Minkowski’s inequality is the triangle inequality for \(L^p\) norms. For \(1\leq p\leq\infty\) and \(f,g\in L^p(X,\mu)\),
$$\|f+g\|_p\leq\|f\|_p+\|g\|_p.$$For vectors in \(\mathbb{R}^n\) it reads \(\big(\sum_i|x_i+y_i|^p\big)^{1/p}\leq\big(\sum_i|x_i|^p\big)^{1/p}+\big(\sum_i|y_i|^p\big)^{1/p}\). It fails for \(0<p<1\).
Minkowski’s Inequality in Three Forms
Hölder’s inequality bounds the integral of a product by two \(L^p\) norms. Minkowski’s inequality answers a different question: how large can the norm of a sum be? Adding functions can increase their \(L^p\) norm, but never by more than the sum of the separate norms. The same statement appears in three common settings.
| Setting | Statement (\(1\leq p<\infty\)) |
|---|---|
| Vectors in \(\mathbb{R}^n\) | \(\big(\sum_{i=1}^n|x_i+y_i|^p\big)^{1/p}\leq\big(\sum_{i=1}^n|x_i|^p\big)^{1/p}+\big(\sum_{i=1}^n|y_i|^p\big)^{1/p}\) |
| Sequences in \(\ell^p\) | The same inequality with \(n\) replaced by \(\infty\), for \(x,y\in\ell^p\) |
| Functions in \(L^p(X,\mu)\) | \(\big(\int_X|f+g|^p\,d\mu\big)^{1/p}\leq\big(\int_X|f|^p\,d\mu\big)^{1/p}+\big(\int_X|g|^p\,d\mu\big)^{1/p}\) |
For \(p=\infty\) the norm is the essential supremum, and the inequality follows directly from \(|f+g|\leq|f|+|g|\) almost everywhere. The sequence and vector forms are the special case of the function form with counting measure. Functions are understood as elements of \(L^p\), so changing a representative on a null set does not change any norm.
This is exactly what makes \(\|\cdot\|_p\) a norm, so that \(L^p\) is a normed vector space. The pointwise triangle inequality \(|f+g|\leq|f|+|g|\) is not enough by itself when \(p>1\), because the power \(p\) and the integral must still be handled. The proof uses Hölder’s inequality to do this; see Proof of Minkowski’s Inequality for the full argument.
Numerical Check in \(\mathbb{R}^3\)
Worked Example: Different Values of \(p\)
Take \(x=(1,2,2)\) and \(y=(2,3,6)\), so \(x+y=(3,5,8)\). The table compares \(\|x+y\|_p\) with \(\|x\|_p+\|y\|_p\).
| \(p\) | \(\|x\|_p\) | \(\|y\|_p\) | \(\|x+y\|_p\) | \(\|x\|_p+\|y\|_p\) |
|---|---|---|---|---|
| 1 | 5 | 11 | 16 | 16 |
| 2 | 3 | 7 | \(\sqrt{98}\approx9.899\) | 10 |
| 3 | 2.571 | 6.308 | 8.724 | 8.879 |
| \(\infty\) | 2 | 6 | 8 | 8 |
In every row the left side is at most the right side. For \(p=2\), \(\|x\|_2=\sqrt{1+4+4}=3\), \(\|y\|_2=\sqrt{4+9+36}=7\) and \(\|x+y\|_2=\sqrt{9+25+64}=\sqrt{98}\), which is slightly below \(10\). For \(p=1\) and \(p=\infty\) equality happens here because all entries are positive and the largest entry of \(x+y\) occurs in the same place as the largest entry of \(y\). That kind of alignment is what the next section describes.
Worked Example: Functions on \([0,1]\)
Let \(X=[0,1]\) with Lebesgue measure, \(p=2\), \(f(x)=x\) and \(g(x)=1\). Then $$ \|f\|_2=\left(\int_0^1x^2\,dx\right)^{1/2}=\frac{1}{\sqrt3},\qquad \|g\|_2=1,\qquad \|f+g\|_2=\left(\int_0^1(1+x)^2\,dx\right)^{1/2}=\sqrt{\frac73}. $$ Numerically \(\sqrt{7/3}\approx1.5275\), while \(\|f\|_2+\|g\|_2\approx0.5774+1=1.5774\). The inequality holds, and it is strict because \(f\) and \(g\) are not multiples of each other.
When Does Equality Hold?
Equality is rare. The answer depends on \(p\).
- \(1<p<\infty\): \(\|f+g\|_p=\|f\|_p+\|g\|_p\) holds exactly when one function is a nonnegative multiple of the other almost everywhere, meaning \(af=bg\) a.e. for some constants \(a,b\geq0\), not both zero. The functions must point in the same “direction”.
- \(p=1\): equality holds exactly when \(|f+g|=|f|+|g|\) almost everywhere, which for real-valued functions means \(f\) and \(g\) never have opposite signs. They need not be multiples of each other.
- \(p=\infty\): there is no simple multiple condition. Equality only requires the two functions to come close to their essential maxima at the same places, as in the last row of the table above.
Worked Example: A Strict Bound on a Two-Point Space
Let \(X=\{1,2\}\) with counting measure and \(p=2\). Set \(f=(1,1)\) and \(g=(1,-1)\). Then \(f+g=(2,0)\), so $$ \|f+g\|_2=2,\qquad\|f\|_2=\sqrt2,\qquad\|g\|_2=\sqrt2. $$ Minkowski’s inequality reads \(2\leq2\sqrt2\approx2.83\), which is strict. Here \(f\) and \(g\) point in different directions, and the cancellation in the second coordinate makes the sum smaller.
Why \(p\geq1\) Is Needed
For \(0<p<1\) the expression \(\|f\|_p=(\int|f|^p)^{1/p}\) is still defined, but the triangle inequality can fail. Take \(X=\{1,2\}\) with counting measure, \(p=\tfrac12\), \(f=(1,0)\) and \(g=(0,1)\). Then $$ \|f+g\|_{1/2}=(1^{1/2}+1^{1/2})^{2}=4,\qquad \|f\|_{1/2}+\|g\|_{1/2}=1+1=2, $$ and \(4>2\). For nonnegative functions and \(0<p<1\) the inequality actually reverses: \(\|f+g\|_p\geq\|f\|_p+\|g\|_p\).
Consequences: Finite Sums and Weighted Averages
Applying the two-function inequality repeatedly gives a bound for any finite collection.
Proof. Induct on \(n\). For \(n=1\) both sides equal \(\|f_1\|_p\). If the claim holds for \(n\), apply the two-function inequality to \(\sum_{k=1}^nf_k\) and \(f_{n+1}\), then use the induction hypothesis on the first term: $$ \left\|\sum_{k=1}^{n+1}f_k\right\|_p \leq\left\|\sum_{k=1}^{n}f_k\right\|_p+\|f_{n+1}\|_p \leq\sum_{k=1}^{n+1}\|f_k\|_p. $$
Proof. Apply the finite-sum bound to the functions \(\alpha_kf_k\) and use \(\|\alpha_kf_k\|_p=\alpha_k\|f_k\|_p\), which holds because \(\alpha_k\geq0\).
Worked Example: Bounding an Approximation Error
Suppose \(u,v,w\in L^p\) with \(\|u-v\|_p\leq0.03\) and \(\|v-w\|_p\leq0.02\). Since \(u-w=(u-v)+(v-w)\), Minkowski’s inequality gives \(\|u-w\|_p\leq0.03+0.02=0.05\). No assumption about the signs or the supports of the two errors is needed.
A Related Result: Minkowski’s Integral Inequality
There is a continuous version in which the sum over \(k\) becomes an integral. If \(F\geq0\) is measurable on a product of \(\sigma\)-finite measure spaces \(X\times Y\) and \(1\leq p<\infty\), then $$ \left(\int_X\left(\int_YF(x,y)\,d\nu(y)\right)^pd\mu(x)\right)^{1/p} \leq\int_Y\left(\int_XF(x,y)^p\,d\mu(x)\right)^{1/p}d\nu(y). $$ In words, the \(L^p\) norm of an integral is at most the integral of the \(L^p\) norms. The finite-sum inequality is the case where \(\nu\) is counting measure on \(\{1,\ldots,n\}\).
Common Mistakes
- Norm bound versus pointwise bound. Minkowski’s inequality does not say that \(|f+g|\) is at most \(\|f\|_p+\|g\|_p\) at every point. It bounds the norm of the sum.
- Infinite sums. The finite-sum bound does not by itself prove an inequality for an infinite series. That needs a limiting argument, such as monotone convergence for nonnegative terms.
- Expecting equality. The inequality is usually strict. Cancellation between the functions, or different shapes, makes the left side smaller.
- Using \(p<1\). The inequality is false in general for \(0<p<1\).
Frequently Asked Questions
What is Minkowski’s inequality?
It is the triangle inequality for the \(L^p\) norm: \(\|f+g\|_p\leq\|f\|_p+\|g\|_p\) for \(1\leq p\leq\infty\).
How is Minkowski’s inequality different from Hölder’s inequality?
Hölder’s inequality bounds an integral of a product, \(\int|fg|\leq\|f\|_p\|g\|_q\) with \(1/p+1/q=1\). Minkowski’s inequality bounds the norm of a sum. The standard proof of Minkowski’s inequality uses Hölder’s inequality.
Does Minkowski’s inequality hold for \(p<1\)?
No. For \(0<p<1\) it can fail, as the example above shows with \(p=\tfrac12\). For nonnegative functions the reverse inequality holds instead.
When is Minkowski’s inequality an equality?
For \(1<p<\infty\), exactly when one function is a nonnegative multiple of the other almost everywhere. For \(p=1\), when \(f\) and \(g\) never have opposite signs.
Why does Minkowski’s inequality matter?
It shows that \(\|\cdot\|_p\) is a norm, so \(L^p\) and \(\ell^p\) are normed spaces. This is the starting point for completeness (Riesz–Fischer), convergence in \(L^p\), and error estimates for approximations.
Check Your Understanding
- For which exponents \(p\) does Minkowski’s inequality hold, and what does it bound?
- Verify the inequality for \(x=(3,4)\), \(y=(0,5)\) and \(p=2\).
- Give an example with \(p=\tfrac12\) where the inequality fails.
- When does equality hold for \(1<p<\infty\)?
- Why does the finite-sum estimate not directly give a bound for an infinite series?