Minkowski's Inequality and Equality Conditions

[AKG]

Definitions and Useful Facts

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

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

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

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

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

||f+g||_{\infty} \leq ||f||_{\infty} + ||g||_{\infty}

(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 z : X \to C where C is the complex circle by:

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

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

|fz| - |f| = |gz| - |g| = |(f+g)z| - |f+g| = 0

hence

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

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:

\mu ( \{x : f(x) &gt; a, |g(x)| &gt; b, ||g(x)| - g(x)| &lt; c|g(x)|\} ) &gt; 0

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?

 

[AKG]

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:

(\forall a &lt; A)(\forall b &lt; B)(\forall \epsilon &gt; 0)(\mu (\{x : |f(x)| &gt; a, |g(x)| &gt; b, |\overline{g(x)} - \overline{f(x)}| &lt; \epsilon \} ) &gt; 0

where \overline{z} = z/|z| for every non-zero complex number z.

 

[StatusX]

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 \{ x | \mbox{ }||f||_\infty-|f(x)|&lt;\epsilon\}, any \epsilon&gt;0, and so outside this range (which can usually be made arbitrarily small), the values of the function are completely irrelevant. Note that your condition |\overline{g(x)} - \overline{f(x)}| &lt; \epsilon is essentially a rewrite of |f(x)|+|g(x)|-|f(x)+g(x)|&lt;\epsilon&#039;, some \epsilon&#039; that goes to zero as \epsilon does (again, up to the case where on of the functions approaches zero), which shows your proposal is correct, if not all that enlightening.

 

[AKG]

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 \{ x | \mbox{ }||f||_\infty-|f(x)|&lt;\epsilon\}, any \epsilon&gt;0, and so outside this range (which can usually be made arbitrarily small), the values of the function are completely irrelevant. Note that your condition |\overline{g(x)} - \overline{f(x)}| &lt; \epsilon is essentially a rewrite of |f(x)|+|g(x)|-|f(x)+g(x)|&lt;\epsilon&#039;, some \epsilon&#039; that goes to zero as \epsilon 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 (\forall a \in (0,A))(\forall b \in (0,B))\dots 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.

 

[StatusX]

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.

 

[AKG]

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?

||f+g||_1 = ||f||_1 + ||g||_1

iff

\int |f+g| = \int |f| + \int |g|

iff

\int |f+g| = \int |f| + |g|

iff

|f+g| = |f| + |g| a.e.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)

 

[StatusX]

Sorry, I had something backwards. It seems weird, but I guess that's right.