# Modulus and inequalities

1. Mar 31, 2010

### Somefantastik

Hello,

I've got

$$|a|^{2} = |a - b + b|^{2}$$

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,

$$|a| = |a - b + b| \leq |a-b| + |b|$$

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

2. Mar 31, 2010

### Staff: Mentor

Well, |a - b + b|2 = |a|2

3. Mar 31, 2010

### mathman

What are you looking for?

4. Mar 31, 2010

### Somefantastik

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. Mar 31, 2010

### JSuarez

Are you looking for something like this?

$$\left|a\right|^2=\left|a^2\right|=\left|a^2-b^2+b^2\right|\leq\left|a^2-b^2\right|+\left|b^2\right|$$

Now, if $\left|a^2-b^2\right|\leq\epsilon$ and $\left|b^2\right|\leq\epsilon$, then $\left|a\right|^2\leq 2\epsilon$

6. Mar 31, 2010

### Somefantastik

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. Mar 31, 2010

### JSuarez

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. Mar 31, 2010

### Somefantastik

I have

$$|f_{n}(t)-f(t)|^{2} \leq\epsilon$$

and $$|f_{n}(t)|^{2} \leq 1$$

I wish to show

$$|f(t)|^{2} \leq 1$$

as well.

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

$$|f(t)|^{2} = |f(t) - f_{n} (t) + f_{n} (t)|^{2} \leq |f_{n}(t) - f(t)|^{2} + |f_{n}(t)|^{2}$$

9. Mar 31, 2010

### JSuarez

No, but you can say:

$$|f(t)|^{2} = |f(t) - f_{n} (t) + f_{n} (t)|^{2} \leq |f_{n}(t) - f(t)|^{2} +2|f_{n} (t)||f_{n}(t) - f(t)|+ |f_{n}(t)|^{2}$$

And you know how to estimate all terms on the right; just keep in mind that $\epsilon$ is arbitrarily small (or at least I think it is).

10. Mar 31, 2010

### Somefantastik

I see you just expanded it...I didn't know you could do that inside absolute value.

11. Mar 31, 2010

### JSuarez

It's just one of those Analysis tricks. After doing a fair number of these proofs, you start to spot them more rapidly.

12. Mar 31, 2010

### Somefantastik

Well thank you so much! And yes, you are right, it did begin with the sup norm.

13. Mar 31, 2010

### JSuarez

14. Apr 1, 2010

### mathman

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. Apr 1, 2010

### ice109

or

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

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

Last edited: Apr 2, 2010