(W+V_0)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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