If [itex]x,y\in\mathbb{R}^n[/itex] are 2 vectors then is the following

[Ted123]

If $x,y\in\mathbb{R}^n$ are 2 vectors then is the following correct:

$\| x-y \| = 0 \iff x-y=0$ ?

If $f,g \in C [a,b]$ are 2 continuous functions on the closed interval $[a,b]$ then with $\displaystyle \| f-g \| = \left( \int^b_a (f-g)^2 \right)^{1/2}$ is the following correct: $$\| f-g \| \geq 0\; ;$$ $$\| f-g \| = 0 \iff f-g=0\; ?$$ (I think this follows from the fact that if a continuous function $h=f-g$ is non-negative and integrates to 0 over an integral $[a,b]$ (with $a<b$) then h is the zero function.)

 

[LCKurtz]

If $x,y\in\mathbb{R}^n$ are 2 vectors then is the following correct:

$\| x-y \| = 0 \iff x-y=0$ ?

You haven't indicated what norm $\|\cdot\|$ represents. But if it is a norm, that property is part of its definition. Perhaps you are to prove it for some proposed norm?

If $f,g \in C [a,b]$ are 2 continuous functions on the closed interval $[a,b]$ then with $\displaystyle \| f-g \| = \left( \int^b_a (f-g)^2 \right)^{1/2}$ is the following correct: $$\| f-g \| \geq 0\; ;$$ $$\| f-g \| = 0 \iff f-g=0\; ?$$ (I think this follows from the fact that if a continuous function $h=f-g$ is non-negative and integrates to 0 over an integral $[a,b]$ (with $a<b$) then h is the zero function.)

Again, those are correct properties of a norm, and your formula is in fact a norm. And your outline of how to prove it is correct. Is your problem to actually do the proof or what?

 

[Ted123]

No it's not a question as such, just trying to prove that the Euclidean metric and $L^2$-metric are indeed metrics (i.e. satisfy the 3 properties).

With the continuous function one, it says to prove $\| f-g \| = 0 \iff f=g$ you need to be careful and can't just state that $\| f-g \| =0 \iff f-g=0$ like with vectors (i.e. you need to use the continuous function property from Analysis)...

 

[LCKurtz]

You haven't indicated what norm $\|\cdot\|$ represents. But if it is a norm, that property is part of its definition. Perhaps you are to prove it for some proposed norm?


Again, those are correct properties of a norm, and your formula is in fact a norm. And your outline of how to prove it is correct. Is your problem to actually do the proof or what?

No it's not a question as such, just trying to prove that the Euclidean metric and $L^2$-metric are indeed metrics (i.e. satisfy the 3 properties).

With the continuous function one, it says to prove $\| f-g \| = 0 \iff f=g$ you need to be careful and can't just state that $\| f-g \| =0 \iff f-g=0$ like with vectors...

So what do you want? Are you stuck on the proofs?

 

[Ted123]

So what do you want? Are you stuck on the proofs?

No I'm OK but why can't you state for the 2nd one that $\| f-g \| = 0 \iff f-g = 0$ ?

 

[LCKurtz]

No I'm OK but why can't you state for the 2nd one that $\| f-g \| = 0 \iff f-g = 0$ ?

Because it relies on the theorem about integrals of continuous functions to which you refer. You are going to use that theorem either by referencing it as a known theorem or proving it. If that theorem is already discussed and "well known" in your class, you might claim that the result is "obvious". Otherwise you might need to give a reference to it from something you learned earlier or include a proof.

 

[Bacle2]

No I'm OK but why can't you state for the 2nd one that $\| f-g \| = 0 \iff f-g = 0$ ?

Find the integrals of f(x)=x on [0,1], and g(x)=x on [0,1]\{1/2}, and g(1/2)=10, and see what you get. Try then f(x)=x , and g(x)=x on [0,1]\{1/2,1/3} , and g(1/2)=g(1/3)=9.

LCKurtz: Would you please suggest on how to include the symbols in the bottom of your post in one's signature file?

 

[LCKurtz]

Find the integrals of f(x)=x on [0,1], and g(x)=x on [0,1]\{1/2}, and g(1/2)=10, and see what you get. Try then f(x)=x , and g(x)=x on [0,1]\{1/2,1/3} , and g(1/2)=g(1/3)=9.

How is that relevant? His space is C[a,b].

LCKurtz: Would you please suggest on how to include the symbols in the bottom of your post in one's signature file?

Click on My PF at the top left of your screen and select to Edit Signature.

 

[Bacle2]

Yes, sorry, I missed the fact that we were working on C[a,b]. Thanks for the reply on the signature.