Norm Axioms

What the Norm Axioms Require

In Norms, a norm was introduced as a size function on a real vector space. The three axioms do more than prescribe a formula: each rules out a different kind of undesirable behavior. Positive definiteness prevents nonzero vectors from having size zero; absolute homogeneity prescribes how size changes under scaling; and the triangle inequality controls how size behaves under addition. In this tutorial we examine these requirements and derive several consequences that hold for every norm, not just the familiar norms on coordinate space or the supremum norm on bounded functions.

Definition: Let $V$ be a real vector space. A function $N:V\to[0,\infty)$ is a norm if, for all $x,y\in V$ and $a\in\mathbb{R}$, it satisfies:

Positive definiteness: $N(x)=0$ if and only if $x=0$.
Absolute homogeneity: $N(ax)=|a|N(x)$.
Triangle inequality: $N(x+y)\leq N(x)+N(y)$.

We write $N(x)$, or $\|x\|$ when the norm is understood, for the size of $x$.

The codomain $[0,\infty)$ already requires every value of $N$ to be nonnegative. Positive definiteness strengthens this by specifying exactly when the value is zero. Absolute homogeneity includes both positive and negative scalars: replacing $x$ by $-x$ cannot change its size. The triangle inequality applies to every pair of vectors, so it also controls sums of many vectors.

Basic Consequences of Homogeneity

A few useful identities follow immediately from the axioms. Substituting the zero scalar into absolute homogeneity gives $N(0x)=|0|N(x)=0$, so $N(0)=0$. Substituting $a=-1$ gives $N(-x)=|-1|N(x)=N(x)$. More generally, multiplying by any scalar changes size by the absolute value of that scalar, not by its sign.

Proposition (Finite-Sum Bound): If $N$ is a norm on $V$, then for any $x_1,\ldots,x_k\in V$, $$ N\left(\sum_{i=1}^k x_i\right)\leq\sum_{i=1}^k N(x_i). $$ More generally, for any scalars $a_1,\ldots,a_k\in\mathbb{R}$, $$ N\left(\sum_{i=1}^k a_i x_i\right)\leq\sum_{i=1}^k |a_i|N(x_i). $$

Proof. The first inequality follows by induction on $k$. It is an equality when $k=1$. If it holds for $k$, the triangle inequality gives $$ N\left(\sum_{i=1}^{k+1}x_i\right) =N\left(\left(\sum_{i=1}^{k}x_i\right)+x_{k+1}\right) \leq N\left(\sum_{i=1}^{k}x_i\right)+N(x_{k+1}) \leq\sum_{i=1}^{k+1}N(x_i). $$ This proves the first inequality for every positive integer $k$. For the second, apply the first inequality to the vectors $a_1x_1,\ldots,a_kx_k$, and then use absolute homogeneity: $$ N\left(\sum_{i=1}^k a_i x_i\right) \leq\sum_{i=1}^k N(a_i x_i) =\sum_{i=1}^k |a_i|N(x_i). $$

Worked Example: A Weighted Coordinate Norm

On $\mathbb{R}^2$, define $N(x,y)=|x|+2|y|$. To check positive definiteness, both terms are nonnegative, and their sum is zero exactly when $|x|=0$ and $|y|=0$. Thus $N(x,y)=0$ exactly when $(x,y)=(0,0)$. For $a\in\mathbb{R}$, $$ N(ax,ay)=|ax|+2|ay|=|a|(|x|+2|y|)=|a|N(x,y). $$ Finally, the absolute-value triangle inequality gives $$ N(x_1+y_1,x_2+y_2) =|x_1+y_1|+2|x_2+y_2| \leq |x_1|+|y_1|+2|x_2|+2|y_2| =N(x_1,x_2)+N(y_1,y_2). $$ All three axioms hold. For instance, if $u=(1,-2)$ and $v=(-3,1)$, then $N(u)=5$, $N(v)=5$, and $N(u+v)=N(-2,-1)=4$, consistent with $N(u+v)\leq N(u)+N(v)$.

Why Each Axiom Matters

A candidate function may resemble a norm while failing one of the requirements. Checking the axioms separately can identify exactly what goes wrong. For example, $P(x,y)=|x|$ is nonnegative and absolutely homogeneous, and it satisfies the triangle inequality because $|x_1+y_1|\leq|x_1|+|y_1|$. But $P(0,1)=0$ even though $(0,1)\neq(0,0)$, so it fails positive definiteness. Such a function is called a seminorm when it satisfies the other norm requirements; it is not a norm on the whole space.

Worked Example: A Function That Fails Positive Definiteness

Consider $P:\mathbb{R}^2\to[0,\infty)$ defined by $P(x,y)=|x|$. For $a\in\mathbb{R}$, $$ P(a x,a y)=|a x|=|a|\,|x|=|a|P(x,y). $$ For $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2$, $$ P(x_1+x_2,y_1+y_2)=|x_1+x_2| \leq |x_1|+|x_2| =P(x_1,y_1)+P(x_2,y_2). $$ However, $P(0,1)=|0|=0$, while $(0,1)$ is nonzero. The failure is not in scaling or addition; it is precisely the failure to distinguish every nonzero vector from zero.

The triangle inequality is an independent restriction. To see why, define $Q:\mathbb{R}^2\to[0,\infty)$ by $$ Q(x,y)= \begin{cases} |x|+|y|,&xy\geq0,\\ \frac{1}{10}(|x|+|y|),&xy<0. \end{cases} $$ This function is positive for every nonzero $(x,y)$ and vanishes at $(0,0)$. Also, multiplying both coordinates by a nonzero scalar does not change the sign of their product, so $Q(ax,ay)=|a|Q(x,y)$; for $a=0$, both sides are zero. Yet for $u=(2,-1)$ and $v=(-1,2)$, the products of the coordinates are negative, while their sum has coordinates with positive product. Direct calculation gives $Q(u)=3/10$, $Q(v)=3/10$, and $Q(u+v)=Q(1,1)=2$. Thus $$ Q(u+v)=2>\frac{3}{10}+\frac{3}{10}=Q(u)+Q(v), $$ so $Q$ is not a norm. Positive definiteness and absolute homogeneity alone do not ensure the triangle inequality.

Worked Example: A Positive, Definite Function That Fails Homogeneity

On $\mathbb{R}$, let $R(t)=t^2$. It is nonnegative, and $R(t)=0$ exactly when $t=0$, so it satisfies positive definiteness. But with $a=2$ and $t=1$, $$ R(2\cdot1)=4\neq 2=|2|R(1). $$ It therefore fails absolute homogeneity. It also fails the triangle inequality: taking $s=t=1$ gives $$ R(s+t)=R(2)=4>2=R(1)+R(1). $$ A formula can satisfy positive definiteness and still fail to be a norm because the remaining axioms impose additional constraints.

The Reverse Triangle Inequality

The triangle inequality bounds the size of a sum from above. A useful companion bounds the difference between two sizes. This estimate follows from the triangle inequality and symmetry $N(-x)=N(x)$; no special formula for the norm is needed.

Theorem (Reverse Triangle Inequality for Norms): For any norm $N$ on $V$ and any $x,y\in V$, $$ |N(x)-N(y)|\leq N(x-y). $$

Proof. Write $x=(x-y)+y$. The triangle inequality gives $$ N(x)\leq N(x-y)+N(y), $$ so $N(x)-N(y)\leq N(x-y)$. Interchanging $x$ and $y$ gives $$ N(y)\leq N(y-x)+N(x). $$ Since $y-x=-(x-y)$, absolute homogeneity gives $N(y-x)=N(x-y)$. Hence $N(y)-N(x)\leq N(x-y)$ as well. The two inequalities together say that $$ -N(x-y)\leq N(x)-N(y)\leq N(x-y), $$ which is equivalent to the claimed absolute-value bound. $\square$

This result says that if two vectors are close in norm, then their sizes are close too. In particular, setting $y=0$ recovers $N(x)=N(x-0)$, while the general inequality compares the sizes of distinct vectors. It is a powerful estimate because it uses only the axioms and applies in any normed vector space.

A Norm Defines a Metric

A norm also supplies a notion of distance. The resulting distance measures the size of the difference between two vectors. The construction works for every real vector space equipped with a norm and yields a metric that is invariant under translating both points by the same vector.

Theorem (Metric Induced by a Norm): If $N$ is a norm on $V$, define $d_N:V\times V\to[0,\infty)$ by $$ d_N(x,y)=N(x-y). $$ Then $d_N$ is a metric: it is nonnegative, equals zero exactly when $x=y$, is symmetric, and satisfies the triangle inequality. Moreover, $d_N(x+z,y+z)=d_N(x,y)$ for all $x,y,z\in V$.

Proof. Nonnegativity follows because the norm takes values in $[0,\infty)$. Positive definiteness of the norm gives $$ d_N(x,y)=0\quad\Longleftrightarrow\quad N(x-y)=0 \quad\Longleftrightarrow\quad x-y=0 \quad\Longleftrightarrow\quad x=y. $$ For symmetry, absolute homogeneity with scalar $-1$ gives $$ d_N(y,x)=N(y-x)=N(-(x-y))=N(x-y)=d_N(x,y). $$ For any $x,y,w\in V$, write $x-w=(x-y)+(y-w)$. The norm triangle inequality then implies $$ d_N(x,w)=N(x-w)\leq N(x-y)+N(y-w)=d_N(x,y)+d_N(y,w). $$ Thus $d_N$ satisfies all the metric axioms. Finally, vector subtraction gives $(x+z)-(y+z)=x-y$, and therefore $$ d_N(x+z,y+z)=N((x+z)-(y+z))=N(x-y)=d_N(x,y). $$ This proves translation invariance and completes the proof. $\square$

Worked Example: Computing Distance from a Norm

Use the weighted norm $N(x,y)=|x|+2|y|$ from earlier in the tutorial. For $p=(2,1)$ and $q=(-1,3)$, their difference is $$ p-q=(2-(-1),1-3)=(3,-2). $$ Thus the induced distance is $$ d_N(p,q)=N(3,-2)=|3|+2|-2|=3+4=7. $$ Reversing the order gives $q-p=(-3,2)$, and $$ d_N(q,p)=|-3|+2|2|=3+4=7, $$ as symmetry requires. If both points are translated by $z=(4,-1)$, their difference remains $$ (p+z)-(q+z)=p-q=(3,-2), $$ so their distance remains $7$.

Convexity of Norm Balls

The triangle inequality and homogeneity also impose geometric structure on the sets defined by norm bounds. For $r\geq0$, the closed ball centered at $0$ with radius $r$ is $\{x\in V:N(x)\leq r\}$. Such a ball is convex: the line segment joining any two of its points stays inside the ball. More generally, the norm itself satisfies a convexity inequality.

Proposition (Convexity Inequality for a Norm): If $N$ is a norm on $V$, $x,y\in V$, and $t\in[0,1]$, then $$ N(tx+(1-t)y)\leq tN(x)+(1-t)N(y). $$ Consequently, every closed norm ball centered at $0$ is convex.

Proof. Since $t$ and $1-t$ are nonnegative, absolute homogeneity gives $N(tx)=tN(x)$ and $N((1-t)y)=(1-t)N(y)$. Applying the triangle inequality, $$ N(tx+(1-t)y)\leq N(tx)+N((1-t)y) =tN(x)+(1-t)N(y). $$ Now suppose $N(x)\leq r$ and $N(y)\leq r$. The inequality just proved yields $$ N(tx+(1-t)y)\leq t r+(1-t)r=r. $$ Thus every point on the segment between $x$ and $y$ lies in the same closed ball. The argument includes $t=0$ and $t=1$, which give the endpoints. $\square$

Worked Example: A Segment Inside a Norm Ball

Again let $N(x,y)=|x|+2|y|$. The vectors $u=(2,0)$ and $v=(0,1)$ both lie in the closed ball of radius $2$, since $$ N(u)=|2|+2|0|=2,\qquad N(v)=|0|+2|1|=2. $$ For $t\in[0,1]$, a point on the segment from $u$ to $v$ is $$ tu+(1-t)v=(2t,1-t). $$ Because $t$ and $1-t$ are nonnegative, $$ N(2t,1-t)=|2t|+2|1-t|=2t+2(1-t)=2. $$ Every point on this segment is in the radius-$2$ ball. In this example the segment lies on the boundary, while the convexity proposition guarantees containment in the ball for every norm.

Using the Axioms as a Checklist

When checking whether a proposed size function is a norm, each axiom should be verified for all relevant vectors and scalars. A calculation for a few examples cannot establish an axiom universally, though a single counterexample can disprove one. In particular, checking only nonnegativity is not enough: a function may vanish at nonzero vectors, fail to scale correctly, or assign a sum a value larger than the sum of the separate sizes.

The axioms also provide a dependable toolkit after a norm has been established. Homogeneity gives control of scalar multiples, the triangle inequality yields bounds for finite sums, and together they imply the reverse triangle inequality, a metric, and convexity of norm balls. These consequences explain why the three axioms are central: they govern both algebraic scaling and the geometry of distance and size.

Takeaway: A norm must be positive definite, absolutely homogeneous, and subadditive. These axioms imply the reverse triangle inequality, make $N(x-y)$ a translation-invariant metric, and ensure that norm balls are convex.

Check Your Understanding

Use the norm axioms and their consequences to answer the following questions.

Which axiom fails for $P(x,y)=|x|$ on $\mathbb{R}^2$, and what nonzero vector demonstrates the failure?
Use the finite-sum bound to estimate $N(3x-2y)$ in terms of $N(x)$ and $N(y)$.
How does the triangle inequality, applied in both orders, give the reverse triangle inequality?
Why does $d_N(x,y)=N(x-y)$ satisfy symmetry?
Use the convexity inequality to explain why the segment between two points in a closed norm ball remains in that ball.

What You'll Learn