Minkowski's Inequality

Controlling the Norm of a Sum

Hölder’s inequality, proved in the previous tutorial, bounds the integral of a product using two $L^p$ norms. A closely related question is how the norm of a sum compares with the norms of its terms. The answer is Minkowski’s inequality: adding functions can increase their $L^p$ norm, but by no more than the sum of their individual norms.

The inequality was stated in Properties of the Lp Norm. We use that result here rather than reproving it. For $1\leq p<\infty$, the norm is $\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}$; for $p=\infty$, it is the essential supremum norm. In either case, functions are understood as elements of $L^p$, so changing representatives on null sets does not change the norm.

Theorem (Minkowski’s Inequality): Let $1\leq p\leq\infty$, and let $f,g\in L^p(X,\mu)$. Then $$ \|f+g\|_p\leq\|f\|_p+\|g\|_p. $$

The statement resembles the triangle inequality for the absolute value of real numbers, $|a+b|\leq|a|+|b|$. But it is a statement about whole functions: after adding the functions pointwise, we measure the size of the result using an integral norm or an essential supremum. Pointwise triangle inequality alone does not immediately give the $L^p$ estimate when $p>1$; the power and integral must also be handled. Minkowski’s inequality supplies precisely that bridge.

Worked Examples

Worked Example: A Strict Bound on a Two-Point Space

Let $X=\{1,2\}$ with counting measure, and use $p=2$. Set $f=(1,1)$ and $g=(1,-1)$. Their sum is $f+g=(2,0)$, so direct calculation gives $$ \|f+g\|_2=\sqrt{2^2+0^2}=2, \qquad \|f\|_2=\sqrt{1^2+1^2}=\sqrt{2}, \qquad \|g\|_2=\sqrt{1^2+(-1)^2}=\sqrt{2}. $$ Thus Minkowski’s inequality reads $2\leq2\sqrt{2}$, which is strict. This example also shows why one should not expect the norm of a sum to equal the sum of the norms. The signs and relative directions of the functions can affect the size of the sum.

Worked Example: A Weighted Average of Indicator Functions

Take $X=[0,1]$ with Lebesgue measure and $p=2$. Let $f=\mathbf{1}_{[0,1/4]}$ and $g=\mathbf{1}_{(1/4,1]}$. The intervals are disjoint and cover $[0,1]$ apart from the choice of endpoint, which has measure zero. Define their average $h=(f+g)/2$. Almost everywhere, $h=1/2$, and therefore $$ \|h\|_2 =\left(\int_0^1\frac14\,dx\right)^{1/2} =\frac12. $$ The individual norms are $$ \|f\|_2=\left(\int_0^{1/4}1\,dx\right)^{1/2}=\frac12, \qquad \|g\|_2=\left(\int_{1/4}^1 1\,dx\right)^{1/2}=\frac{\sqrt{3}}{2}. $$ The average of those norms is $(1+\sqrt{3})/4$, which exceeds $1/2$. The weighted-average estimate below predicts exactly this type of upper bound: $$ \left\|\frac{f+g}{2}\right\|_2 \leq\frac{\|f\|_2+\|g\|_2}{2} =\frac{1+\sqrt{3}}{4}. $$

Worked Example: Bounding an Approximation Error

Suppose $u,v,w\in L^p(X,\mu)$, where $1\leq p\leq\infty$, and an approximation to $u$ is formed by adding two errors: $u-w=(u-v)+(v-w)$. Applying Minkowski’s inequality gives $$ \|u-w\|_p =\|(u-v)+(v-w)\|_p \leq\|u-v\|_p+\|v-w\|_p. $$ For instance, if $\|u-v\|_p\leq0.03$ and $\|v-w\|_p\leq0.02$, then $\|u-w\|_p\leq0.05$. The estimate does not require the errors to have the same sign or to be supported on the same region. It provides a reliable bound even when their detailed relationship is unknown.

Finite Sums of Functions

The two-function form can be applied repeatedly. This gives a useful estimate for any finite collection of $L^p$ functions. It is important that the collection is finite here: this result alone does not justify passing to an infinite series.

Theorem (Minkowski’s Inequality for Finite Sums): Let $1\leq p\leq\infty$, let $n$ be a positive integer, and let $f_1,\ldots,f_n\in L^p(X,\mu)$. Then $$ \left\|\sum_{k=1}^n f_k\right\|_p \leq\sum_{k=1}^n\|f_k\|_p. $$

Proof. We use induction on $n$. When $n=1$, both sides are $\|f_1\|_p$, so the inequality holds. Suppose it holds for some positive integer $n$. Since $L^p$ is linearly closed, $\sum_{k=1}^n f_k$ belongs to $L^p$. Apply the two-function Minkowski inequality to that sum and $f_{n+1}$: $$ \left\|\sum_{k=1}^{n+1}f_k\right\|_p \leq \left\|\sum_{k=1}^{n}f_k\right\|_p+\|f_{n+1}\|_p. $$ By the induction hypothesis, the first term on the right is at most $\sum_{k=1}^{n}\|f_k\|_p$. Hence $$ \left\|\sum_{k=1}^{n+1}f_k\right\|_p \leq\sum_{k=1}^{n}\|f_k\|_p+\|f_{n+1}\|_p =\sum_{k=1}^{n+1}\|f_k\|_p. $$ This proves the assertion for $n+1$, and induction completes the proof.

This estimate is useful whenever a function is assembled from several components. For example, if $r=\sum_{k=1}^n r_k$ and each $r_k$ is an error term, then $\|r\|_p\leq\sum_{k=1}^n\|r_k\|_p$. The estimate may not be sharp, because cancellation between the terms is ignored. Its strength is that it remains valid without requiring any information about such cancellation.

Weighted Averages

A convex combination is a finite weighted average with nonnegative weights that add to one. Minkowski’s inequality also bounds the norm of such an average. This is a useful way to combine several functions while keeping track of how much each one contributes.

Theorem (Norm Bound for Convex Combinations): Let $1\leq p\leq\infty$, let $f_1,\ldots,f_n\in L^p(X,\mu)$, and let $\alpha_1,\ldots,\alpha_n\geq0$ satisfy $\sum_{k=1}^n\alpha_k=1$. Then $$ \left\|\sum_{k=1}^n\alpha_k f_k\right\|_p \leq\sum_{k=1}^n\alpha_k\|f_k\|_p. $$

Proof. Each $\alpha_k f_k$ belongs to $L^p$. Apply the finite-sum inequality to these functions: $$ \left\|\sum_{k=1}^n\alpha_k f_k\right\|_p \leq\sum_{k=1}^n\|\alpha_k f_k\|_p. $$ By homogeneity of the $L^p$ norm and $\alpha_k\geq0$, each term satisfies $\|\alpha_k f_k\|_p=\alpha_k\|f_k\|_p$. Substitution gives the desired inequality. This proof also covers zero weights, since $\|0f_k\|_p=0$.

The condition that the weights sum to one is part of the interpretation as an average, but the same argument works for any nonnegative weights: it gives $\|\sum_k\alpha_k f_k\|_p\leq\sum_k\alpha_k\|f_k\|_p$. The normalization to one simply expresses the bound in a particularly useful form.

What the Inequality Does—and Does Not—Say

Minkowski’s inequality is an upper bound, not an equality formula. The first worked example has strict inequality because the sum is smaller than the total of the separate norms. In other situations, the terms may reinforce one another and equality can occur. In either case, the estimate is valid without deciding in advance whether cancellation or reinforcement takes place.

A common pitfall is to confuse a bound on the norm with a pointwise bound on the functions. Minkowski’s inequality does not say that $|f+g|$ is bounded by the number $\|f\|_p+\|g\|_p$ at almost every point. Rather, it bounds the $L^p$ norm of the pointwise sum. The pointwise estimate $|f+g|\leq|f|+|g|$ is a separate scalar triangle inequality; Minkowski’s inequality then controls the norm of the resulting sum.

Another useful distinction is between applying the estimate to finitely many terms and asserting it for an infinite sum. The finite-sum result follows by induction, but an infinite-series claim needs additional hypotheses and an argument that controls the limit. The finite estimate is often the first step in such arguments, not a replacement for them.

Key takeaway: Minkowski’s inequality bounds the $L^p$ norm of a sum by the sum of the $L^p$ norms. Repeated application controls finite sums, and homogeneity gives a corresponding bound for weighted averages.

Check Your Understanding

Use the two-function inequality and its consequences to answer the following questions.

For $f,g\in L^p$, what does Minkowski’s inequality bound, and for which exponents $p$ does it hold?
How does the induction step prove the finite-sum version from the two-function inequality?
Why must the weights be nonnegative in the stated convex-combination estimate?
If $\|u-v\|_p\leq a$ and $\|v-w\|_p\leq b$, what upper bound follows for $\|u-w\|_p$?
Why does the finite-sum estimate by itself not establish an inequality for an infinite series?

What You'll Learn