Minkowski's Inequality for Integrals/Equality

Theorem

Let $f, g$ be (Darboux) integrable functions.

Let $p \in \R$ such that $p > 1$.

Then equality in Minkowski's Inequality for Integrals, that is:

$\ds \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1/p} = \paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p}$

holds if and only if, for all $x \in \closedint a b$:

$\map g x = c \map f x$

for some $c \in \R_{>0}$.

Proof

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Also presented as

Some sources present the for Minkowski's inequality for integrals as:

$\ds \paren {\int_a^b \size {\map f x + \map g x}^p \rd x}^{1/p} = \paren {\int_a^b \size {\map f x}^p \rd x}^{1 / p} + \paren {\int_a^b \size {\map g x}^p \rd x}^{1 / p}$

holds if and only if, for all $x \in \closedint a b$:

$\dfrac {\map f x} {\map g x} = c$

for some $c \in \R_{>0}$.

This may confuse those who ask the question as to what happens if $\map g x = 0$.

The immediate response here is that if $\map g x = 0$, then $\dfrac {\map f x} {\map g x} \ne c$ unless it is the case that $\map f x = 0$ also.

However, this would then suggest that the is undefined at such a point.

Hence the given $\map g x = c \map f x$ is in general preferred.

Source of Name

This entry was named for Hermann Minkowski.

Sources

1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.2$ Inequalities: Minkowski's Inequality for Integrals: $3.2.13$
1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 36$: Inequalities: Minkowski's Inequality for Integrals: $36.15$
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 37$: Inequalities: Minkowski's Inequality for Integrals: $37.15.$