0
$\begingroup$

Very basic question, but one I would like a formal understanding as to why.

For a normed linear space $(X,|| \cdot ||)$, where $X$ is over the field $\rm l\!F$ and $||\cdot ||:X\rightarrow\rm l\!F$.

Want I want to understand formally is why $||a||=|a|$, for $a\in \rm l\!F$. Is this wrong instead in general?

$\endgroup$
3
  • 1
    $\begingroup$ This is just the norm of $a$ when you think of $a$ as a vector in the one dimensional vector space $F$ over $F$. $\endgroup$ Commented Oct 19, 2022 at 21:10
  • $\begingroup$ The only meaning I can give to that is: any absolute value $|~|:\mathbb F\to\mathbb R$ may be considered as a norm on the $\mathbb F$-vector space $\mathbb F.$ $\endgroup$ Commented Oct 19, 2022 at 21:14
  • $\begingroup$ I agree now after reading both comments. Norm of a only has meaning if a itself is an element in X. So it's false in general and only true in very specific occasions. $\endgroup$
    – Waaal
    Commented Oct 19, 2022 at 21:29

1 Answer 1

0
$\begingroup$

As the comments brought to my attention, it is not true in general that the norm of a scalar is well defined. Strictly speaking, we need $\mathbb{F}\subset X$ for the norm of a scalar to make sense.

But there are also some cases, where one can see the scalars as elements of $X$ that act in the same way for scalar multiplication. For example, $C(\mathbb{R})$ the set for continuous real valued functions. $(C(\mathbb{R}),||\cdot||_{max})$ is a normed linear space and in this case one can relate the scalars to constant functions. Still, $||a||$ is undefined for $a\in\mathbb{R}$, but for a$(x)=a$ we have $||$a$||=|a|$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .