Metric Induced by a Norm

From Measuring Vectors to Measuring Distances

In Examples of Norms, we saw several ways to measure the size of a vector or function. A norm also provides a way to measure the separation between two elements: measure the size of their difference. In Norm Axioms, the Theorem (Metric Induced by a Norm) established that this construction gives a metric. Here we examine what that metric tells us about the geometry of the space, especially the shape of its balls and how distances behave under translations and scaling.

Let $V$ be a real vector space with norm $N$. The induced metric is $d_N(x,y)=N(x-y)$. The subtraction matters: distance is determined not by the separate sizes $N(x)$ and $N(y)$, but by how large their difference is. In particular, every point has distance zero from itself, and the distance from the origin to $x$ is exactly $N(x)$.

Definition: For $x\in V$ and $r\geq0$, the open ball of radius $r$ centered at $x$, and the closed ball of radius $r$ centered at $x$, in the induced metric are $$ B_{d_N}(x,r)=\{y\in V:d_N(x,y)<r\},\qquad \overline{B}_{d_N}(x,r)=\{y\in V:d_N(x,y)\leq r\}. $$ Equivalently, these are the sets of $y$ for which $N(y-x)<r$ and $N(y-x)\leq r$, respectively.

We will use the open and closed unit balls centered at the origin: $$ U=\{v\in V:N(v)<1\},\qquad \overline{U}=\{v\in V:N(v)\leq1\}. $$ These sets record the geometry of the norm. Every ball of positive radius is obtained from one of them by translation and dilation.

Metric Balls as Translated and Scaled Unit Balls

Theorem: Let $N$ be a norm on $V$, and let $d_N(x,y)=N(x-y)$. For $x\in V$ and $r>0$, $$ B_{d_N}(x,r)=x+rU,\qquad \overline{B}_{d_N}(x,r)=x+r\overline{U}, $$ where $x+rA=\{x+ru:u\in A\}$ for a set $A\subseteq V$. At radius zero, $B_{d_N}(x,0)=\varnothing$ and $\overline{B}_{d_N}(x,0)=\{x\}$.

Proof. Fix $r>0$. An element $y\in V$ belongs to $B_{d_N}(x,r)$ exactly when $N(y-x)<r$. By homogeneity of the norm, $$ N\left(\frac{y-x}{r}\right)=\frac{1}{r}N(y-x). $$ Therefore $N(y-x)<r$ holds exactly when $(y-x)/r\in U$. This is equivalent to $y=x+ru$ for some $u\in U$, so $B_{d_N}(x,r)=x+rU$.

The same calculation with a non-strict inequality gives $$ N(y-x)\leq r \quad\Longleftrightarrow\quad N\left(\frac{y-x}{r}\right)\leq1 \quad\Longleftrightarrow\quad y\in x+r\overline{U}. $$ Thus $\overline{B}_{d_N}(x,r)=x+r\overline{U}$. When $r=0$, no nonnegative number is strictly less than zero, so the open ball is empty. The inequality $N(y-x)\leq0$ holds exactly when $N(y-x)=0$, which, by positive definiteness of the norm, is equivalent to $y=x$. Hence the closed ball is $\{x\}$. $\square$

This description also explains why the geometry of a metric ball depends on the chosen norm. The ball is always a translated and scaled copy of the norm's unit ball, but different norms can have different unit-ball shapes. The shape is not an extra feature added to the metric: it is already encoded in how the norm measures differences.

Worked Example: A Ball in a Polynomial Space

On $\mathcal{P}_2$, use the norm $$ N(a_0+a_1x+a_2x^2)=|a_0|+2|a_1|+|a_2|. $$ Take $p(x)=2+x-x^2$ and $q(x)=-1+3x+2x^2$. Their difference is $$ p(x)-q(x)=3-2x-3x^2, $$ so the induced distance is $$ d_N(p,q)=N(p-q)=|3|+2|-2|+|-3|=3+4+3=10. $$ For comparison, $N(p)=|2|+2|1|+|-1|=5$, and $N(q)=|-1|+2|3|+|2|=9$. The distance is the norm of the difference, not the sum of the norms: here $10$ is less than $5+9$, as the triangle inequality guarantees.

The open ball of radius $4$ centered at $p$ consists precisely of the polynomials $s\in\mathcal{P}_2$ whose difference from $p$ satisfies $N(s-p)<4$. In coefficient terms, if $s(x)=b_0+b_1x+b_2x^2$, this condition is $$ |b_0-2|+2|b_1-1|+|b_2+1|<4. $$ Thus a metric ball in the polynomial space can be described directly by a condition on the coefficients of the difference.

How Translations and Scaling Affect Distance

Theorem: Let $N$ be a norm on a real vector space $V$, with induced metric $d_N$. For all $x,y,z\in V$ and $c\in\mathbb{R}$, $$ d_N(x+z,y+z)=d_N(x,y),\qquad d_N(cx,cy)=|c|\,d_N(x,y). $$ In particular, translation by any fixed vector preserves distances. If $c\neq0$, scaling by $c$ multiplies every distance by $|c|$; if $|c|=1$, it preserves distances.

Proof. By the definition of the induced metric and cancellation of $z$, $$ d_N(x+z,y+z)=N\bigl((x+z)-(y+z)\bigr)=N(x-y)=d_N(x,y). $$ For scalar multiplication, homogeneity gives $$ d_N(cx,cy)=N(cx-cy)=N\bigl(c(x-y)\bigr)=|c|N(x-y)=|c|d_N(x,y). $$ If $c\neq0$, the map $x\mapsto cx$ is a bijection of $V$, and the formula shows exactly how it changes all distances. When $|c|=1$, those distances are unchanged. When $c=0$, all points are sent to the origin and every resulting distance is zero, which agrees with the formula. $\square$

Translation invariance is useful when comparing balls: translating both the center and all the points in a ball leaves every distance unchanged. Scaling behaves differently. A dilation by a positive factor $c$ changes the radius by the same factor; a negative factor also reflects the vectors through the origin, while changing distances by $|c|$.

Worked Example: Translation Does Not Change a Function Distance

On $C[0,1]$, use the norm $\|f\|_1=\int_0^1|f(x)|\,dx$. Let $f(x)=x$ and $g(x)=x^2$. Since $x-x^2\geq0$ for $0\leq x\leq1$, $$ d_N(f,g)=\int_0^1|x-x^2|\,dx =\int_0^1(x-x^2)\,dx =\frac12-\frac13 =\frac16. $$ Now translate both functions by the same $h(x)=1-x$. Their difference does not change: $$ (f+h)-(g+h)=f-g. $$ Consequently $d_N(f+h,g+h)=1/6$, exactly as translation invariance predicts. Adding the same function to both inputs can change their individual norms without changing the distance between them.

Worked Example: Scaling a Distance in the Supremum Norm

On $C[0,1]$, use the supremum norm $\|f\|_\infty=\sup_{0\leq x\leq1}|f(x)|$. Let $f(x)=x^2$ and let $g(x)=1/2$. Since $x^2$ ranges from $0$ to $1$, the largest value of $|x^2-1/2|$ is $1/2$, attained at both endpoints. Therefore $$ d_N(f,g)=\|f-g\|_\infty=\frac12. $$ If both functions are multiplied by $3$, their distance becomes $$ d_N(3f,3g)=\|3f-3g\|_\infty =3\|f-g\|_\infty =\frac32. $$ The factor of $3$ changes the distance by exactly the absolute value of the scaling factor.

Symmetry and Convexity of Balls

There are further geometric consequences of the norm axioms. A ball is symmetric about its center: if $y$ is in a ball centered at $x$, then the point on the opposite side of $x$, namely $2x-y$, is in the same ball. This follows from $N(-(y-x))=N(y-x)$. Balls are also convex. This means that if two points belong to a ball, then every point on the line segment between them belongs to that ball.

Proposition: Every open or closed ball in a norm-induced metric is symmetric about its center and convex.

Proof. First consider symmetry. If $y$ belongs to the open ball centered at $x$ with radius $r$, then $N(y-x)<r$. Since $$ (2x-y)-x=-(y-x), $$ homogeneity gives $N((2x-y)-x)=N(y-x)<r$. Thus $2x-y$ belongs to the same open ball. Replacing $<r$ with $\leq r$ proves the statement for a closed ball.

For convexity, let $y_1,y_2$ belong to the same open ball centered at $x$, with radius $r$, and let $t\in[0,1]$. Then $N(y_1-x)<r$ and $N(y_2-x)<r$. The Convexity Inequality for a Norm, established in Norm Axioms, gives $$ N\bigl(t y_1+(1-t)y_2-x\bigr) \leq tN(y_1-x)+(1-t)N(y_2-x)<tr+(1-t)r=r. $$ The strict inequality also holds when $t=0$ or $t=1$, because the corresponding endpoint is already in the open ball. Therefore the entire segment lies in the open ball. For a closed ball, the same estimate with $N(y_i-x)\leq r$ gives $$ N\bigl(t y_1+(1-t)y_2-x\bigr)\leq tr+(1-t)r=r, $$ so the segment lies in the closed ball as well. $\square$

Worked Example: A Diamond-Shaped Ball in the Plane

In $\mathbb{R}^2$, let $N(u,v)=|u|+|v|$. The induced distance between $(a,b)$ and $(c,d)$ is $$ d_N\bigl((a,b),(c,d)\bigr)=|a-c|+|b-d|. $$ For example, the distance between $(1,-2)$ and $(-2,1)$ is $$ |-2-1|+|1-(-2)|=|-3|+|3|=6. $$ The closed ball of radius $2$ centered at $(1,-2)$ is $$ \{(u,v):|u-1|+|v+2|\leq2\}. $$ Its boundary has vertices $(3,-2)$, $(1,0)$, $(-1,-2)$, and $(1,-4)$, so the ball has a diamond shape. The symmetry and convexity of this region also follow from the proposition, without needing to inspect its boundary point by point.

Why the Induced Metric Depends on the Norm

A vector space can carry more than one norm, and those norms can assign different distances to the same pair of elements. For instance, in $\mathbb{R}^2$, the distance from $(0,0)$ to $(1,1)$ is $2$ for the norm $N(u,v)=|u|+|v|$, but is $\sqrt{2}$ for the Euclidean norm $N(u,v)=\sqrt{u^2+v^2}$. Both formulas give metrics through the Metric Induced by a Norm Theorem, but their balls have different shapes and their numerical distances differ.

A useful distinction is that the metric measures separation, while the norm measures size relative to the origin. The identity $d_N(x,0)=N(x)$ connects the two, and translation invariance explains how the same measurement applies around any center. When using a norm-induced metric, first identify the norm and then apply it to the difference of the two points. Applying the norm separately to the points does not, in general, compute their distance.

Takeaway: A norm induces a metric by measuring differences. Its balls are translated and scaled unit balls, translations preserve distances, and scalar multiplication scales distances by the absolute value of the scalar.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

Why does the formula for a ball as a translated and scaled unit ball require a separate argument when the radius is zero?
For the norm $N(u,v)=|u|+|v|$, what is the induced distance between $(2,1)$ and $(-1,3)$?
What happens to the distance between two points when both are translated by the same vector?
Why is the image of an open ball under multiplication by a positive scalar $c$ an open ball whose radius is multiplied by $c$?
Which norm property ensures that reflecting a point across the center of a ball keeps it in that ball?

What You'll Learn