Minkowski's Inequality


by AKG
Tags: inequality, minkowski
AKG
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#1
Dec28-06, 01:18 AM
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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?
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AKG
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Dec28-06, 03:43 AM
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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.
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Dec29-06, 10:28 AM
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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.

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Dec29-06, 11:02 AM
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Minkowski's Inequality


Quote Quote by StatusX View Post
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.
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.
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Dec29-06, 03:36 PM
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Quote Quote by AKG View Post
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?
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.
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Dec29-06, 04:25 PM
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Quote Quote by StatusX View Post
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.
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)
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Dec29-06, 06:45 PM
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Sorry, I had something backwards. It seems weird, but I guess that's right.


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