Question

Let X=C([0,1]), the space of continous functions on the interval [0,1], and let \(||f||_\infty = sup\{|f(x)|~~:~~x\in[0,1]\}\). Show that \(|| ~~||_\infty\) is a vector space norm.
I do not know how to start.
Please, help

Answer 1

@SithsAndGiggles

Answer 2

Ooo I also need to know how to do this one :c

Answer 3

I think I have to have || || satisfies:
1)|| f(x)|| > 0 for all x, =0 if f(x) =0
2) ||c f(x) ||= |c| ||f(x)|| for every c in R
3)||f(x)+g(x)|| \(\leq\) ||f(x) ||+||g(x)||

Answer 4

the last one is hardest part.

Answer 5

@ikram002p @phi

Answer 6

i dont get the notations hmm
i'll read again >.<

Answer 7

i dont get any clue sorry

Answer 8

Thanks for your time :)

Answer 9

Have you established any of the conditions yet?

Answer 10

:) not yet. I do not know how to start.

Answer 11

The definition of a norm states that for a vector space \(X\), a function \(\|\cdot\|:X\to\mathbb{R}\) must satisfy, for all \(k\in\mathbb{C}\) and \(x,y\in X\),
\[\begin{array}{cl}
(1)&\|x\|=0~~\iff~~x=0\\
(2)&\|kx\|=|k|~\|x\|\\
(3)&\|x+y\|\le\|x\|+\|y\|
\end{array}\]
Notice that if \(x\not=0\), if the second axiom holds then you can establish non-negativity.

Answer 12

The first condition is fairly simple to manage:
\[\|f\|_\infty=0~~\implies~~\sup|f(x)|=0~~\implies~~0\le|f(x)|\le0\]
which means \(f(x)=0\).

Answer 13

I suppose you'll have to show the other direction as well, but that should be easy enough for you.

Answer 14

hold on, is it not that sup |f(x) | in [0,1] = 1?

Answer 15

No, we are assuming the norm is equal to 0, and by definition of the given norm, you have
\[\|f\|_\infty=\sup|f(x)|=0\]

Answer 16

Ok,

Answer 17

Can you show why the second axiom holds? Use properties of the supremum.

Answer 18

Nah, I go around again. ha!!

Answer 19

The property I'm referring to is that for \(k\ge0\), you have \(\sup kx=k\sup x\).

Answer 20

but in book, that is definition,

Answer 21

They just say: ||.||_infty : M-->R is a norm

Answer 22

Assume \(k>0\), then
\[\|kf\|=\sup|kf(x)|=|k|\sup|f(x)|=|k|~\|f\|\]

Answer 23

Is that all I have to do?

Answer 24

For the second axiom, yes.

Answer 25

How about the last axiom?

Answer 26

I suspect it might be more complicated than what I have here, but it should give you a rough idea of how to approach it. Let \(f,g\in X\), then
\[\begin{align*}\|f+g\|&=\sup|f(x)+g(x)|\\
&\le\sup\bigg(|f(x)|+|g(x)|\bigg)\\
&=\sup|f(x)|+\sup|g(x)|\\
&=\|f\|+\|g\|
\end{align*}\]

Answer 27

Refer to Prop. 2.16 for a proof regarding the first step: https://www.math.ucdavis.edu/~hunter/m125b/ch2.pdf

Answer 28

Thank you so much.

Answer 29

yw