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Homework Help: Sup norm and inner product on R2

  1. May 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the sup norm on R2 is not derived from an inner product on R2. Hint: suppose <x,y> is an inner product on R2 (not the dot product) and has the property that |x|=<x,y>0.5. Compute <x±y, x±y> and apply to the case x=e1, y=e2.

    2. Relevant equations
    I've noticed that the notation can vary for the sup norm - in this case |x| is the sup norm.

    3. The attempt at a solution
    My understanding of the sup norm is fragile; in fact, my understanding of linear algebra in general is full of gaps, which is why I'm trying to work through questions in the first chapter of Munkres' Analysis on Manifolds. It looks as though this is set up to be a proof by contradiction, so assuming the hint is true I decided to expand and get to the point where the hint would be useful:
    <x+y, x+y> = <x, x> + <y, y> + 2<x, y>
    Using the hint, <x, y> = |x|2, so:
    <x+y, x+y> = ||x||2 + ||y||2 + 2|x|2
    ||x+y||2 = ||x||2 + ||y||2 + 2|x|2
    Using the fact that x=e1 and y=e2, this gives 2 = 4. Since the expansion is correct under the definition of inner product, the assumption given in the hint must be incorrect. So, this means that the sup norm does not follow from an inner product, or at least this particular inner product; case closed! Or is it?

    My concern is that it doesn't seem to address the full question, since at a glance it only appears to address the case when x=e1 and y=e2, so hypothetically, using this argument, could it be possible another x, y satisfies? Or is the fact that we're using a basis for R2 enough that the argument is generalized by its usage?
  2. jcsd
  3. May 22, 2012 #2
    I'm gonna guess you maybe misread the hint. It should probably say ##<x,x>=|x|^{\frac{1}{2}}##.

    Now I could be wrong, but I'm pretty basis vectors need not have norm 1. So your argument doesn't fly. You're on the right track, though. Go back and look up the definition of the sup norm.
  4. May 22, 2012 #3
    Thanks for the reply! I checked the hint once again, but no, everything is exactly as I've written it as far as the question and the hint are concerned.

    I used the part of the hint where x=e1 (so x1=1, x2=0) and y=e2. In the book it says that the sup norm is calculated as follows:
    |x|= max{|x1|,...,|xn|} which in this case would be 1.

    Hah, I realized that I didn't square my value for ||x+y||, though, so on that front my argument completely falls apart.
    Last edited: May 22, 2012
  5. May 22, 2012 #4
    Even if you copied it right, the hint as stated is wrong. It should be
    \langle x, x \rangle = \|x\|^2
    for any ##x \in \mathbb{R}^2##. (You got it slightly backwards, gopher_p!)

    I think you're supposed to look at ##\langle e_1 + e_2, e_1 + e_2 \rangle##.
  6. May 22, 2012 #5
    Well I really think it's a typo then. (1) The definition of "norm determined by inner product" is ##\|x\|=<x,x>^{\frac{1}{2}}##, (2) it's not needed for the problem.

    Yes, you're right. My bad on that.

    Actually, I think you need to be even more careful about computing ##\|e_1+e_2\|##. Remember it's the sup norm.
  7. May 22, 2012 #6
    From my understanding, I think the hint is asking us to assume this property is true (it's clearly not, of course) and in showing that it doesn't hold, it should prove that the sup norm isn't derived from the inner product.

    Revising my argument, if I expanded <x+y, x+y>, I would get
    <x, x> + <y, y> + 2<x, y> which by the hint should be equivalent to <x, x> + <y, y> + 2|x|2
    and in this case we'd once again arrive at the 4=2 situation...except with correct math.

    The question seems kind of ridiculous to me, though - does following the hint and trying to find an algebraic error like I did really provide all the proof that's needed to answer the question?
  8. May 22, 2012 #7
    Perhaps if I give you the "lemma" related to this problem you'll see how this problem is supposed to work out.

    Let ##V## be a vector space over a field ##F## and ##<\cdot,\cdot>:V\times V\rightarrow F## an inner product. Show that for all ##x,y\in V##, ##<x+y,x+y>+<x-y,x-y>=2<x,x>+2<y,y>##.
  9. May 22, 2012 #8


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    If it does, then either your < , > isn't an inner product or your | | isn't a norm. Just consider a non-zero x, and y orthogonal to x. Then you have 0≠|x|=0. So it's definitely a typo.

    Just think about what the statement you want to prove means. It means that the assumption that < , > is an inner product such that ##\|x\|_\infty^2=\langle x,x\rangle## (where ##\|x\|_\infty## denotes the sup norm of x) will lead to a contradiction*. So it's actually pretty obvious that the hint contains a typo.

    *) This will be a contradiction derived using the definition of the sup norm. Note that the contradiction 0≠0 that we derived from the mistyped condition could be found without using the definition of the sup norm.
  10. May 23, 2012 #9
    Thanks for clarifying about the hint.

    I see now what you were saying, gopher_p. Using the parallelogram law, then, (where I'm still using |x| to denote the sup norm...must learn Latex!)

    |x+y|2+|x−y|2 = 2|x|2 +2|y|2

    But x=e1 and y=e2, so the parallelogram law clearly isn't satisfied and once again gives 2=4, but without referencing the hint. Because the law isn't satisfied, the sup norm can't be derived from an inner product.

    I'm curious to know how the contradiction 0≠0 is obtained without using the definition of the sup norm.
  11. May 23, 2012 #10
    Well, we don't need the parallelogram law to solve the problem I gave. We just need properties of inner products. And we don't need norms either. See if you can prove my claim just using the axioms of inner products on vector spaces.
  12. May 23, 2012 #11


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    As I said, if ##\|\ \|## denotes a norm, and ##\langle\ ,\ \rangle## an inner product, and you choose x≠0, and y orthogonal to x, the (incorrect) equality ##\|x\|=\sqrt{\langle x,y\rangle}## becomes "something non-zero"=0. The result ##0\neq \|x\|=0## implies that ##0\neq 0##.

    So you have to use that ##\|\ \|## is a norm, but you don't have to use that it's specifically the sup norm.
  13. May 23, 2012 #12
    Here is a completely different way to approach the problem. According to bilinearity of the inner product: <(x,y),(x,y)> = ax^2+2bxy+cy^2, where a = <(1,0),(1,0)> etc. Now assume that for some choice of a,b,c that formula is equal to |(x,y)|^2 for all (x,y) in R2.

    So lets set x=1 and let y vary between -1 and 1. What does the equation say then and what can you deduce about the coefficients a,b,c?
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