# Proof of an inner product space

1. Oct 3, 2008

### cabin5

1. The problem statement, all variables and given/known data
Prove that the normed linear space $$l_{\infty}^{2}$$ is not an inner product space.

2. Relevant equations
parallelogram law;
$$\left\|x+y\right\|^2+\left\|x-y\right\|^2=2\left\|x\right\|^2+2\left\|y\right\|^2$$

3. The attempt at a solution
Well, I tried to apply parallelogram law to the $$l_{\infty}^{2}$$ space where

$$x=(\alpha^1,\alpha^2)$$ and $$y=(\beta^1,\beta^2)\in l_{\infty}^{2}$$ .

$$\left\|x\right\|=max\left\{\left|\alpha^1\right|,\left|\alpha^2\right|\right\} and \left\|y\right\|=max\left\{\left|\beta^1\right|,\left|\beta^2\right|\right\}$$

If one puts these norms into the parallelogram law equation, one gets a fuzzy expression on both sides of the equation, therefore it is important to put out expressions inside the max{} function which I could not achieve to do.

Is there another method to solve this problem or am I misapplying the law to $$l_{\infty}^{2}$$ space?

Last edited: Oct 3, 2008
2. Oct 3, 2008

### Dick

You only need a counterexample to show it's not an inner product space. Start putting some actual numbers in for the alphas and betas.

3. Oct 3, 2008

### cabin5

Is it a mathematically correct method?

4. Oct 3, 2008

### Dick

What would not be 'mathematical' or 'correct' about it? If I claim all primes are odd, and somebody points out 2 is even, that's enough to prove me wrong. Just do it.

5. Oct 3, 2008

### HallsofIvy

Staff Emeritus
Yes, as Dick said, proving that a general statement is NOT true by counterexample is a perfectly correct method.

6. Oct 3, 2008

### cabin5

thanks for the post!

7. Oct 3, 2008

### cabin5

well,
I tried the parallelogram law for x=(0,3) and y=(2,5) and it perfectly worked on both sides of the equation.
Should I choose the complex field for that space?

What's wrong with that?

Last edited: Oct 3, 2008
8. Oct 3, 2008

### Dick

You must have put some effort in to find a case where it works. Almost all other cases don't work. Like x=(1,0) and y=(0,1). It has to work for all cases or your norm doesn't come from an inner product.

9. Oct 3, 2008

### cabin5

Finally, It worked :)
I thought any ordered pair would work as a counterexample.

Thanks a lot!

10. Oct 5, 2008

### HallsofIvy

Staff Emeritus
No, and ordered pair won't work because R2 DOES make a inner product space. This question was about $l_\infty^2$. Your counterexample must be from that space. What are the vectors in that space?

11. Oct 5, 2008

### Dick

I took it to be R^2 with the max norm. You think it's bounded functions on R^2, right? It's still pretty easy to find a counterexample with bounded functions.

12. Oct 5, 2008

### cabin5

I have no clue whether one must use a bounded function or not in order to prove that.

13. Oct 5, 2008

### Dick

The question is, what is the definition of the space l^2_infinity? Can you tell us what it is?

14. Oct 12, 2008

### cabin5

The definition of $$l_{\infty}^{n}$$ :
On the linear space $$V_{n}(F)$$ with the infinity norm defined by
$$\left\|x\right\|_p=\left[\sum^{\infty}_{i=1}\left|\alpha^{i}\right|^p\right]^{1/p}$$

where $$x=(\alpha^i)$$.
The corresponding linear space to this norm is denoted by $$l_{\infty}^{n}$$.

Last edited: Oct 12, 2008
15. Oct 12, 2008

### HallsofIvy

Staff Emeritus
No, I was assuming infinite sequences {an} such that {a2n} was summable.

16. Oct 12, 2008

### cabin5

so $$l_{\infty}^{2}$$ defines that norm which is basically the square of root total sum of square of each element of 2 vectors defined over R^2 field.

Eventually , I think that you're example is correct, but besides I have no idea about whether functions defined in $$l_{\infty}^{2}$$ is bounded or not (since I didn't take any real analysis course during undergrad.)

17. Oct 12, 2008

### Dick

I don't think that's what the problem is supposed to be about.

18. Oct 12, 2008

### Dick

I think what you just described is l^2_2. The 'infinity' usually designates a max norm (supremum) rather than a power root norm.

19. Oct 12, 2008

### cabin5

oh, ****!

you're right , It was supposed to be max norm! I miswrote the definition.