Modulus and inequalities

  • #1

Main Question or Discussion Point

Hello,

I've got

[tex] |a|^{2} = |a - b + b|^{2} [/tex]

What can I do with this guy? Usually when the square isn't there I use the triangle inequality and things fall out pretty quick, for example,

[tex] |a| = |a - b + b| \leq |a-b| + |b| [/tex]

Is there something like that I can do with the orginial guy?
 

Answers and Replies

  • #2
33,173
4,858
Hello,

I've got

[tex] |a|^{2} = |a - b + b|^{2} [/tex]

What can I do with this guy? Usually when the square isn't there I use the triangle inequality and things fall out pretty quick, for example,

[tex] |a| = |a - b + b| \leq |a-b| + |b| [/tex]

Is there something like that I can do with the orginial guy?
Well, |a - b + b|2 = |a|2
 
  • #3
mathman
Science Advisor
7,760
415
What are you looking for?
 
  • #4
What are you looking for?
I know a property concerning |a-b|2 (it's less than some epsilon) and I know what |b|2 is, so, if I can break this down into an inequality, I can say something about |a|2. Does that make sense? this is part of a bigger problem concerning metric spaces.
 
  • #5
402
1
Are you looking for something like this?

[tex]\left|a\right|^2=\left|a^2\right|=\left|a^2-b^2+b^2\right|\leq\left|a^2-b^2\right|+\left|b^2\right|[/tex]

Now, if [itex]\left|a^2-b^2\right|\leq\epsilon[/itex] and [itex]\left|b^2\right|\leq\epsilon[/itex], then [itex]\left|a\right|^2\leq 2\epsilon[/itex]
 
  • #6
No, my a is a function in C[0,1] so that statement may not be true. I'm looking for a way to massage out that to which I responded to mathman.
 
  • #7
402
1
Well, my guess is that you are working with the sup norm, and want to prove something related to uniform convergence, but I think you must give more details. After all, you stated that you had information about |a2-b2|, |b2| and needed to relate these to |a2|. Given only this, I can't add anything else.
 
  • #8
I have

[tex]|f_{n}(t)-f(t)|^{2} \leq\epsilon [/tex]

and [tex] |f_{n}(t)|^{2} \leq 1 [/tex]

I wish to show

[tex] |f(t)|^{2} \leq 1 [/tex]

as well.

As much as I'd like to, I cannot say

[tex] |f(t)|^{2} = |f(t) - f_{n} (t) + f_{n} (t)|^{2} \leq |f_{n}(t) - f(t)|^{2} + |f_{n}(t)|^{2} [/tex]
 
  • #9
402
1
No, but you can say:

[tex]
|f(t)|^{2} = |f(t) - f_{n} (t) + f_{n} (t)|^{2} \leq |f_{n}(t) - f(t)|^{2} +2|f_{n} (t)||f_{n}(t) - f(t)|+ |f_{n}(t)|^{2}
[/tex]

And you know how to estimate all terms on the right; just keep in mind that [itex]\epsilon[/itex] is arbitrarily small (or at least I think it is).
 
  • #10
I see you just expanded it...I didn't know you could do that inside absolute value.
 
  • #11
402
1
It's just one of those Analysis tricks. After doing a fair number of these proofs, you start to spot them more rapidly.:smile:
 
  • #12
Well thank you so much! And yes, you are right, it did begin with the sup norm.
 
  • #13
402
1
Glad to help.:wink:
 
  • #14
mathman
Science Advisor
7,760
415
I see you just expanded it...I didn't know you could do that inside absolute value.
When you expand the square of the sum or difference of two terms, you get two squares and a product. Taking absolute values for each term can only make the sum of the three terms bigger.
 
  • #15
1,707
5
or

[tex] |f_n(t) - f(t)|^2 \leq \epsilon \implies [/tex]
[tex] |f_n(t) - f(t)| \leq \epsilon' \implies [/tex] by reverse triangle inequality
[tex] |f(t)| - |f_n(t)| \leq \epsilon' \implies [/tex]
[tex] |f(t)| \leq 1 + \epsilon' \implies [/tex]
[tex] |f(t)|^2 \leq 1 [/tex]

edit: this is true for one n if the OP's statements still hold. as in if [itex] |f_n(t) - f(t) | \leq \epsilon [/itex] for all epsilon without incrementing n.
 
Last edited:

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