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Suppose the seminorm on X is real valued, $U=X\bigcap\lbrace x:\vert x\vert =0\rbrace$, Y=X/U, and $\pi:X\rightarrow Y$ is the canonical projection.

I want to show that $\vert\bullet\vert\circ\pi^{-1}$ is a norm on Y.

But I do not know how to prove that $\vert\bullet\vert\circ\pi^{-1}(y_1+y_2)\le \vert\bullet\vert\circ\pi^{-1}(y_1)+\vert\bullet\vert\circ\pi^{-1}(y_2)$, but it is one of the necessary condition, right?

  • The inverse of the projection map isn’t necessarily well-defined. If it were then you’d just take advantage of the fact that the semi-norm has the triangle inequality property that you’re looking for in building the norm. Moreover, I would like to mention that the existence of a semi-norm is not enough to build a norm on a general topological vector space. – user328442 Dec 17 '18 at 14:19

1 Answers1

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You have a seminorm $|\cdot|$ on $X$ and define $$ |\pi(x)|=|x| $$ over $Y=X/U$, where $U=\{x\in X:|x|=0\}$ ($U$ is known to be a subspace of $X$). Every element of $Y$ is of the form $\pi(x)$ for some $x\in X$.

First of all, you have to prove that $|\cdot|$ is well defined on $Y$. If $\pi(x)=\pi(x')$, then $x-x'\in U$, so $$ |x|\le |x-x'|+|x'|=|x'| $$ and, similarly, $|x'|\le|x|$. Thus $|x|=|x'|$ and the proof is complete.

The proof of the seminorm properties is essentially trivial; for the triangle inequality, $$ |\pi(x)+\pi(y)|=|\pi(x+y)|=|x+y|\le|x|+|y|=|\pi(x)|+|\pi(y)| $$ and similarly for the other properties. We have a norm because $$ |\pi(x)|=0 \iff |x|=0 \iff x\in U \iff \pi(x)=\pi(0) $$ so $|\pi(x)|=0$ implies $\pi(x)=\pi(0)$.

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