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Bessel's Inequality

  1. Jul 26, 2008 #1
    The problem statement, all variables and given/known data
    Let [itex]v \in \mathbb{R}^n[/itex], let [itex]\{u_1, \ldots, u_k\}[/itex] be an orthonormal subset of [itex]\mathbb{R}^n[/itex] and let [itex]c_i[/itex] be the coefficient of the projection of v to the span of [itex]u_i[/itex]. Show that [itex]\|v\|^2 \ge c_1^2 + \cdots + c_k^2[/itex].

    The attempt at a solution
    [itex]c_i = v \cdot u_i[/itex] and [itex]\| v \|^2 = v \cdot v[/itex] so I can write the inequality as

    [tex]v \cdot (v - (u_1 + \cdots + u_k)) \ge 0[/tex]

    This means the angle between v and [itex]v - (u_1 + \cdots + u_k)[/itex] is less than 90 degrees. This is all I've been able to conjure. I'm trying to reverse-engineer the inequality back to something I know is true. Is this a good approach? Is there a better approach?
  2. jcsd
  3. Jul 26, 2008 #2


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    Couldn't you just complete {u1,...,uk} to an orthonormal basis {u1,...,uk,uk+1,...,un}? So ||v||^2=c1^2+...cn^2.
  4. Jul 26, 2008 #3
    Ah! Good point. I think that'll work. All that remains to be proved is that [itex]v = c_1u_1 + \cdots c_nu_n[/itex].
  5. Jul 27, 2008 #4
    Quick question: the basis doesn't have to be orthonormal does it?
  6. Jul 27, 2008 #5


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    Do you really have to prove that? v is some linear combination of the basis vectors u_i. So c_i must be v.u_i since they are orthonormal. Isn't that what an orthonormal basis is all about?
  7. Jul 27, 2008 #6
    Never mind. I understand why it has to be orthonormal: If it isn't, I couldn't write v as a linear combination of the u's using the c's as the coefficients.
  8. Jul 27, 2008 #7


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    The only thing that might have to be proved is that you can complete an orthonormal subset to an orthonormal basis. But that's Gram-Schmidt.
  9. Jul 27, 2008 #8
    I know I can expand the set of u's to a basis, then orthogonalize it using Gram-Schmidt, and then normalize the result. This will yield an orthonormal basis with the original u's.
  10. Jul 27, 2008 #9


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    Right. So not much to prove really. That one was easy.
  11. Jul 27, 2008 #10
    Thank you for your help.
  12. Mar 7, 2011 #11
    Can you do it these two proof???
    I tried but i don't know these proofs...

    (b) Prove Parseval’s Indentity: For any w ∈ span(S), we have

    ||w||^2 = |w · u1 |^2 + |w · u2 |^2+ · · · + |w · uk |^2 .

    (c) Prove Bessel’s Inequality: For any x ∈ R^n we have

    ||x||^2 ≥ |x · u1 |^2 + |x · u2 |^2 + · · · + |x · uk |^2 ,

    and this is an equality if and only if x ∈ span(S).
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