Equivalent Norms

Comparing Two Norms on the Same Space

A vector space can carry different norms, and those norms can assign different sizes to the same vector. The central question is whether the differences are controlled uniformly across the whole space. If they are, the norms may give different numerical values while agreeing on the basic notions of closeness and convergence. This is the idea behind equivalent norms.

Let $N_1$ and $N_2$ be norms on the same real vector space $V$. It is important that the comparison use constants that work for every vector, not constants chosen separately for each vector.

Definition: The norms $N_1$ and $N_2$ on $V$ are equivalent if there are constants $c,C>0$ such that $$ cN_1(x)\leq N_2(x)\leq CN_1(x) \qquad\text{for every }x\in V. $$ The constants may depend on the pair of norms, but not on $x$.

The upper bound says that $N_2$ cannot grow arbitrarily large compared with $N_1$. The lower bound prevents $N_2$ from becoming arbitrarily small compared with $N_1$ on nonzero vectors. Both directions matter. A one-sided inequality can control some behavior, but does not by itself establish equivalence.

Equivalence is an equivalence relation on the set of norms on $V$. For clarity, if $N_1,N_2,N_3$ are norms and $N_1$ is equivalent to $N_2$ and $N_2$ to $N_3$, then the constants in the two comparisons can be combined. This elementary fact lets us group norms according to whether they give the same scale up to fixed factors.

Theorem (Equivalence Is an Equivalence Relation): Equivalence of norms on a fixed vector space is reflexive, symmetric, and transitive.

Proof. For reflexivity, $N(x)=1\cdot N(x)$ for every $x$, so $N$ is equivalent to itself. For symmetry, suppose $cN_1(x)\leq N_2(x)\leq CN_1(x)$, where $c,C>0$. Rearranging both inequalities gives $$ \frac{1}{C}N_2(x)\leq N_1(x)\leq \frac{1}{c}N_2(x) \qquad\text{for every }x\in V. $$ Thus $N_2$ is equivalent to $N_1$.

For transitivity, suppose $c_1N_1(x)\leq N_2(x)\leq C_1N_1(x)$ and $c_2N_2(x)\leq N_3(x)\leq C_2N_2(x)$ for every $x$. Substitution yields $$ c_2c_1N_1(x)\leq N_3(x)\leq C_2C_1N_1(x). $$ Both combined constants are positive, so $N_1$ and $N_3$ are equivalent. This proves all three properties. $\square$

Equivalent Norms and Identity Maps

There is a useful way to read the two inequalities in the definition: each norm controls the other. In the language of bounded linear maps, this means that the identity map from one normed space to the other is bounded in both directions. The identity here sends each vector to itself; only the norm used to measure its input and output changes.

Theorem (Identity-Map Criterion for Equivalent Norms): Two norms $N_1$ and $N_2$ on $V$ are equivalent if and only if both identity maps $I:(V,N_1)\to(V,N_2)$ and $I:(V,N_2)\to(V,N_1)$ are bounded.

Proof. Suppose first that the norms are equivalent, so $cN_1(x)\leq N_2(x)\leq CN_1(x)$ for every $x$. The upper inequality gives $$ N_2(I(x))=N_2(x)\leq CN_1(x), $$ so the identity map from $(V,N_1)$ to $(V,N_2)$ is bounded. The lower inequality can be rewritten as $N_1(x)\leq c^{-1}N_2(x)$, which proves that the identity map in the other direction is bounded.

Conversely, suppose both identity maps are bounded. By the Boundedness Criterion for Linear Maps, there are finite constants $C>0$ and $D>0$ such that $$ N_2(I(x))\leq CN_1(x) \quad\text{and}\quad N_1(I(x))\leq DN_2(x) \qquad\text{for every }x\in V. $$ The second inequality implies $D^{-1}N_1(x)\leq N_2(x)$. Combining the bounds gives $$ D^{-1}N_1(x)\leq N_2(x)\leq CN_1(x) \qquad\text{for every }x\in V. $$ Thus the norms are equivalent. $\square$

This criterion connects the definition to the results on bounded linear maps and continuity of linear maps from earlier in the course. In particular, by the Continuity-Boundedness Equivalence for Linear Maps, two norms are equivalent exactly when each identity map is continuous. A single continuous identity map supplies only one of the two needed inequalities.

Worked Example: Adding a Point Evaluation to the Supremum Norm

On $C[0,1]$, define $$ N_1(f)=\|f\|_\infty, \qquad N_2(f)=\|f\|_\infty+|f(1/3)|. $$ The second expression is a norm: it is nonnegative and absolutely homogeneous; its triangle inequality follows from the triangle inequalities for the supremum norm and absolute value; and $N_2(f)=0$ implies $\|f\|_\infty=0$, hence $f=0$. For every $f\in C[0,1]$, the definition of the supremum norm gives $|f(1/3)|\leq\|f\|_\infty$. Therefore $$ N_1(f)\leq N_2(f)\leq 2N_1(f). $$ The norms are equivalent, with constants $c=1$ and $C=2$. The additional point evaluation changes the norm value in general, but only by a uniformly controlled amount.

What Equivalent Norms Preserve

The two-sided bounds immediately compare distances, since the distance induced by a norm is $d_N(x,y)=N(x-y)$. They also compare balls. If $N_1$ and $N_2$ satisfy the definition with constants $c,C>0$, then $$ B_{N_1}(x,r/C)\subseteq B_{N_2}(x,r)\subseteq B_{N_1}(x,r/c) \qquad (r>0). $$ For the first inclusion, $N_1(y-x)<r/C$ implies $N_2(y-x)\leq CN_1(y-x)<r$. For the second, $N_2(y-x)<r$ and $cN_1(y-x)\leq N_2(y-x)$ imply $N_1(y-x)<r/c$. Consequently, the two norms generate the same open sets: every ball for either norm contains a ball for the other around the same center.

Equivalent norms also agree on which sequences converge, which sequences are Cauchy, and which sets are bounded. The uniform constants are what make these conclusions work simultaneously for all terms of a sequence or all points of a set.

Theorem (Sequential and Boundedness Invariance): Suppose $N_1$ and $N_2$ are equivalent norms on $V$. A sequence $(x_n)$ converges to $x$ in $N_1$ if and only if it converges to $x$ in $N_2$. It is Cauchy in $N_1$ if and only if it is Cauchy in $N_2$. A subset of $V$ is bounded in one norm if and only if it is bounded in the other.

Proof. Choose $c,C>0$ such that $cN_1(z)\leq N_2(z)\leq CN_1(z)$ for every $z\in V$. If $x_n\to x$ in $N_1$, then $$ N_2(x_n-x)\leq CN_1(x_n-x)\longrightarrow 0, $$ so $x_n\to x$ in $N_2$. If $x_n\to x$ in $N_2$, then $N_1(x_n-x)\leq c^{-1}N_2(x_n-x)\to0$, proving the reverse implication.

For the Cauchy property, if $(x_n)$ is Cauchy in $N_1$, then for every $\varepsilon>0$ there is an index $n_0$ such that $N_1(x_n-x_m)<\varepsilon/C$ whenever $n,m\geq n_0$. It follows that $$ N_2(x_n-x_m)\leq CN_1(x_n-x_m)<\varepsilon. $$ Thus it is Cauchy in $N_2$. Conversely, if it is Cauchy in $N_2$, then $N_1(x_n-x_m)\leq c^{-1}N_2(x_n-x_m)$, so it is Cauchy in $N_1$.

Finally, if a set $A$ is bounded in $N_1$, there is an $M\geq0$ such that $N_1(x)\leq M$ for every $x\in A$. Then $N_2(x)\leq CM$ for every $x\in A$, so it is bounded in $N_2$. If $A$ is bounded in $N_2$, the inequality $N_1(x)\leq c^{-1}N_2(x)$ proves it is bounded in $N_1$. This establishes all three claims. $\square$

The earlier theorem Completeness Is Invariant Under Equivalent Norms is another consequence of this perspective: equivalent norms preserve Cauchy sequences and convergence, so completeness does not depend on which of two equivalent norms is used. The equivalence does not say that norm values are identical, or that the constants are close to $1$. It says only that one fixed pair of positive constants controls all comparisons.

Worked Example: Two Weighted Norms on Finite Sequences

Let $c_{00}$ be the vector space of real sequences with only finitely many nonzero terms. Define $$ N_1(x)=\sum_{n=1}^{\infty}|x_n|, \qquad N_2(x)=\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)|x_n|. $$ Both sums are finite for $x\in c_{00}$. Each formula is a norm: positivity and homogeneity follow term by term, the triangle inequality follows by applying $|x_n+y_n|\leq|x_n|+|y_n|$ in every term, and a zero sum forces every coordinate to be zero.

For every $n\geq1$, $1\leq1+1/n\leq2$. Multiplying these inequalities by $|x_n|$ and summing the finitely many nonzero terms gives $$ N_1(x)\leq N_2(x)\leq2N_1(x). $$ Hence the two norms are equivalent. For example, if $x=(2,-1,0,\ldots)$, then $N_1(x)=3$ and $N_2(x)=2(2)+\tfrac32(1)=\tfrac{11}{2}$; the comparison holds because $3\leq\tfrac{11}{2}\leq6$.

A One-Sided Bound Is Not Enough

It is tempting to see one useful inequality and conclude that two norms are equivalent. The next example shows why the reverse bound can fail. The functions have increasingly narrow peaks: their supremum remains fixed while their integral becomes small. This separates uniform size from average size.

Worked Example: The Supremum and Integral Norms Are Not Equivalent

On $C[0,1]$, consider $N_\infty(f)=\|f\|_\infty$ and $N_1(f)=\int_0^1|f(x)|\,dx$. Since the interval has length $1$, every $f\in C[0,1]$ satisfies $$ N_1(f)\leq N_\infty(f). $$ Indeed, $|f(x)|\leq\|f\|_\infty$ for each $x$, and integrating gives the inequality.

For each positive integer $n$, let $f_n(x)=x^n$. Since $0\leq x^n\leq1$ on $[0,1]$, and $f_n(1)=1$, we have $N_\infty(f_n)=1$. Direct integration gives $$ N_1(f_n)=\int_0^1x^n\,dx=\frac{1}{n+1}. $$ If the norms were equivalent, there would be a constant $c>0$ such that $cN_\infty(f)\leq N_1(f)$ for every $f$. Applying this to $f_n$ would give $c\leq1/(n+1)$ for every positive integer $n$, which is impossible for a fixed positive $c$. Thus the norms are not equivalent. The one-sided estimate $N_1\leq N_\infty$ cannot replace the missing lower bound.

Why Equivalence Matters

Equivalent norms provide flexibility. One norm may be easier for calculation, while another may make a particular estimate or geometric feature more transparent. If the norms are equivalent, switching between them does not change the open sets, convergent sequences, Cauchy sequences, or bounded sets. When using such a switch, however, keep track of the comparison constants: estimates in one norm may acquire a factor when translated to the other.

Finite-dimensional spaces provide an important source of equivalent norms. The Equivalence of Norms in Finite Dimensions theorem from earlier in the course states that every norm on a finite-dimensional vector space is equivalent to the Euclidean norm after coordinates are chosen. The present discussion explains what that conclusion means and why its two-sided bounds are useful; the forthcoming proof will establish the finite-dimensional result itself. In infinite-dimensional spaces, as the function-space example shows, different norms need not be equivalent.

Takeaway: Equivalent norms bound each other above and below by positive constants. Their identity maps are bounded in both directions, and they induce the same convergence, Cauchy behavior, bounded sets, and open sets. A one-sided bound does not guarantee equivalence.

Check Your Understanding

Use the two-sided bounds and examples in this tutorial to answer the following questions.

Why must the constants in the definition of equivalent norms be independent of the vector?
How does boundedness of the two identity maps imply the two inequalities required for equivalent norms?
If $N_2(x)\leq CN_1(x)$, which direction of convergence follows immediately, and what additional bound is needed for the converse?
Why does the sequence $f_n(x)=x^n$ rule out equivalence of the supremum and integral norms on $C[0,1]$?
What does equivalence of norms imply about the open sets they induce, and what does it not imply about their numerical values?

What You'll Learn