I need to show that if $p_i$ is a seminorm, then $\|\cdot\|$ is a norm s.t $\|x\|:=\max p_i(x)$, for every $x\in V$, where $V$ is a locally convex topological vector space.

I have shown that homogeneity and triangle inequality hold true, but I am unable to show that $||x|| = 0 \iff x=0$.

If $x=0$ then $\|x\| = 0$. But when $\|x\|=0$ $\Rightarrow max(p_i(x))=0$. For this I am only able to say since $p_i(x)$ is non negative that $p_i(x)=0$. And here is where I am stuck. As far as I know it is not true in seminorms that this implies $x=0$. What is it that I am missing?

  • $\begingroup$ Do you mean that $\{p_i\}_i$ is a family of seminorms? $\endgroup$ – popoolmica Sep 6 '19 at 20:44
  • $\begingroup$ I think that's what the professor assumed, otherwise I don't see why he'd use this notation. The exercise continued with: " prove that the locally convex topology of V is equal to the ||.|| topology". But I was stuck way before that. ;-( $\endgroup$ – Iniciador Sep 6 '19 at 20:48
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    $\begingroup$ It is not true. For example, in the plane $p(x) = |x_1|$ is a semi norm but not a norm. You need more conditions so that for any $x$ there is some seminorm $q$ such that $q(x) > 0$. $\endgroup$ – copper.hat Sep 6 '19 at 20:53
  • $\begingroup$ @copper.hat That makes sense, maybe I missed something when he was stating the exercise... $\endgroup$ – Iniciador Sep 6 '19 at 20:55
  • $\begingroup$ @Iniciador As norms, seminorms are always assumed to attain positive values only (if you are unsure about this, look it up in some book on functional analysis). Hence, your argument above shows that $\Vert \cdot \Vert$ is indeed a norm. It is not harder than that. The other statement you want to show is not too difficult, too :) $\endgroup$ – Adriano Sep 6 '19 at 23:23

Some assumptions are missing in your statement. If $(p_i)$ is just some family of semi-norm then the statement is obviously false. (Take a single semi-norm which is not a norm). So you have to assume that $p_i$'s generate the topology of $V$. This means that sets of the form $\{x:p_{i_1}(x) <r_1,p_{i_2}(x) <r_2,..,p_{i_N}(x) <r_N\}$ ( $N$, $i_j$'s arbitrary and $r_j$'s $>0$) form base for the neighborhoods of $0$. Under this assumption it follows that $p_i(x)=0$ for all $i$ implies that $x$ belongs to every neighborhood of $0$. Assuming that ths space is Hausdorff this implies $x=0$.


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