Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Schwarz inequality

  1. Aug 2, 2009 #1
    The couple of proofs that I've seen of Schwarz's inequality

    [tex]\left|\mathbf{u} \cdot \mathbf{v} \right| \leq ||\mathbf{u}|| \: ||\mathbf{v}||[/tex]

    both begin with statements to the effect that it's trivially true when u or v = 0. How so? In Bowen and Wang's Introduction to Vectors and Tensors, the inequality is said to be true for any inner product space, defined as a vector space (V,F,s), where V is an additive abelian group over a set of vectors, F a field, and s the function of scalar multiplication relating V and F, along with a function

    [tex]f : V \rightarrow F[/tex]

    called the inner product, defined by the following axioms:

    [tex](1)\; f\left(\mathbf{u},\mathbf{v} \right) = \overline{ f\left(\mathbf{v},\mathbf{u} \right) }[/tex]

    [tex](2)\; \lambda f\left(\mathbf{u},\mathbf{v} \right) = f\left(\lambda \mathbf{u},\mathbf{v} \right)[/tex]

    [tex](3)\; f\left(\mathbf{u_{1}} + \mathbf{u_{2}},\mathbf{v} \right) = f\left(\mathbf{u_{1}},\mathbf{v} \right) + f\left(\mathbf{u_{2}},\mathbf{v} \right)[/tex]

    [tex](4)\; f\left(\mathbf{u},\mathbf{u} \right) \geq 0 \; and \; f(\mathbf{u},\mathbf{u}) = 0 \Leftrightarrow \mathbf{u} = \mathbf{0}[/tex]

    They give an example of the vector space [tex]\mathbb{C}^{n}[/tex], where elements of V are n-tuples of complex numbers. (Is n-tuple synonymous with "ordered n-tuple"?) And they defined an inner product for this space:

    [tex]\mathbf{u} \cdot \mathbf{v} = \sum_{i=1}^{n} u_{i} \overline{v}_{i}[/tex]

    No doubt the Schwarz inequality is true for such an inner product space when u or v = 0, since it just comes down to multiplication of complex numbers, but I don't know where to begin showing that it must be true generally for any inner product space as defined by those four axioms. Since the bar in the first axiom denotes the complex conjugate, presumable the field has to be the complex numbers, a subfield thereof, or something analogous? In that case, I can see that the right-hand side of the Schwarz inequality must equal 0. The left-hand sight must be positive, since it's the modulus of a complex number, so this trivial part of the proof amounts to showing that [tex]\left|\mathbf{u} \cdot \mathbf{v} \right|[/tex] = 0 if u or v = 0. Does it depend somehow on the first axiom?

    [tex]f\left( \mathbf{0},\mathbf{v} \right) \; \overline{ f\left( \mathbf{0},\mathbf{v} \right)} = f\left( \mathbf{v},\mathbf{0} \right) \overline{ f\left( \mathbf{v},\mathbf{0} \right)}[/tex]

    And then does it follow from some property of complex numbers, some basic algebra, or from something in the definition of a function, or something to do with the linearity of the dot product operation? It certainly feels as if it ought to be 0, but how to prove it only with these axioms. I'm new to the subject, and what may be trivial to the authors just isn't jumping out at me yet.
  2. jcsd
  3. Aug 2, 2009 #2


    User Avatar
    Science Advisor
    Gold Member

    I'm guessing your issue is with the LHS. Hint:
    \mathbf{0} = 0 \mathbf{0}

    and use the axioms you have to show
  4. Aug 2, 2009 #3
    Ooh, thanks, I think I've got it now! By Axiom 2 of the inner product space,

    [tex]\mathbf{0} \cdot \mathbf{v} = 0 \left(\mathbf{0} \cdot \mathbf{v}\right) = 0[/tex]

    and by Axioms 1 and 2,

    [tex]\mathbf{v} \cdot \mathbf{0} = \overline{ \mathbf{0} \cdot \mathbf{v}} = \overline{ 0 \left(\mathbf{0} \cdot \mathbf{v}\right)} = 0 \overline{ \left(\mathbf{0} \cdot \mathbf{v}\right)} = 0[/tex]

    because, by the distributive properties of a field,

    [tex]0a = \left(b-b \right)a = ba - ba = 0[/tex]
    [tex]a0 = a \left(b-b \right) = ab - ab = 0[/tex]
  5. Aug 3, 2009 #4
    So, moving on to the actual proof, for the case where neither u nor v = 0, Bowen and Wang proceed by asking us to consider the vector

    [tex](\mathbf{u\cdot \mathbf{v}})\mathbf{v} - (\mathbf{v}\cdot \mathbf{u})\mathbf{u}[/tex]

    and employ Axiom 4, "which requires that every vector have a nonzero length, hence"

    [tex]\left ( \left \| \mathbf{u}^{2} \right \| \left \| \mathbf{v}^{2} \right \| - (\mathbf{u\cdot v})( \overline{\mathbf{u\cdot v}})\right )\left \| \mathbf{u}^{2} \right \|\geq 0[/tex]

    I can see how this inequality would imply the Schwarz inequality, but how is it established? Employing Axiom 4 tells me that the product of the squared lengths of u and v is positive, as is the complex norm of the inner product of these two vectors. Subtracting the latter from the former is therefore to subtract one positive real number from another. But the result of such an operation isn’t necessarily positive, is it? And if the result wasn’t positive, the inequality wouldn’t hold. So what am I not taking into account? Of course, if the Schwarz inequality was true, this inequality would have to be true too, but it’s the Schwarz inequality that this is meant to be a proof of.
  6. Aug 3, 2009 #5


    User Avatar
    Science Advisor
    Gold Member

    Your book may have a typo (and it may not!). There are many ways to prove the Schwartz inequality, but to use the same basic approach as the book try looking at
    f(a\mathbf{u} + b\mathbf{v}, a\mathbf{u} + b\mathbf{v})
    and make a clever choice for [tex]a[/tex] and [tex]b[/tex].
  7. Aug 3, 2009 #6


    User Avatar
    Science Advisor
    Gold Member

    There's a typo. You should consider

    [tex]\mathbf{x} = \left\langle\mathbf{u} ,\mathbf{\color{red}{u}}}\right\rangle\mathbf{v} - \left\langle\mathbf{v} , \mathbf{u}\right\rangle\mathbf{u}[/tex]​

    and the fact that [itex]\left\langle\mathbf{x} , \mathbf{x}\right\rangle \geq 0[/itex].
    Last edited: Aug 3, 2009
  8. Aug 4, 2009 #7
    Thanks to you both! I reckon I've got it at last - phew. The typo was my mistake. The book had it right.

    Let x = [tex]\left(\mathbf{u \cdot u} \right) \mathbf{v} - \left(\mathbf{v \cdot u} \right) \mathbf{u}[/tex].

    [tex]\mathbf{x \cdot x} \geq 0[/tex]

    [tex]\mathbf{x \cdot x} = \left[ \left(\mathbf{u \cdot u} \right) \mathbf{v} - \left(\mathbf{v \cdot u} \right) \mathbf{u} \right] \mathbf{ \cdot x}[/tex]

    [tex]= \left(\mathbf{u \cdot u} \right) \mathbf{v \cdot x} - \left(\mathbf{v \cdot u} \right) \mathbf{u \cdot x}[/tex]

    [tex]= ||\mathbf{u}^{2}|| \; \left( \mathbf{v \cdot x}\right) - \left(\mathbf{v \cdot u} \right) \left( \mathbf{u \cdot x} \right)[/tex]

    [tex]= ||\mathbf{u}^{2}|| \; \overline{\mathbf{x \cdot v}} - \left(\mathbf{v \cdot u} \right) \overline{ \mathbf{x \cdot u}}[/tex]

    [tex]= ||\mathbf{u}^{2}|| \; \left[ \overline{||\mathbf{u}||^{2} \; ||\mathbf{v}||^{2} - \left(\mathbf{v \cdot u} \right) \left(\mathbf{u \cdot v} \right)}\right] - \left(\mathbf{v \cdot u} \right) \left[ \overline{||\mathbf{u}||^{2}\left(\mathbf{v \cdot u} \right) - ||\mathbf{u}||^{2}\left(\mathbf{v \cdot u} \right)}\right][/tex]

    [tex]= ||\mathbf{u}||^{2}\left(||\mathbf{u}||^{2} \; ||\mathbf{v}||^{2} - |\mathbf{u \cdot v}|^{2}\right) \geq 0[/tex]

    [tex]\Rightarrow ||\mathbf{u}||^{2} \; ||\mathbf{v}||^{2} - |\mathbf{u \cdot v}|^{2} \geq 0[/tex]

    [tex]\Rightarrow ||\mathbf{u}||^{2} \; ||\mathbf{v}||^{2} \geq |\mathbf{u \cdot v}|^{2}[/tex]

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Schwarz inequality
  1. Umm, cauchy-schwarz (Replies: 8)