Proof of an inner product space

[cabin5]

Homework Statement

Prove that the normed linear space $$l_{\infty}^{2}$$ is not an inner product space.

Homework Equations

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

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

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

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?

 

[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.

 

[cabin5]

Is it a mathematically correct method?

 

[Dick]

Is it a mathematically correct method?

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.

 

[HallsofIvy]

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

 

[cabin5]

thanks for the post!

 

[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?

 

[Dick]

You must have put some effort into 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.

 

[cabin5]

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

Thanks a lot!

 

[HallsofIvy]

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?

 

[Dick]

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?

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.

 

[cabin5]

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

 

[Dick]

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

 

[cabin5]

Sorry for so late reply:

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 $$l_{\infty}^{n}$$.

 

[HallsofIvy]

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.

No, I was assuming infinite sequences {a_n} such that {a²_n} was summable.

 

[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.)

 

[Dick]

No, I was assuming infinite sequences {a_n} such that {a²_n} was summable.

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

 

[Dick]

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.)

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

 

[cabin5]

oh, ****!

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