# Inner Product Sum Inequality

1. Nov 20, 2012

### JonoPUH

1. The problem statement, all variables and given/known data
Let V be a real inner product space, and let v1, v2, ... , vk be a set of orthonormal vectors.
Prove
Ʃ (from j=1 to k)|<x,vj><y,vj>| ≤ ||x|| ||y||

When is there equality?

2. Relevant equations

3. The attempt at a solution

I've tried using the two inequalities given to us in lectures, namely Cauchy-Schwarz Inequality which states

|<v,w>| ≤ ||v|| ||w||

But surely, using this inequality, we get Ʃ (from j=1 to k)|<x,vj><y,vj>| ≤ k(||x|| ||v|| ||y|| ||v|| = k( ||x|| ||y||) since the v are orthonormal!

I understand this is an inequality, and so obviously the inequality above is a better approximation than the one I've just shown, but I'm not sure where to go.

The other inequality is Bessel's Inequality which states

||v||2 ≥ Ʃ|<v, ei>|2 if ei is a set of orthonormal elements.

Thanks

Last edited: Nov 20, 2012
2. Nov 20, 2012

### micromass

Staff Emeritus
The space $\mathbb{R}^k$ is an inner product space for the usual inner product. What does the Cauchy-Schwarz inequality say in this special case $\mathbb{R}^k$??

3. Nov 21, 2012

### JonoPUH

Ok, so according to Wikipedia (I haven't been taught this in lectures), the Cuachy-Schwarz inequality over ℝn is:

(Ʃ xiyi)2 ≤ Ʃxi2 Ʃyi2

Do I replace multiplication with inner products? I've tried that, but I must be doing something wrong.

(Ʃ <x,vj>)2 ≤ Ʃ<x,x> Ʃ<vj,vj> = k||x||2Ʃ<vj,vj> = k||x||2 x k since ||vj||=1 ?
But then where should I go from here, if here is where I should be?
Sorry

4. Nov 21, 2012

### micromass

Staff Emeritus
The version of Cauchy-Schwarz that is most standard is actually

$$\sum_{k=1}^n |\alpha_k\beta_k|\leq \sqrt{\sum_{k=1}^n |\alpha_k|^2}\sqrt{\sum_{k=1}^n |\beta_k|^2}$$

Now, apply this on your original problem

$$\sum_{k=1}^n |<x,v_k><y,v_k>|$$

5. Nov 21, 2012

### JonoPUH

Thank you so much! I think I have it, although it seems very easy, which always seems suspicious to me in maths. Here goes:

Ʃ |<x,vj><y,vj>| ≤ √(Ʃ<x,vj>2) √(Ʃ<xy,vj>2)

Then by Bessel's Inequality

√Ʃ<x,vj>2√Ʃ<xy,vj>2 ≤ √||x||2 √||y||2

So Ʃ |<x,vj><y,vj>| ≤ ||x|| ||y|| as required!

Last edited: Nov 21, 2012
6. Nov 21, 2012

### micromass

Staff Emeritus
That's it!!

7. Nov 21, 2012

### JonoPUH

Thank you very much! You've made my night!