Seminorm proof of a function

Question

I have an example in a book which is not very clear to me :

let $E$ vector space made of numerical functions (or complex) $f$ defined on a set $A$. $\forall a \in A, N_a : f \rightarrow |f(a)|$ is a seminorm

but the problem here is that when i'm trying to verify that $N(x)=0$ then $x=0$ it's not that clear because :

if $N_a(f)=|f(a)|=0 \iff \forall a \in A, f(a)=0 \iff f=0$ which is the definition of a norm and not a seminorm.

it seems to be obvious for the author but i'm not convinced of it yet.

thank you in advance for your help !

Answer 1

If $V,W$ are vector spaces and $|\cdot|$ is a norm (ort seminorm) on $W$, and $T\colon V\to W$ is linear, then $v\mapsto |Tv|$ is a seminorm on $V$

Answer 2

$N_a(f)=0$ does not imply $f(x)=0$ for all $x\in A$. It only implies $f(a)=0$.

Answer 3

Perhaps an example will help. Let $A = \{7, 13\}$. Then $E$ is the vector space of complex-valued functions on $\{7, 13\}$. Let $a = 7$; then $N_7$ is a seminorm on $E$. Letting $a=13$, it's also true that $N_{13}$ is a seminorm on $E$.

For instance, letting $f(7) = i, f(13) = 0$, we have $N_7(f) = |f(7)| = |i| = 1$, whereas $N_{13}(f) = |f(13)| = 0$. Since $f$ is non-zero yet $N_{13}(f) = 0$, it's clear that $N_{13}$ cannot be a norm, and from the triangle inequality it's a seminorm.