Worked Applications Of Minkowski's Inequality

Using Minkowski as an Estimate

Minkowski’s inequality turns a sum of functions into a bound involving the norms of its summands. In applications, the bound is often not the exact norm: computing both quantities shows how much information the inequality retains. We will compare them for functions that overlap, functions whose nonzero sets are disjoint, and functions that are proportional. The examples also show that equality can occur, but should not be expected automatically.

For a measure space $(X,\mathcal{M},\mu)$, and a real-valued measurable function $f$, write $\|f\|_p=(\int_X |f|^p\,d\mu)^{1/p}$ for $1\le p<\infty$, when this quantity is finite. The space $L^p(\mu)$ consists of measurable functions with finite $p$-norm, identified when equal $\mu$-a.e. For $p=\infty$, $\|f\|_\infty$ is the essential supremum of $|f|$; $L^\infty(\mu)$ consists of functions with finite essential supremum, again identified up to equality $\mu$-a.e.

Theorem (Minkowski’s inequality): On any measure space $(X,\mathcal{M},\mu)$, if $f,g\in L^p(\mu)$ are real-valued and $1\le p\le\infty$, then $f+g\in L^p(\mu)$ and $\|f+g\|_p\le\|f\|_p+\|g\|_p$. For $1<p<\infty$, this is Minkowski’s inequality proved earlier in the course; the $p=1$ and $p=\infty$ cases follow from the pointwise triangle inequality and the definitions of these norms.

The inequality supplies an upper bound without requiring us to know the shape of $f+g$. When the functions have a simple form, however, it is useful to compute $\|f+g\|_p$ directly and compare. Equality is not a consequence of membership in $L^p$; it depends on how the summands relate.

Overlapping Functions: A Strict Bound

Worked Example: The Constant Function Plus the Identity

On $X=[0,1]$ with the Lebesgue sigma-algebra and Lebesgue measure $m$, take $p=2$, $f(x)=1$, and $g(x)=x$. Both functions are real-valued and belong to $L^2(m)$. Compute the exact norm of their sum and compare it with Minkowski’s bound.

First, the individual norms are

\|f\|_2=\left(\int_0^1 1\,dx\right)^{1/2}=1, \qquad \|g\|_2=\left(\int_0^1 x^2\,dx\right)^{1/2} =\left(\frac13\right)^{1/2}=\frac1{\sqrt3}.

Minkowski therefore gives $\|f+g\|_2\le 1+1/\sqrt3$. To find the exact value, square the sum inside the integral and integrate each term:

\|f+g\|_2 =\left(\int_0^1(1+x)^2\,dx\right)^{1/2} =\left(\int_0^1(1+2x+x^2)\,dx\right)^{1/2} =\left(1+1+\frac13\right)^{1/2} =\sqrt{\frac73}.

The bound is strict. Both sides are positive, and their squares differ by

\left(1+\frac1{\sqrt3}\right)^2-\frac73 =\left(\frac43+\frac2{\sqrt3}\right)-\frac73 =\frac2{\sqrt3}-1>0,

where the last inequality follows from $2/\sqrt3>1$, since both sides are positive and $4/3>1$. Check: squaring the exact answer gives $7/3$, which is the integral of $1+2x+x^2$; squaring the Minkowski bound gives $4/3+2/\sqrt3$, a strictly larger value.

This calculation illustrates a useful distinction: Minkowski estimates the norm of the sum using only the separate norms, whereas the exact integral also reflects the interaction between the functions. In this example, the summands are not proportional, and the estimate leaves a positive gap.

Disjoint Nonzero Sets: An Exact Formula

Proposition: Let $1\le p<\infty$, and let $f,g\in L^p(\mu)$ be real-valued on a measure space $(X,\mathcal{M},\mu)$. Suppose $\{x:f(x)\ne0\}\cap\{x:g(x)\ne0\}$ is a $\mu$-null set. Then $$ \|f+g\|_p^p=\|f\|_p^p+\|g\|_p^p. $$ Consequently, $\|f+g\|_p=(\|f\|_p^p+\|g\|_p^p)^{1/p}$.

Proof. Outside the null intersection of the nonzero sets, at most one of $f(x)$ and $g(x)$ is nonzero. If $f(x)\ne0$, then $g(x)=0$ and $|f(x)+g(x)|^p=|f(x)|^p=|f(x)|^p+|g(x)|^p$. If $g(x)\ne0$, the same equality holds with the roles reversed. If both are zero, both sides are zero. Thus $|f+g|^p=|f|^p+|g|^p$ $\mu$-a.e. Integrating this equality is valid because the functions are nonnegative and integrable; the integral is unchanged by alteration on a null set. Additivity of the integral gives the stated identity. Taking nonnegative $p$th roots gives the norm formula. $\square$

The hypothesis concerns the sets where the functions are nonzero, up to a null set. It does not assert that their topological supports—the closures of those sets—are disjoint. This distinction matters at shared boundary points, even though a single boundary point has Lebesgue measure zero in the example below.

Worked Example: A Disjoint-Sets Estimate in L³

On $X=[0,5]$ with Lebesgue measure, let $f=2\chi_{[0,1)}$ and $g=\chi_{[1,5]}$. These are real-valued measurable functions in $L^3(m)$. The sets where they are nonzero are $[0,1)$ and $[1,5]$, which are disjoint, including at $x=1$. Compute the exact norm and the Minkowski bound.

The first nonzero set has measure $1-0=1$, and the second has measure $5-1=4$. Therefore

\|f\|_3=(2^3\cdot1)^{1/3}=2, \qquad \|g\|_3=(1^3\cdot4)^{1/3}=\sqrt[3]{4}.

The disjoint-sets formula gives

\|f+g\|_3=(2^3\cdot1+1^3\cdot4)^{1/3} =\sqrt[3]{12}.

Minkowski gives the larger upper bound $2+\sqrt[3]{4}$. To check strictness without decimal approximations, put $a=\sqrt[3]{4}>0$. Then

$$ (2+a)^3=8+12a+6a^2+a^3 =12+12a+6a^2>12. $$

Taking cube roots preserves the strict inequality, so $\sqrt[3]{12}<2+\sqrt[3]{4}$. Check: cubing the exact norm gives $12$, equal to the integral of $|f+g|^3$, while cubing the bound gives $12+12a+6a^2>12$.

For disjoint nonzero sets, the exact formula combines the $p$th powers of the norms, not the norms themselves. Since $p>1$ makes $(u^p+v^p)^{1/p}$ smaller than $u+v$ when $u,v>0$, the strict gap in this example is expected.

Equality and Endpoint Comparisons

Worked Example: Proportional Functions Give Equality

On $[0,1]$ with Lebesgue measure, take $p=2$, $h(x)=x$, $f(x)=2h(x)$, and $g(x)=3h(x)$. These are finite-valued real measurable functions in $L^2(m)$. Compute both sides of Minkowski’s inequality.

Since $\int_0^1x^2\,dx=1/3$, we have $\|h\|_2=1/\sqrt3$. Hence

\|f\|_2=\frac2{\sqrt3}, \qquad \|g\|_2=\frac3{\sqrt3}, \qquad \|f\|_2+\|g\|_2=\frac5{\sqrt3}.

Pointwise, $f+g=5x$. Direct integration gives

\|f+g\|_2 =\left(\int_0^1(5x)^2\,dx\right)^{1/2} =\left(25\cdot\frac13\right)^{1/2} =\frac5{\sqrt3}.

Thus equality holds. Check: the square of the exact norm is $25/3$, and the square of the sum of the individual norms is $(2/\sqrt3+3/\sqrt3)^2=25/3$. More generally, if $h$ is a finite-valued real measurable function in $L^p(\mu)$, and $f=ah,\ g=bh$ for constants $a,b\ge0$, then $f+g=(a+b)h$; direct homogeneity of the integral shows equality in Minkowski. Requiring $h$ to be finite-valued avoids undefined products at points where an extended-valued representative might equal infinity.

The endpoint $p=\infty$ can also be used for quick estimates. The essential supremum ignores changes on null sets, but it still measures the largest values that occur outside every null exceptional set.

Worked Example: An L-Infinity Bound with Equality

On $[0,1]$ with Lebesgue measure, let $f(x)=1$ and $g(x)=1$, both real-valued. Each has essential supremum $1$, and $f+g=2$ everywhere. Therefore

\|f+g\|_\infty=2=\|f\|_\infty+\|g\|_\infty.

This verifies equality directly. As a contrasting calculation on the same space, take $u(x)=x$ and $v(x)=1-x$. Their essential suprema are both $1$: for every $K<1$, the sets where $x>K$ and where $1-x>K$ have positive measure, while both functions are at most $1$. But $u+v=1$ everywhere, so $\|u+v\|_\infty=1<2=\|u\|_\infty+\|v\|_\infty$. Check: the pointwise sum is exactly $1$, while each individual essential supremum is $1$; thus the strict bound is $1\le2$, not equality.

Adding More Than Two Functions

Corollary: If $1\le p\le\infty$ and $f_1,\dots,f_N\in L^p(\mu)$ are real-valued on a measure space $(X,\mathcal{M},\mu)$, where $N$ is a positive integer, then $$ \left\|\sum_{j=1}^N f_j\right\|_p\le\sum_{j=1}^N\|f_j\|_p. $$

Proof. For $N=1$, the assertion is equality. Suppose it holds for some positive integer $N$, and set $F_N=\sum_{j=1}^N f_j$. The induction hypothesis gives $F_N\in L^p(\mu)$. Minkowski applied to $F_N$ and $f_{N+1}$ then gives $\|F_N+f_{N+1}\|_p\le\|F_N\|_p+\|f_{N+1}\|_p\le\sum_{j=1}^{N+1}\|f_j\|_p$. It also gives $F_N+f_{N+1}\in L^p(\mu)$. Induction proves the claim for every positive integer $N$. $\square$

This iterative estimate is useful when the summands are individually easy to bound but their sum is difficult to integrate directly. It also makes the limitation plain: a sum of separate upper bounds need not be sharp, as the strict examples above demonstrate.

A Common Mistake

A common error is to treat disjoint nonzero sets as if the norm of the sum were the sum of the norms. In the L³ example, that incorrect calculation would give $\|f+g\|_3=2+\sqrt[3]{4}$. But disjointness gives $\|f+g\|_3^3=\|f\|_3^3+\|g\|_3^3=8+4=12$, so the correct value is $\sqrt[3]{12}$. The incorrect answer is the Minkowski upper bound, not the exact norm. Also, disjointness of the sets where functions are nonzero is not the same as disjointness of their topological supports: closures can meet even when the nonzero sets do not.

Check Your Understanding

For each question, state the measure space and exponent, and distinguish an exact norm from an upper bound.

On $[0,1]$ with Lebesgue measure, compute $\|1+x\|_2$ and compare it with $\|1\|_2+\|x\|_2$.
If $f,g\in L^p(\mu)$, $1\le p<\infty$, and their nonzero sets intersect only on a null set, what is $\|f+g\|_p^p$?
On $[0,1]$, let $f(x)=x$ and $g(x)=4x$. Verify directly that equality holds in Minkowski for $p=2$.
Give an example on $[0,1]$ where the $L^\infty$ Minkowski bound is strict, and compute both sides.
Explain why a shared endpoint of the closures of two nonzero sets does not by itself contradict disjointness of the nonzero sets.