Minkowski's Inequality

  1. AKG

    AKG 2,585
    Science Advisor
    Homework Helper

    Definitions and Useful Facts

    If [itex]f : X \to \mathbb{C}[/itex] is a measurable function, define the essential supremum of f to be:

    [tex]||f||_{\infty} = \inf \{a \in [0,\infty ] : \mu (\{x : |f(x)| > a\}) = 0\}[/tex]

    where [itex]\mu[/itex] is a measure, and we adopt the convention [itex]\inf \emptyset = \infty[/itex]. Note that

    [tex]||f||_{\infty} \in \{a \in [0,\infty ] : \mu (\{x : |f(x)| > a\}) = 0\}[/tex]

    If f has finite essential supremum, we say f is an [itex]L^{\infty}[/itex] function. The set of [itex]L^{\infty}[/itex] functions forms a Banach space and [itex]||.||_{\infty}[/itex] defines a norm on this space. So if f and g are [itex]L^{\infty}[/itex] functions, then so is f+g, and the following inequality holds:

    [tex]||f+g||_{\infty} \leq ||f||_{\infty} + ||g||_{\infty}[/tex]

    (Note: We will treat two functions as identical if the subset of the domain on which they differ has measure 0. All the terms defined above remain well-defined upon adopting this convention.)

    Problem

    When does equality hold in the above inequality?

    Attempt

    Define [itex]z : X \to C[/itex] where C is the complex circle by:

    [tex]z(x) = \frac{|f(x)|}{f(x)} \mbox{ if } f(x) \neq 0;\ z(x) = 1\mbox{ if } f(x) = 0[/tex]

    Then fz is a non-negative real-valued function, and

    [tex]|fz| - |f| = |gz| - |g| = |(f+g)z| - |f+g| = 0[/tex]

    hence

    [tex]||fz||_{\infty} - ||f||_{\infty} = ||gz||_{\infty} - ||g||_{\infty} = ||(f+g)z||_{\infty} - ||f+g||_{\infty} = 0[/tex]

    So assume w.l.o.g. that f is a non-negative real-valued function. Let A and B denote the essential suprema of f and g respectively. Right now my rough idea is that we get the desired equality iff for all a < A, for all b < B, and for all c > 0, the following holds:

    [tex]\mu ( \{x : f(x) > a, |g(x)| > b, ||g(x)| - g(x)| < c|g(x)|\} ) > 0[/tex]

    It basically says that equality holds iff there is a sizeable region of the domain where f is close to its maximum, |g| is close to its maximum, and g is close to being a positive real. Is this right? Is there a nicer way to put it?
     
    Last edited: Dec 28, 2006
  2. jcsd
  3. AKG

    AKG 2,585
    Science Advisor
    Homework Helper

    Wanting g(x) to be close to being a positive real, where f is is non-negative real-valued function, is the same as wanting g(x)/|g(x)| to be close to f(x)/|f(x)|, where f is now just an arbitrary function. And this is important because g(x)/|g(x)| and f(x)/|f(x)| are close iff f(x) and g(x) point in pretty much the same direction (thinking of the numbers f(x) and g(x) as arrows/vectors in the complex plane) iff |f(x) + g(x)| is close to |f(x)| + |g(x)|. So it might be neater to propose that equality holds iff:

    [tex](\forall a < A)(\forall b < B)(\forall \epsilon > 0)(\mu (\{x : |f(x)| > a, |g(x)| > b, |\overline{g(x)} - \overline{f(x)}| < \epsilon \} ) > 0[/tex]

    where [itex]\overline{z} = z/|z|[/itex] for every non-zero complex number z.
     
    Last edited: Dec 28, 2006
  4. StatusX

    StatusX 2,567
    Homework Helper

    Up to accounting for the trivial case where one of the functions is a.e. zero, I think that's the best you can do. There's no nice condition as for other Lp norms, like one function being a scalar multiple of the other, since clearly the only part of the domain that matters for this condition is the set [itex]\{ x | \mbox{ }||f||_\infty-|f(x)|<\epsilon\}[/itex], any [itex]\epsilon>0[/itex], and so outside this range (which can usually be made arbitrarily small), the values of the function are completely irrelevant. Note that your condition [itex]|\overline{g(x)} - \overline{f(x)}| < \epsilon[/itex] is essentially a rewrite of [itex]|f(x)|+|g(x)|-|f(x)+g(x)|<\epsilon'[/itex], some [itex]\epsilon'[/itex] that goes to zero as [itex]\epsilon[/itex] does (again, up to the case where on of the functions approaches zero), which shows your proposal is correct, if not all that enlightening.
     
  5. AKG

    AKG 2,585
    Science Advisor
    Homework Helper

    Okay thanks. I think the condition for Lp norms for 1 < p < oo is that there is some constant non-negative real r such that f = rg or g = rf. For p = 1, r can vary with x, so the condition is that there's some non-negative real valued function r on X such that for each x in X, either f(x) = g(x)r(x) or f(x)r(x) = g(x) (and r can "switch sides" as x varies). Is this right?

    Also, I can easily account for the a.e. 0 case by changing the quantifiers to say [itex](\forall a \in (0,A))(\forall b \in (0,B))\dots[/itex] so if one of the functions is a.e. 0, then a or b will quantify over the empty set, making the thing trivially true.
     
    Last edited: Dec 29, 2006
  6. StatusX

    StatusX 2,567
    Homework Helper

    Yea, that's right, except I don't see how r could vary with position. Otherwise all real functions would have the same L1 norm.
     
  7. AKG

    AKG 2,585
    Science Advisor
    Homework Helper

    Why would that be?

    [tex]||f+g||_1 = ||f||_1 + ||g||_1[/tex]

    iff

    [tex]\int |f+g| = \int |f| + \int |g|[/tex]

    iff

    [tex]\int |f+g| = \int |f| + |g|[/tex]

    iff

    [tex]|f+g| = |f| + |g| a.e.[/tex]


    iff there exist non-negative real-valued functions r and q such that they are never both zero for the same x and such that fr = gq a.e. (EDITED)
     
    Last edited: Dec 30, 2006
  8. StatusX

    StatusX 2,567
    Homework Helper

    Sorry, I had something backwards. It seems weird, but I guess that's right.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?