Prove Minkowski's Inequality

Why the \(L^p\) Norm Needs a Triangle Inequality

Hölder’s inequality controls the integral of a product by two norms. Minkowski’s inequality uses that product estimate to show that adding functions cannot make their \(L^p\) norm exceed the sum of their separate norms. This is the triangle inequality for \(L^p\), and it is an essential part of treating the \(L^p\) norm as a measure of distance.

Throughout, \((X,\mathcal A,\mu)\) is a measure space. We use the definitions of \(L^p(X)\), \(\|f\|_p\), and \(\|f\|_\infty\) from the previous tutorial, Prove Hölder’s Inequality. Functions in \(L^p(X)\) are understood up to equality almost everywhere. For \(1<p<\infty\), let \(q=p/(p-1)\), so \(1/p+1/q=1\). We will prove the inequality for \(1\leq p\leq\infty\), treating the endpoint cases separately.

Theorem (Minkowski’s Inequality): Let \(f,g\in L^p(X)\). If \(1\leq p<\infty\), then \(f+g\in L^p(X)\) and \( \displaystyle \|f+g\|_p\leq\|f\|_p+\|g\|_p. \) If \(p=\infty\), then \(f+g\in L^\infty(X)\) and \( \displaystyle \|f+g\|_\infty\leq\|f\|_\infty+\|g\|_\infty. \)

For \(p=1\) and \(p=\infty\), the proof follows directly from the pointwise triangle inequality. When \(1<p<\infty\), the main step is applying Hölder’s inequality to \(f\) and a suitable power of \(f+g\). Before doing so, we must know that \(f+g\) is in \(L^p(X)\); this guarantees that the power used in Hölder’s inequality has a finite norm.

A Preliminary Bound for Sums

For nonnegative \(a,b\) and \(p\geq1\), the elementary estimate \( (a+b)^p\leq 2^{p-1}(a^p+b^p) \) ensures the needed integrability. It follows from the convexity of \(t\mapsto t^p\) on \([0,\infty)\): applying the midpoint convexity inequality gives \( ((a+b)/2)^p\leq(a^p+b^p)/2 \). The function is convex because its derivative \(pt^{p-1}\) is nondecreasing on \((0,\infty)\), with continuity at zero. Multiplying the midpoint inequality by \(2^p\) yields the stated estimate.

Since \(|f+g|\leq |f|+|g|\) almost everywhere, the estimate with \(a=|f(x)|\) and \(b=|g(x)|\) gives, for \(1\leq p<\infty\), \( |f+g|^p\leq 2^{p-1}(|f|^p+|g|^p) \) almost everywhere. The right-hand side has finite integral because \(f,g\in L^p(X)\). Thus \(f+g\in L^p(X)\). This preliminary argument establishes membership only; it is not yet the sharper norm estimate in Minkowski’s inequality.

Proof for \(1<p<\infty\)

Set \(h=f+g\). By the preliminary bound, \(h\in L^p(X)\). If \(\|h\|_p=0\), the desired inequality holds immediately, since \( 0=\|h\|_p\leq\|f\|_p+\|g\|_p \). Suppose instead that \(\|h\|_p>0\). At almost every point, the pointwise triangle inequality gives \( |h|^p=|h|\,|h|^{p-1}\leq(|f|+|g|)|h|^{p-1} \). We can integrate this bound and use Hölder’s inequality on each term.

Proof of Minkowski’s Inequality for \(1<p<\infty\): Because \(h\in L^p(X)\) and \((p-1)q=p\), we have \( \displaystyle \big\||h|^{p-1}\big\|_q =\left(\int_X|h|^{(p-1)q}\,d\mu\right)^{1/q} =\left(\int_X|h|^p\,d\mu\right)^{1/q} =\|h\|_p^{p-1}. \) The middle equality uses \((p-1)q=p\); the last uses \(p/q=p-1\). By monotonicity of the integral and Hölder’s inequality, \( \displaystyle \int_X|h|^p\,d\mu \leq\int_X|f|\,|h|^{p-1}\,d\mu+\int_X|g|\,|h|^{p-1}\,d\mu \leq(\|f\|_p+\|g\|_p)\|h\|_p^{p-1}. \) The integrals on the right are finite by Hölder’s inequality, so the inequalities are justified. Since \(\int_X|h|^p\,d\mu=\|h\|_p^p\) and \(\|h\|_p>0\), divide both sides by \(\|h\|_p^{p-1}\) to obtain \( \displaystyle \|f+g\|_p=\|h\|_p\leq\|f\|_p+\|g\|_p. \) This proves the finite-exponent case for \(1<p<\infty\). \(\square\)

The conjugate exponent is chosen to match the power of \(h\): raising \(|h|^{p-1}\) to the \(q\)-th power gives \(|h|^p\). As a result, Hölder’s inequality contributes precisely \(\|h\|_p^{p-1}\), which can be divided out. The zero-norm case is separate because division by \(\|h\|_p^{p-1}\) would otherwise be invalid.

The Endpoint Cases

Proof for \(p=1\): The pointwise triangle inequality gives \(|f+g|\leq|f|+|g|\) almost everywhere. Integrating and using the monotonicity and additivity of the integral for nonnegative functions, \( \displaystyle \|f+g\|_1=\int_X|f+g|\,d\mu \leq\int_X|f|\,d\mu+\int_X|g|\,d\mu =\|f\|_1+\|g\|_1. \) The right-hand side is finite, so \(f+g\in L^1(X)\). \(\square\)

For \(p=\infty\), the definition of essential supremum gives \(|f|\leq\|f\|_\infty\) almost everywhere and \(|g|\leq\|g\|_\infty\) almost everywhere. To see why the bound holds even if the infimum in the definition is not initially known to be attained, for each positive integer \(n\) we have \(|f|\leq\|f\|_\infty+1/n\) outside a null set. The union of these countably many null sets is null; off that union, letting \(n\) increase gives \(|f|\leq\|f\|_\infty\). The same reasoning applies to \(g\).

Proof for \(p=\infty\): Outside the union of the two null sets just described, both bounds hold. Hence \( \displaystyle |f+g|\leq|f|+|g|\leq\|f\|_\infty+\|g\|_\infty. \) This is an almost-everywhere bound by a finite constant. Therefore \(f+g\in L^\infty(X)\), and taking the essential supremum gives \( \displaystyle \|f+g\|_\infty\leq\|f\|_\infty+\|g\|_\infty. \) This proves the remaining endpoint case. \(\square\)

Worked Applications

Worked Example: Proportional Functions Give Equality

Let \(X=[0,1]\) with Lebesgue measure, let \(p=3\), and define \(f(x)=1\) and \(g(x)=2\). Their norms are \( \displaystyle \|f\|_3=\left(\int_0^1 1^3\,dx\right)^{1/3}=1,\qquad \|g\|_3=\left(\int_0^1 2^3\,dx\right)^{1/3}=2. \) Since \(f+g=3\) everywhere, \( \displaystyle \|f+g\|_3=\left(\int_0^1 3^3\,dx\right)^{1/3}=3. \) Thus \(\|f+g\|_3=3=\|f\|_3+\|g\|_3\). This example illustrates that the inequality can be an equality when the two functions point in the same direction, here because both are nonnegative constant multiples of the same function.

Worked Example: A Strict Inequality on the Unit Interval

On \([0,1]\), take \(p=2\), \(f(x)=1\), and \(g(x)=x\). We have \( \displaystyle \|f\|_2=\left(\int_0^1 1\,dx\right)^{1/2}=1,\qquad \|g\|_2=\left(\int_0^1 x^2\,dx\right)^{1/2}=\frac{1}{\sqrt3}. \) Also, \(f+g=1+x\), so \( \displaystyle \|f+g\|_2 =\left(\int_0^1(1+x)^2\,dx\right)^{1/2} =\left(\int_0^1(1+2x+x^2)\,dx\right)^{1/2} =\sqrt{\frac73}. \) The asserted bound is \(\sqrt{7/3}\leq1+1/\sqrt3\). Both sides are positive, so squaring is valid. The squared right-hand side is \(4/3+2/\sqrt3\), and \( \displaystyle \frac73\leq\frac43+\frac{2}{\sqrt3} \) is equivalent to \(1\leq2/\sqrt3\). This holds because both sides are positive and \(1\leq4/3\) after squaring. The inequality is strict.

Worked Example: The Supremum Norm of a Sum of Vectors

A finite set with counting measure turns the \(L^\infty\) norm into the maximum absolute coordinate. Take \(f=(2,-1,0)\) and \(g=(-1,3,1)\). Then \( \displaystyle \|f\|_\infty=\max\{2,1,0\}=2,\qquad \|g\|_\infty=\max\{1,3,1\}=3. \) Their coordinatewise sum is \(f+g=(1,2,1)\), giving \( \displaystyle \|f+g\|_\infty=\max\{1,2,1\}=2\leq5=\|f\|_\infty+\|g\|_\infty. \) The coordinate calculations show why cancellation in some entries can make the bound strict.

Worked Example: Disjoint Supports

On \([0,2]\), let \(f\) be the indicator function of \([0,1]\), let \(g\) be the indicator function of \((1,2]\), and take \(1\leq p<\infty\). The two functions are nonzero on disjoint sets, and each set has measure one. Consequently, \( \displaystyle \|f\|_p=\left(\int_0^2|f|^p\,dx\right)^{1/p}=1,\qquad \|g\|_p=\left(\int_0^2|g|^p\,dx\right)^{1/p}=1. \) Their sum equals one almost everywhere on \([0,2]\), so \( \displaystyle \|f+g\|_p=\left(\int_0^2|f+g|^p\,dx\right)^{1/p}=2^{1/p}. \) Since \(2^{1/p}\leq2\) for \(p\geq1\), Minkowski’s inequality gives the correct bound in this example. For \(p>1\), the inequality is strict; at \(p=1\), it is equality.

What the Inequality Accomplishes

Minkowski’s inequality establishes the triangle inequality for the \(L^p\) norm. In particular, it bounds the distance between two functions, measured by \(\|f-g\|_p\), in a way that is compatible with addition. This is why the result is more than an estimate for a single integral: it supplies the geometric structure needed to discuss distances and convergence in \(L^p\) spaces.

A common proof error is to apply Hölder’s inequality to \(|f|\) and \(|f+g|^{p-1}\) before verifying that \(f+g\in L^p(X)\). The preliminary bound resolves this issue without assuming the conclusion. Another error is to divide by \(\|f+g\|_p^{p-1}\) without first considering whether that norm is zero. Keeping both steps explicit makes the argument valid in all cases.

Verify membership.
For finite \(p\), use the preliminary power bound to establish \(f+g\in L^p(X)\).

Separate the zero case.
If \(\|f+g\|_p=0\), the triangle inequality is immediate.

Match Hölder’s exponents.
For \(1<p<\infty\), pair \(|f|\) and \(|g|\) with \(|f+g|^{p-1}\) using the conjugate exponent \(q\).

Divide only after estimating.
The resulting bound contains \(\|f+g\|_p^{p-1}\), which can be divided out in the positive-norm case.

Check Your Understanding

Use the proof and examples to answer these questions.

Why is a preliminary estimate needed to show that \(f+g\in L^p(X)\) before applying Hölder’s inequality?
For \(1<p<\infty\), why is the conjugate exponent applied to \(|f+g|^{p-1}\) a useful choice?
Which case must be handled before dividing by \(\|f+g\|_p^{p-1}\), and why?
How does the proof of Minkowski’s inequality for \(p=1\) differ from the proof for \(1<p<\infty\)?
What almost-everywhere bounds lead directly to the triangle inequality for the \(L^\infty\) norm?

What You'll Learn