The Compactness Argument Behind Equivalence
The previous tutorial introduced equivalent norms and explained what their two-sided bounds preserve. It also noted that finite-dimensional spaces are a setting in which every pair of norms is equivalent. The Equivalence of Norms in Finite Dimensions theorem was stated earlier in the course; here we give its proof and isolate the central technique: on a finite-dimensional space, compactness turns pointwise positivity into a uniform positive lower bound.
The argument begins by choosing a basis. That choice gives coordinates and a Euclidean norm, even if the norm we ultimately want to study is quite different. We first compare an arbitrary norm to this coordinate norm. Then we compare any two norms through the same coordinate norm.
Bounding a Norm Above and Below
Let \(V\) be a real vector space of finite dimension \(n\geq1\), and choose a basis \(e_1,\ldots,e_n\). Write each \(x\in V\) uniquely as \(x=\sum_{i=1}^n a_i e_i\), and define its coordinate Euclidean norm by \(|x|_2=(\sum_{i=1}^n a_i^2)^{1/2}\). This is a norm on \(V\). The coordinates identify its unit sphere with the usual unit sphere in \(\mathbb{R}^n\), which is compact by the Heine–Borel Theorem in \(\mathbb{R}^n\).
Proof. For \(x=\sum_{i=1}^n a_i e_i\), the triangle inequality and absolute homogeneity give $$ N(x)\leq\sum_{i=1}^n |a_i|N(e_i) \leq\left(\sum_{i=1}^n a_i^2\right)^{1/2} \left(\sum_{i=1}^n N(e_i)^2\right)^{1/2}. $$ The last inequality is the Cauchy–Schwarz inequality for finite sums. Thus the upper bound holds with \(C=(\sum_{i=1}^n N(e_i)^2)^{1/2}\), which is finite and positive.
This upper bound also proves that \(N\) is continuous with respect to the coordinate Euclidean norm. Indeed, the Reverse Triangle Inequality for Norms gives $$ |N(x)-N(y)|\leq N(x-y)\leq C|x-y|_2. $$ Now let \(S=\{x\in V:|x|_2=1\}\). It is nonempty because \(n\geq1\), and it is compact by the coordinate identification above. The function \(x\mapsto N(x)\) is continuous, so the Extreme-Value Theorem implies that it attains a minimum \(m\) on \(S\). Every point of \(S\) is nonzero, and a norm is positive on every nonzero vector; therefore this attained minimum satisfies \(m>0\).
If \(x\neq0\), then \(u=x/|x|_2\) belongs to \(S\). By absolute homogeneity, $$ N(x)=|x|_2N(u)\geq m|x|_2. $$ For \(x=0\), the same inequality reads \(0\geq0\), so it holds there as well. This proves the lower bound and completes the proof. \(\square\)
The compactness step is essential. Positivity of \(N(x)\) at each nonzero \(x\) alone does not say that the values of \(N\) have a positive lower bound across all vectors of Euclidean length one. Compactness ensures that the minimum is attained; because every point on that sphere is nonzero, the attained minimum cannot be zero.
Proof of Equivalence of Norms
We now use the proposition to prove the finite-dimensional result. If the space has dimension zero, it consists only of the zero vector. Any two norms on it satisfy the definition of equivalence, for example with both constants equal to \(1\). So suppose the dimension is positive, choose a basis, and let \(|\cdot|_2\) be its coordinate Euclidean norm.
Proof. Apply the proposition to \(N_1\) and \(N_2\). There are positive constants \(m_1,C_1,m_2,C_2\) such that for every \(x\in V\), $$ m_1|x|_2\leq N_1(x)\leq C_1|x|_2, \qquad m_2|x|_2\leq N_2(x)\leq C_2|x|_2. $$ The upper bound for \(N_1\) implies \(|x|_2\geq N_1(x)/C_1\), and the lower bound for \(N_2\) therefore gives $$ N_2(x)\geq m_2|x|_2\geq \frac{m_2}{C_1}N_1(x). $$ The lower bound for \(N_1\) implies \(|x|_2\leq N_1(x)/m_1\), so the upper bound for \(N_2\) gives $$ N_2(x)\leq C_2|x|_2\leq \frac{C_2}{m_1}N_1(x). $$ Both comparison constants are positive and independent of \(x\). Thus $$ \frac{m_2}{C_1}N_1(x)\leq N_2(x)\leq \frac{C_2}{m_1}N_1(x) \qquad\text{for every }x\in V, $$ which is exactly the definition of equivalence. \(\square\)
The proof does not require either norm to have a special formula. Its structure is the same in every finite dimension: bound a norm above using a basis; use that estimate to prove continuity; then use compactness of the unit sphere to obtain a positive lower bound. The finite-dimensional result follows by comparing each norm to one coordinate norm.
Worked Example: A Weighted Norm on \(\mathbb{R}^3\)
Define \(N(x_1,x_2,x_3)=|x_1|+4|x_2|+2|x_3|\). This is a norm: nonnegativity and absolute homogeneity hold term by term, and the triangle inequality follows by applying the triangle inequality for absolute value to each coordinate. If \(N(x_1,x_2,x_3)=0\), then each nonnegative term is zero, so every coordinate is zero.
For the Euclidean norm \(|x|_2=(x_1^2+x_2^2+x_3^2)^{1/2}\), we have $$ |x|_2\leq |x_1|+|x_2|+|x_3| \leq |x_1|+4|x_2|+2|x_3|=N(x). $$ The first inequality follows by squaring the nonnegative quantities: \((|x_1|+|x_2|+|x_3|)^2\geq x_1^2+x_2^2+x_3^2\). In the other direction, Cauchy–Schwarz gives $$ N(x)\leq \sqrt{1^2+4^2+2^2}\,|x|_2=\sqrt{21}\,|x|_2. $$ For instance, at \(x=(1,-1,2)\), \(N(x)=1+4+4=9\) and \(|x|_2=\sqrt6\); indeed, \(\sqrt6\leq9\leq\sqrt6\,\sqrt{21}=\sqrt{126}\). The uniform estimates apply to every vector, not just this test vector.
Worked Example: A Norm Defined by Two Linear Expressions
On \(\mathbb{R}^2\), set \(u=x+2y\), \(v=3x-y\), and define \(N(x,y)=|u|+|v|\). This is a norm because it is the sum of absolute values of linear expressions, and \(N(x,y)=0\) forces \(u=v=0\). To check the latter implication, solve for the coordinates: $$ x=\frac{u+2v}{7},\qquad y=\frac{3u-v}{7}. $$ Thus \(u=v=0\) implies \(x=y=0\).
The Cauchy–Schwarz inequality in \(\mathbb{R}^2\) gives $$ |u|=|x+2y|\leq\sqrt{5}\,(x^2+y^2)^{1/2}, \qquad |v|=|3x-y|\leq\sqrt{10}\,(x^2+y^2)^{1/2}. $$ Adding yields \(N(x,y)\leq(\sqrt5+\sqrt{10})|(x,y)|_2\). For a lower bound, the coordinate formulas imply $$ |x|+|y|\leq\frac{|u|+2|v|+3|u|+|v|}{7} =\frac{4|u|+3|v|}{7}\leq |u|+|v|=N(x,y). $$ Since \(|(x,y)|_2\leq |x|+|y|\), we obtain \(|(x,y)|_2\leq N(x,y)\). These calculations give explicit equivalence constants; the general theorem would guarantee their existence even without finding them directly.
Worked Example: Two Norms on the Space of Affine Functions
Let \(W\) be the space of affine functions on \([0,1]\), so each \(f\in W\) has the form \(f(t)=a+bt\). Consider the supremum norm \(\|f\|_\infty=\sup_{t\in[0,1]}|f(t)|\) and the endpoint norm \(N(f)=|f(0)|+|f(1)|\). The endpoint expression is a norm on \(W\): if it is zero, then \(f(0)=f(1)=0\), and an affine function with these two values is identically zero. Homogeneity and the triangle inequality follow from those properties of absolute value.
For each \(t\in[0,1]\), an affine function satisfies \(f(t)=(1-t)f(0)+tf(1)\). Hence $$ |f(t)|\leq(1-t)|f(0)|+t|f(1)| \leq |f(0)|+|f(1)|=N(f). $$ Taking the supremum gives \(\|f\|_\infty\leq N(f)\). Conversely, each endpoint value is bounded by the supremum norm, so $$ N(f)=|f(0)|+|f(1)|\leq2\|f\|_\infty. $$ Thus the norms are equivalent on this two-dimensional function space. The argument depends on restricting to affine functions: endpoint values determine a member of \(W\), whereas they do not determine an arbitrary continuous function.
Why Finite Dimension Matters
The proof gives more than a comparison of formulas. It identifies why finite dimension is decisive: the Euclidean unit sphere is compact, so the continuous positive function \(N\) has a strictly positive minimum there. In an infinite-dimensional space, the unit sphere need not be compact, and positivity at each individual point does not by itself provide such a uniform minimum. Accordingly, one cannot use this proof to conclude that all norms on an infinite-dimensional space are equivalent. The supremum and integral norms on \(C[0,1]\), compared in the previous tutorial, provide an example where equivalence fails.
The constants in the finite-dimensional proof depend on the norms and on the chosen coordinate norm. They need not be close to \(1\), and the theorem does not claim that the norms assign the same numerical value to any nonzero vector. Rather, each norm controls the other up to fixed positive factors. By the Sequential and Boundedness Invariance theorem from the previous tutorial, those bounds ensure that the norms agree on convergence, Cauchy behavior, and boundedness.
Check Your Understanding
Use the compactness argument and the examples above to answer the following questions.
- How does the triangle inequality give an upper bound for an arbitrary norm in terms of coordinates?
- Why does the upper bound imply continuity of the norm in the coordinate Euclidean norm?
- Where are compactness and positivity used to prove the lower bound?
- How do two pairs of bounds against the same coordinate norm imply equivalence of two norms?
- Why does the endpoint norm in the example determine an affine function but not every continuous function?
- Which step of the proof may fail in an infinite-dimensional space, and why?