Proving the Inner Product Sum Inequality: Exploring the Equality Condition

[JonoPUH]

Homework Statement

Let V be a real inner product space, and let v₁, v₂, ... , v_k be a set of orthonormal vectors.
Prove
Ʃ (from j=1 to k)|<x,v_j><y,v_j>| ≤ ||x|| ||y||

When is there equality?

Homework Equations

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,v_j><y,v_j>| ≤ 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, e_i>|² if e_i is a set of orthonormal elements.

Thanks

 

[micromass]

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 $\mathbb{R}^k$??

 

[JonoPUH]

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

(Ʃ x_iy_i)² ≤ Ʃx_i² Ʃy_i²

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


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

 

[micromass]

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

(Ʃ x_iy_i)² ≤ Ʃx_i² Ʃy_i²

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


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

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

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

 

[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,v_j>²) √(Ʃ<xy,v_j>²)

Then by Bessel's Inequality

√Ʃ<x,v_j>²√Ʃ<xy,v_j>² ≤ √||x||² √||y||²

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

 

[micromass]

That's it!

 

[JonoPUH]

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