Inner product space and norm.

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Chris*

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So, I have (X,<,>) is an inner product space. With the norm of x∈X defined by ||x||2=<x,x>.

The problem asks to prove or disprove:
If x,y∈X then ||x+y||∙||x-y||≤||x||2+||y||2.

I'm not even sure where to begin on this. This was an aside in a lecture, so I don't have much exposure to inner product spaces.

I tried just plugging numbers from ℝ and ℂ into the inequality using the standard dot product. Assuming I did it correctly, the inequality was true. But that doesn't tell me anything about an arbitrary space.

Does anyone have any suggestions on how to prove (or disprove) this?
 
Chris* said:
So, I have (X,<,>) is an inner product space. With the norm of x∈X defined by ||x||2=<x,x>.

The problem asks to prove or disprove:
If x,y∈X then ||x+y||∙||x-y||≤||x||2+||y||2.

I'm not even sure where to begin on this. This was an aside in a lecture, so I don't have much exposure to inner product spaces.

I tried just plugging numbers from ℝ and ℂ into the inequality using the standard dot product. Assuming I did it correctly, the inequality was true. But that doesn't tell me anything about an arbitrary space.

Does anyone have any suggestions on how to prove (or disprove) this?
Click to expand...
Writing the norm in terms of the inner product, you are asked to prove that \(\displaystyle \sqrt{<x+y, x+y>}\sqrt{<x- y, x- y>}\le <x, x>+ <y, y>\). Now you need to know some properties of the inner product- in particular that it is linear: <x+ y, x+ y>= <x,x+ y>+ <y, x+ y>= <x, x>+ <x, y>+ <y, x>+ <y, y>. Whether <x, y>+ <y, x> is the same as 2<x, y> depends upon whether you are working with a vector space over the real numbers, where <x, y>= <y, x> or over the complex numbers where <x, y>= <y, x>*, the "*" indicating complex conugation.
 
Hmm. I think that may be one of the reasons I'm so confused. The definition in class given was
Let V be a linear space. A function <,>:VxV→ℝ is an inner product if ∀x,y∈V and α,β∈ℝ
(i)<x,x>≥0, and <x,x>=0 ⟺ x=0
(ii)<x,y>=<y,x>
(iii)<αx+βy,z>= α<x,z> + β<y,z>

(V,<,>) is called an inner product space.

I guess I should I assume V is a linear space over the real numbers...
Going with that, would this then be correct?
\(\displaystyle ||x+y||\cdot||x-y||\)
\(\displaystyle =\sqrt{<x+y,x+y>}\sqrt{<x-y,x-y>}\)
\(\displaystyle =\sqrt{<x+y,x+y><x-y,x-y>}\)
\(\displaystyle =\sqrt{(<x,x>+2<x,y>+<y,y>)(<x,x>-2<x,y>+<y,y>)}\)
\(\displaystyle =\sqrt{(<x,x>^2+2<x,x><y,y>+<y,y>^2-4<x,y>^2}\) By distribution and simplification.
\(\displaystyle =\sqrt{(<x,x>+<y,y>)^2-4<x,y>^2}\)
\(\displaystyle \leq\sqrt{(<x,x>+<y,y>)^2}\) Since 4<x,y>^2 is a positive number.
\(\displaystyle =<x,x>+<y,y>\)
\(\displaystyle =||x||^2+||y||^2\)
 
Chris* said:
Hmm. I think that may be one of the reasons I'm so confused. The definition in class given was
Let V be a linear space. A function <,>:VxV→ℝ is an inner product if ∀x,y∈V and α,β∈ℝ
(i)<x,x>≥0, and <x,x>=0 ⟺ x=0
(ii)<x,y>=<y,x>
(iii)<αx+βy,z>= α<x,z> + β<y,z>

(V,<,>) is called an inner product space.
Click to expand...
Okay, so your space is over the real numbers- that's simpler. And your "(iii) <αx+βy,z>= α<x,z> + β<y,z>" is exactly what I said- the inner product is linear.

I guess I should I assume V is a linear space over the real numbers...
Going with that, would this then be correct?
\(\displaystyle ||x+y||\cdot||x-y||\)
\(\displaystyle =\sqrt{<x+y,x+y>}\sqrt{<x-y,x-y>}\)
\(\displaystyle =\sqrt{<x+y,x+y><x-y,x-y>}\)
\(\displaystyle =\sqrt{(<x,x>+2<x,y>+<y,y>)(<x,x>-2<x,y>+<y,y>)}\)
\(\displaystyle =\sqrt{(<x,x>^2+2<x,x><y,y>+<y,y>^2-4<x,y>^2}\) By distribution and simplification.[/tex] I don't see how this follows.
\(\displaystyle =\sqrt{(<x,x>+<y,y>)^2-4<x,y>^2}\)
\(\displaystyle \leq\sqrt{(<x,x>+<y,y>)^2}\) Since 4<x,y>^2 is a positive number.
\(\displaystyle =<x,x>+<y,y>\)
\(\displaystyle =||x||^2+||y||^2\)
Click to expand...
This is NOT what you were asked to prove and is not true!
 
It's not? Sorry, I guess I really don't understand this. :-(

Am I not allowed to use distribution?

\(\displaystyle (<x,x>+2<x,y>+<y,y>)(<x,x>-2<x,y>+<y,y>)\)
\(\displaystyle =<x,x>^2+2<x,y><x,x>+<x,x><y,y>-2<x,y><x,x>\)
\(\displaystyle -4<x,y>^2-2<x,y><y,y>+<x,x><y,y>+2<x,y><y,y>+<y,y>^2\)
\(\displaystyle =<x,x>^2+2<x,x><y,y>+<y,y>^2-4<x,y>^2\) Here, I factored the first three terms.
\(\displaystyle =(<x,x>+<y,y>)^2-4<x,y>^2)\)
 
Chris* said:
It's not? Sorry, I guess I really don't understand this.
Am I not allowed to use distribution?
\(\displaystyle (<x,x>+2<x,y>+<y,y>)(<x,x>-2<x,y>+<y,y>)\)
\(\displaystyle =<x,x>^2+2<x,y><x,x>+<x,x><y,y>-2<x,y><x,x>\)
\(\displaystyle -4<x,y>^2-2<x,y><y,y>+<x,x><y,y>+2<x,y><y,y>+<y,y>^2\)
\(\displaystyle =<x,x>^2+2<x,x><y,y>+<y,y>^2-4<x,y>^2\) Here, I factored the first three terms.
\(\displaystyle =(<x,x>+<y,y>)^2-4<x,y>^2)\)
Click to expand...

That is correct.
If you look at \(\displaystyle (<x,x>+2<x,y>+<y,y>)(<x,x>-2<x,y>+<y,y>)\) as \(\displaystyle ([<x,x>+<y,y>]+2<x,y>)([<x,x>+<y,y>]-2<x,y>)\)
then that is the product of the sum&difference of same two expressions.
That gives \(\displaystyle [<x,x>+<y,y>]^2-4<x,y>^2\).
 
Yes, I see that. Though, I'm still confused as to what isn't true and why what I proved isn't what was asked.
 
Just a little tip for latex "\langle x,y \rangle" or "\left< x, y \right>" both produce \(\displaystyle \langle x,y \rangle\). Will make for nicer-looking inner products ;)
 
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