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Proof of Cauchy-Schwarz Inequality

  • Thread starter Vespero
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  • #1
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Homework Statement



Let V be a vector space with inner product <x,y> and norm ||x|| = <x,x>^1/2.
Prove the Cauchy-Schwarz inequality <x,y> <= ||x|| ||y||.

Hint given in book: If x,y != 0, set c = 1/||x|| and d = 1/||y|| and use the fact that
||cx ± dy|| >= 0.

Here, the inner product is not necessarily the dot product, nor the norm the Euclidean norm. All that is necessary is that the two fulfill properties of inner products and norms.

For inner products, the given properties are
1. <x,y> = <y,x>
2. <x + y, z> = <x,y> + <x,z>
3. <cx,y> = c<x,y> = <x,cy>
4. <x,x> > 0 if x != 0.

For norms, the given properties are
1. ||x|| > 0 if x != 0
2. ||cx|| = |c|||x||
3. ||x + y|| <= ||x|| + ||y||

Homework Equations



The inequality ||cx ± dy|| >= 0 is given as true.

The Attempt at a Solution



I have attempted to solve the problem by starting with the Cauchy-Schwarz inequality and working towards the inequality given in the hint as true:

<x,y> >= ||x|| ||y||
-2<x,y> >= -2||x|| ||y||
-2cd<x,y> >= -2
1 - 2cd<x,y> + 1 >= 0
(c^2)||x||^2 - 2cd<x,y> + (d^2)||d||^2 >= 0
(c^2)<x,x> -cd<x,y> - cd<y,x> + (d^2)<y,y> >= 0
<cx,cx> + <cx, -dy> + <-dy, cx> + <dy,dy> >= 0
<cx - dy, cx - dy> >= 0
||cx - dy||^2 >= 0

Taking the square root here yields
||cx - dy|| >= 0 or ||cx - dy|| <= 0

but the norm of a vector is a non-negative number, so the right inequality simplifies to
||cx - dy|| = 0
The two inequalities can then be combined to yield
||cx - dy|| >= 0, which was given as a fact at the beginning of the problem.

Do you see any flaws in my logic, or any steps that could be simplified or added to make the proof more clear?
 

Answers and Replies

  • #2
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The path is essentially correct, but it really should go the other way around. Going from inequality to something obvious is good for devising the proof, but it's kind of "backward logic". Rewriting the steps in the opposite we can get nice chain of logical consequences.
That's just my point of view. But maybe I care about such things too much.
 
  • #3
28
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The path is essentially correct, but it really should go the other way around. Going from inequality to something obvious is good for devising the proof, but it's kind of "backward logic". Rewriting the steps in the opposite we can get nice chain of logical consequences.
That's just my point of view. But maybe I care about such things too much.
Thanks for the reply, losiu. I actually began the proof the other way around, but ran into trouble, but I think I understand why. The hint say to use the fact that ||cx ± dy|| >= 0. The ± sign was throwing me off, as when I reached the end of the proof, I had something like
±<x,y> >= -||x|| ||y||.
The minus sign yields the Cauchy-Schwarz inequality, but the plus sign yields
<x,y> >= -||x|| ||y||
and I'm not sure what to do about it.

Is there a way to resolve this problem? Should the ± still be present by the end of the problem, or is there a way to get rid of it? Should I just take the case ||cx - dy|| >= 0,
which eventually directly yields the Cauchy-Schwarz, with no sign problem?
 
  • #4
133
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The "minus" version is, as you noticed, sufficient for the proof. To be honest, I have no idea why ± is present in the hint. Just take the || cx - dy ||, factor everything and that's it.
 

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