Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite
1

The following question is from Kreyszig, Introductory Functional Analysis.

$(M2) d(x,y)=0$ iff $x=y.$

$(M3) d(x,y)=d(y,x)$.

$(M4) d(x,y)\le d(x,z)+d(z,y)$.

Show that (M3) and (M4) can be obtained from (M2) and

$d(x,y)\le d(z,x)+d(z,y)$.

If I assume (M3), I can show (M4) follows from (M3) and $d(x,y)\le d(z,x)+d(z,y)$. I just can't see how to get (M3) from (M2) and $d(x,y)\le d(z,x)+d(z,y)$. Any tips?

share|improve this question
Try setting $z=y$ which will give one side of the inequality. Swapping variables gives the other. – copper.hat Jan 9 at 6:37

2 Answers

up vote 4 down vote accepted

If you choose $z$ carefully in the inequality you can get a comparison between $d(x, y)$ and $d(y, x)$. Then swapping $x$ and $y$ you can get another comparison from which you should be able to deduce $(M3)$.

share|improve this answer
Thanks. I initially did that but it seemed to me that I had only shown it to be true for those special choices of z. Or am I missing something? – jim Jan 9 at 6:41
1  
I'm not sure what you mean by that. There is no $z$ in $(M3)$. – Michael Albanese Jan 9 at 6:42
Also, doing it that way doesn't require using (M2). So I thought there must be way that requires (M2)? – jim Jan 9 at 6:43
1  
I'm pretty sure you do need $(M2)$. Note that in the inequality you have two terms on the right hand side, but by making a specific choice for $z$, and using $(M2)$, you only get one. – Michael Albanese Jan 9 at 6:45
Of course, you're right (about no z term in M3)! And yes, I do remember using M2 now. Thank you very much. I understand now. – jim Jan 9 at 6:50
up vote 2 down vote

Set $z=y$, we get $d(x, y) \leq d(y, x) + d(y, y) = d(y, x)$. Flip $x$ and $y$ to get equality.

share|improve this answer
Try to give a hint, not the whole idea. – AD. Jan 9 at 6:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.