Minkowski's Inequality and Equality Conditions

  • Thread starter AKG
  • Start date
  • Tags
    Inequality
In summary, the conversation discusses the definition of the essential supremum of a measurable function and its relation to L^{\infty} functions. It also explores the case when equality holds in the inequality ||f+g||_{\infty} \leq ||f||_{\infty} + ||g||_{\infty}. The conclusion is that equality holds when there is a sizeable region of the domain where both f and g are close to their maximum values and have similar directions in the complex plane. The conversation also briefly touches on the conditions for Lp norms, where p is between 1 and infinity.
  • #1
AKG
Science Advisor
Homework Helper
2,567
4
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:
Physics news on Phys.org
  • #2
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:
  • #3
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.
 
  • #4
StatusX said:
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.
 
Last edited:
  • #5
AKG said:
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.
 
  • #6
StatusX said:
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)
 
Last edited:
  • #7
Sorry, I had something backwards. It seems weird, but I guess that's right.
 

1. What is Minkowski's Inequality?

Minkowski's Inequality is a mathematical inequality that states the relationship between the sum of two numbers raised to a power and the individual numbers raised to the same power. It is often used to prove other inequalities and is named after the mathematician Hermann Minkowski.

2. What is the significance of Minkowski's Inequality?

Minkowski's Inequality is significant because it provides a way to compare the size of two numbers, even when they are raised to different powers. This makes it a valuable tool in many areas of mathematics, including analysis, geometry, and number theory.

3. What are the equality conditions for Minkowski's Inequality?

The equality conditions for Minkowski's Inequality are when the two numbers are equal, or when one number is 0 and the other is any real number. In these cases, the inequality becomes an equality, meaning that the two numbers are of the same size.

4. How is Minkowski's Inequality used in other areas of mathematics?

Minkowski's Inequality is used in many different areas of mathematics. In analysis, it is used to prove the triangle inequality, which is a fundamental property of norms and metrics. In geometry, it is used to prove the Harnack's Inequality, which is used to study the curvature of surfaces. In number theory, it is used to prove the Cauchy–Schwarz inequality, which plays a central role in many other inequalities.

5. Are there any generalizations of Minkowski's Inequality?

Yes, there are several generalizations of Minkowski's Inequality, including the Hölder's Inequality and the Young's Inequality. These generalizations extend the concept of Minkowski's Inequality to include more than two numbers and different types of powers, making it a powerful tool in various mathematical contexts.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
792
  • Calculus and Beyond Homework Help
Replies
8
Views
343
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
163
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
444
  • Topology and Analysis
Replies
4
Views
204
Replies
4
Views
639
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus
Replies
9
Views
2K
Back
Top