Finite-Dimensional Normed Spaces

In this note, we explore the topological and geometric properties of finite-dimensional normed spaces. The entire theory of finite-dimensional normed spaces can be built systematically upon a single, powerful foundational result: the Linear Independence Lemma.

We will first state and prove this lemma, and then use it as our primary tool to prove:

1. Equivalence of Norms: All norms on a finite-dimensional vector space are equivalent.
2. Completeness: Every finite-dimensional normed space is complete (a Banach space).
3. Compactness (Heine-Borel): A subset is compact if and only if it is closed and bounded.
4. Riesz's Characterization: The closed unit ball is compact if and only if the space is finite-dimensional.

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1. The Core Tool: Linear Independence Lemma

The following lemma establishes a lower bound on the norm of a linear combination of linearly independent vectors in terms of the sum of the absolute values of the coefficients.

Geometric Intuition

In Euclidean space, a sum of linearly independent vectors cannot "shrink" to zero unless the individual coefficients also shrink to zero. The lemma generalizes this to any normed space: the norm of the sum$\left\| \sum a_i x_i \right\|$is bounded below by a positive multiple of the sum of the magnitudes of the coefficients$\sum |a_i|$.

Lemma (Linear Independence Lemma)

Let$\{x_1, \dots, x_n\}$be a linearly independent set of vectors in a normed space$X$(of any dimension). Then there exists a constant$c > 0$such that for any choice of scalars$a_1, \dots, a_n \in \mathbb{K}$(where$\mathbb{K} = \mathbb{R}$or$\mathbb{C}$), we have:

$$\left\| \sum_{i=1}^n a_i x_i \right\| \ge c \sum_{i=1}^n |a_i|.$$

Proof

Let$s_n = \sum_{i=1}^n |a_i|$.

• If$s_n = 0$, then$a_1 = \dots = a_n = 0$, and the inequality holds trivially for any$c > 0$since$0 \ge 0$.
• If$s_n > 0$, dividing both sides by$s_n$allows us to normalize the coefficients. Let$b_i = a_i / s_n$, so that:
  $$\sum_{i=1}^n |b_i| = 1.$$
  In this case, the desired inequality becomes:
  $$\left\| \sum_{i=1}^n b_i x_i \right\| \ge c.$$
  Hence, it suffices to prove that there exists a constant$c > 0$such that for all scalars$\{b_i\}_{i=1}^n$satisfying$\sum_{i=1}^n |b_i| = 1$, we have:
  $$\left\| \sum_{i=1}^n b_i x_i \right\| \ge c.$$

We proceed by contradiction. Suppose that no such$c > 0$exists. Then, for every$m \in \mathbb{N}$, we can find a sequence of scalars$\{b_i^{(m)}\}_{i=1}^n$satisfying:

$$\sum_{i=1}^n \left| b_i^{(m)} \right| = 1 \quad \text{for all } m \in \mathbb{N},$$

such that:

$$y_m = \sum_{i=1}^n b_i^{(m)} x_i \quad \text{satisfies} \quad \|y_m\| \to 0 \quad \text{as } m \to \infty.$$

Since$\sum_{i=1}^n |b_i^{(m)}| = 1$, each coordinate coefficient sequence$(b_i^{(m)})_{m=1}^\infty$is bounded since$|b_i^{(m)}| \le 1$for all$i \in \{1, \dots, n\}$. By the classical Bolzano-Weierstrass theorem, any bounded sequence of scalars has a convergent subsequence. We can extract convergent subsequences for each coordinate one by one:

1. First Coordinate: Since the sequence of scalars$(b_1^{(m)})_{m=1}^\infty$is bounded, there exists a subsequence of indices$\{m_{1,k}\}_{k=1}^\infty \subset \mathbb{N}$and a scalar$\beta_1 \in \mathbb{K}$such that:
   $$\lim_{k \to \infty} b_1^{(m_{1,k})} = \beta_1.$$
2. Second Coordinate: Consider the subsequence of the second coordinate$(b_2^{(m_{1,k})})_{k=1}^\infty$. Since it is also bounded, we can extract a further subsequence of indices$\{m_{2,k}\}_{k=1}^\infty \subset \{m_{1,k}\}_{k=1}^\infty$and a scalar$\beta_2 \in \mathbb{K}$such that:
   $$\lim_{k \to \infty} b_2^{(m_{2,k})} = \beta_2.$$
   Note that since$\{m_{2,k}\}$is a subsequence of$\{m_{1,k}\}$, we still have$\lim_{k \to \infty} b_1^{(m_{2,k})} = \beta_1$.
3. General Coordinate: Inductively, for each$j = 2, \dots, n$, we extract a subsequence of indices$\{m_{j,k}\}_{k=1}^\infty \subset \{m_{j-1,k}\}_{k=1}^\infty$such that:
   $$\lim_{k \to \infty} b_j^{(m_{j,k})} = \beta_j.$$
   At each step, convergence of the previously established coordinates is preserved.

Let the final subsequence of indices be denoted by$\{s_k\}_{k=1}^\infty = \{m_{n,k}\}_{k=1}^\infty$. For this subsequence of vectors:

$$y_{s_k} = \sum_{i=1}^n b_i^{(s_k)} x_i,$$

we have:

$$\lim_{k \to \infty} b_i^{(s_k)} = \beta_i \quad \text{for every } i = 1, \dots, n.$$

We define the limit vector:

$$y = \sum_{i=1}^n \beta_i x_i.$$

By the triangle inequality:

$$\|y_{s_k} - y\| = \left\| \sum_{i=1}^n \left( b_i^{(s_k)} - \beta_i \right) x_i \right\| \le \sum_{i=1}^n \left| b_i^{(s_k)} - \beta_i \right| \|x_i\|.$$

Since$b_i^{(s_k)} \to \beta_i$for each$i$as$k \to \infty$, the right-hand side converges to$0$. Thus,$y_{s_k} \to y$in norm. By the continuity of the norm (which follows from the reverse triangle inequality$|\|a\| - \|b\|| \le \|a - b\|$):

$$\|y\| = \lim_{k \to \infty} \|y_{s_k}\|.$$

Since$\|y_m\| \to 0$as$m \to \infty$, we must have$\|y_{s_k}\| \to 0$, which implies:

$$\|y\| = 0 \implies y = 0 \implies \sum_{i=1}^n \beta_i x_i = 0.$$

Because$\{x_1, \dots, x_n\}$is a linearly independent set, the only linear combination that equals the zero vector is the trivial one. Thus:

$$\beta_i = 0 \quad \text{for all } i = 1, \dots, n.$$

However, taking the limit as$k \to \infty$of the normalization condition$\sum_{i=1}^n |b_i^{(s_k)}| = 1$yields:

$$\sum_{i=1}^n |\beta_i| = 1.$$

This is a contradiction, since we concluded all$\beta_i = 0$, which would sum to$0$. Therefore, there must exist some$c > 0$such that the inequality holds.$\blacksquare$

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2. Application 1: Equivalence of Norms

Definition & Proposition (Equivalent Norms & Topology)

Two norms$\|\cdot\|_a$and$\|\cdot\|_b$on a vector space$X$are said to be equivalent if there exist constants$k, K > 0$such that:

$$k \|x\|_a \le \|x\|_b \le K \|x\|_a \quad \text{for all } x \in X.$$

It is straightforward to verify that this defines an equivalence relation on the set of all norms on$X$:

• Reflexivity: Taking$k = K = 1$shows$\|\cdot\|_a$is equivalent to itself.
• Symmetry: If$k \|x\|_a \le \|x\|_b \le K \|x\|_a$, then$\frac{1}{K} \|x\|_b \le \|x\|_a \le \frac{1}{k} \|x\|_b$.
• Transitivity: If$\|\cdot\|_a \sim \|\cdot\|_b$and$\|\cdot\|_b \sim \|\cdot\|_c$, then by combining the inequalities, we get$\|\cdot\|_a \sim \|\cdot\|_c$.

The algebraic definition of norm equivalence perfectly captures topological equivalence, as shown by the following proposition:

Proposition (Equivalence of Norms and Topology)

Let$\|\cdot\|_a$and$\|\cdot\|_b$be two norms on a vector space$X$. The following statements are equivalent:

1. Algebraic Equivalence: There exist constants$k, K > 0$such that$k \|x\|_a \le \|x\|_b \le K \|x\|_a$for all$x \in X$.
2. Topological Equivalence: A subset$U \subset X$is open under the$\|\cdot\|_a$-induced metric if and only if it is open under the$\|\cdot\|_b$-induced metric. (That is, the induced topologies are identical).
3. Convergence Equivalence: A sequence$(x_n)_{n=1}^\infty$in$X$converges to$x \in X$in$(X, \|\cdot\|_a)$if and only if it converges to$x$in$(X, \|\cdot\|_b)$.

Proof

We will prove the cycle of implications$(1) \implies (2) \implies (3) \implies (1)$.

• $(1) \implies (2)$: Suppose algebraic equivalence holds. Let$U \subset X$be open in$(X, \|\cdot\|_a)$. We show$U$is open in$(X, \|\cdot\|_b)$. Let$x_0 \in U$. Since$U$is open in$(X, \|\cdot\|_a)$, there exists an$a$-ball of radius$\epsilon > 0$centered at$x_0$contained in$U$:

  $$B_a(x_0, \epsilon) = \{x \in X : \|x - x_0\|_a < \epsilon\} \subset U.$$
  We seek a$b$-ball$B_b(x_0, \delta) = \{x \in X : \|x - x_0\|_b < \delta\} \subset U$. Using the algebraic equivalence, we have$\|x - x_0\|_a \le \frac{1}{k} \|x - x_0\|_b$. If we define$\delta = k\epsilon > 0$, then for any$x \in B_b(x_0, \delta)$, we have:
  $$\|x - x_0\|_a \le \frac{1}{k} \|x - x_0\|_b < \frac{1}{k} \delta = \epsilon \implies x \in B_a(x_0, \epsilon) \subset U.$$
  Thus,$B_b(x_0, \delta) \subset U$, meaning$U$is open in$(X, \|\cdot\|_b)$. By symmetry, using the inequality$\|x - x_0\|_b \le K \|x - x_0\|_a$with$\delta = \epsilon/K$, we conclude that any set open in$(X, \|\cdot\|_b)$is also open in$(X, \|\cdot\|_a)$. Hence, the topologies are identical.

• $(2) \implies (3)$: Sequence convergence is a topological property defined purely in terms of open sets. A sequence$x_n \to x$if and only if for every open set$V$containing$x$, there exists$N \in \mathbb{N}$such that$x_n \in V$for all$n \ge N$. Since the topologies are identical, the open sets containing$x$are identical, and hence convergence and limits are identical in both spaces.

• $(3) \implies (1)$: We prove the contrapositive. Suppose$(1)$does not hold. Then either we cannot find$K > 0$such that$\|x\|_b \le K \|x\|_a$for all$x$, or we cannot find$k > 0$such that$\|x\|_b \ge k \|x\|_a$for all$x$. Without loss of generality, assume there is no constant$K > 0$satisfying$\|x\|_b \le K \|x\|_a$. Then, for every$n \in \mathbb{N}$, there exists a vector$z_n \in X$such that:

  $$\|z_n\|_b > n \|z_n\|_a.$$
  Clearly$z_n \neq 0$. We define the sequence:
  $$x_n = \frac{z_n}{n \|z_n\|_a}.$$
  Under the$\|\cdot\|_a$norm, we have:
  $$\|x_n\|_a = \frac{\|z_n\|_a}{n \|z_n\|_a} = \frac{1}{n} \to 0 \quad \text{as } n \to \infty.$$
  Thus,$x_n \to 0$in$(X, \|\cdot\|_a)$. However, under the$\|\cdot\|_b$norm, we have:
  $$\|x_n\|_b = \frac{\|z_n\|_b}{n \|z_n\|_a} > \frac{n \|z_n\|_a}{n \|z_n\|_a} = 1.$$
  Since$\|x_n\|_b > 1$for all$n$, the sequence$(x_n)$cannot converge to$0$in$(X, \|\cdot\|_b)$. This violates the equivalence of convergence, which completes the proof.$\blacksquare$

Corollary (Cauchy Sequence Equivalence)

If$\|\cdot\|_a$and$\|\cdot\|_b$are equivalent norms on$X$, then a sequence$(x_n)$is Cauchy in$(X, \|\cdot\|_a)$if and only if it is Cauchy in$(X, \|\cdot\|_b)$.

Proof

Let$(x_n)$be Cauchy in$(X, \|\cdot\|_a)$. For any$\epsilon > 0$, choose$N \in \mathbb{N}$such that$\|x_n - x_m\|_a < \epsilon/K$for all$n, m \ge N$. Then:

$$\|x_n - x_m\|_b \le K \|x_n - x_m\|_a < K \left(\frac{\epsilon}{K}\right) = \epsilon \quad \text{for all } n, m \ge N.$$

Thus,$(x_n)$is Cauchy in$(X, \|\cdot\|_b)$. The converse follows by symmetry.$\blacksquare$

Theorem (Equivalence of Norms on Finite-Dimensional Spaces)

On any finite-dimensional vector space$X$, all norms are equivalent.

Proof

Let$X$be an$n$-dimensional vector space with basis$\{e_1, \dots, e_n\}$. For any$x \in X$, write$x = \sum_{i=1}^n \alpha_i e_i$. Define the reference$l^1$-norm on$X$:

$$\|x\|_1 = \sum_{i=1}^n |\alpha_i|.$$

We will show that any arbitrary norm$\|\cdot\|$on$X$is equivalent to$\|\cdot\|_1$:

1. Upper Bound: By the triangle inequality and homogeneity of the norm:
   $$\|x\| = \left\| \sum_{i=1}^n \alpha_i e_i \right\| \le \sum_{i=1}^n |\alpha_i| \|e_i\| \le C \sum_{i=1}^n |\alpha_i| = C \|x\|_1,$$
   where$C = \max_{1 \le i \le n} \|e_i\| > 0$is a constant.
2. Lower Bound: Since$\{e_1, \dots, e_n\}$is a basis, it is a linearly independent set in the normed space$(X, \|\cdot\|)$. By the Linear Independence Lemma, there exists a constant$c > 0$such that:
   $$\|x\| = \left\| \sum_{i=1}^n \alpha_i e_i \right\| \ge c \sum_{i=1}^n |\alpha_i| = c \|x\|_1.$$

Combining these yields$c \|x\|_1 \le \|x\| \le C \|x\|_1$. Since equivalence of norms is transitive, any two arbitrary norms on$X$must be equivalent to each other.$\blacksquare$

Examples in$\mathbb{R}^n$

Consider the standard$l^1$,$l^2$(Euclidean), and$l^\infty$norms on$\mathbb{R}^n$:

$$\|x\|_1 = \sum_{i=1}^n |x_i|, \quad \|x\|_2 = \left( \sum_{i=1}^n |x_i|^2 \right)^{1/2}, \quad \|x\|_\infty = \max_{1 \le i \le n} |x_i|.$$

These norms are equivalent, as demonstrated by the following sharp inequalities:

• $\|x\|_\infty \le \|x\|_2 \le \|x\|_1 \le n \|x\|_\infty$
• $\|x\|_2 \le \sqrt{n} \|x\|_\infty$
• $\|x\|_1 \le \sqrt{n} \|x\|_2$(by the Cauchy-Schwarz inequality)

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3. Application 2: Completeness of Finite-Dimensional Spaces

A complete normed space is called a Banach space. In infinite-dimensional spaces, completeness is a rare and prized property. In finite dimensions, completeness is guaranteed.

Theorem (Completeness)

Every finite-dimensional normed space$X$is complete.

Proof

Let$\{e_1, \dots, e_n\}$be a basis of$X$. Let$(x^{(k)})_{k=1}^\infty$be a Cauchy sequence in$X$. Write:

$$x^{(k)} = \sum_{i=1}^n \alpha_i^{(k)} e_i.$$

By the Linear Independence Lemma, there exists a constant$c > 0$such that for any$k, m \ge 1$:

$$\|x^{(k)} - x^{(m)}\| = \left\| \sum_{i=1}^n \left( \alpha_i^{(k)} - \alpha_i^{(m)} \right) e_i \right\| \ge c \sum_{i=1}^n \left| \alpha_i^{(k)} - \alpha_i^{(m)} \right|.$$

Since$(x^{(k)})$is Cauchy, for any$\epsilon > 0$there is$N \in \mathbb{N}$such that$\|x^{(k)} - x^{(m)}\| < c \epsilon$for all$k, m \ge N$. Thus:

$$c \sum_{i=1}^n \left| \alpha_i^{(k)} - \alpha_i^{(m)} \right| < c \epsilon \implies \sum_{i=1}^n \left| \alpha_i^{(k)} - \alpha_i^{(m)} \right| < \epsilon \quad \text{for all } k, m \ge N.$$

Consequently, for each coordinate index$i \in \{1, \dots, n\}$, we have:

$$\left| \alpha_i^{(k)} - \alpha_i^{(m)} \right| < \epsilon \quad \text{for all } k, m \ge N.$$

This shows that for each$i$, the sequence of scalars$(\alpha_i^{(k)})_{k=1}^\infty$is Cauchy in the scalar field$\mathbb{K}$($\mathbb{R}$or$\mathbb{C}$). Since the scalar field$\mathbb{K}$is complete, there exist scalars$\alpha_1, \dots, \alpha_n \in \mathbb{K}$such that$\lim_{k\to\infty} \alpha_i^{(k)} = \alpha_i$.

We define the limit vector$x = \sum_{i=1}^n \alpha_i e_i \in X$. By the triangle inequality:

$$\|x^{(k)} - x\| = \left\| \sum_{i=1}^n \left( \alpha_i^{(k)} - \alpha_i \right) e_i \right\| \le \sum_{i=1}^n \left| \alpha_i^{(k)} - \alpha_i \right| \|e_i\|.$$

Since$\alpha_i^{(k)} \to \alpha_i$for each$i$as$k \to \infty$, the right-hand side converges to$0$. Thus,$(x^{(k)})$converges to$x \in X$, showing$(X, \|\cdot\|)$is complete.$\blacksquare$

Corollary (Subspaces)

Every finite-dimensional subspace$Y$of a normed space$X$is closed in$X$.

Proof

Since$Y$is a finite-dimensional normed space under the norm restricted from$X$,$Y$is complete by the previous theorem. Because any complete subset of a metric space is closed,$Y$must be closed in$X$.$\blacksquare$

Counterexample in Infinite Dimensions

If the subspace is infinite-dimensional, it is not necessarily closed. For example, let$X = C([0,1])$with the supremum norm$\|f\|_\infty = \sup_{t\in[0,1]} |f(t)|$. Let$Y = \mathcal{P}([0,1])$be the subspace of all polynomials on$[0, 1]$.$Y$is infinite-dimensional. The sequence of polynomials:

$$p_k(t) = \sum_{j=0}^k \frac{t^j}{j!} \in Y$$

converges uniformly on$[0, 1]$to the function$f(t) = e^t$because it is the Taylor series expansion. Since$e^t$is continuous,$f \in C([0,1])$. However,$f$is not a polynomial, so$f \notin Y$. Thus,$Y$is not closed in$X$.

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4. Application 3: Compactness (Heine-Borel Theorem)

In general metric spaces, a subset is compact if and only if it is complete and totally bounded. In finite dimensions, this simplifies to the classical Heine-Borel theorem.

Theorem (Heine-Borel)

Let$X$be a finite-dimensional normed space. A subset$K \subset X$is compact if and only if it is closed and bounded.

Proof

• $(\implies)$Let$K \subset X$be compact.
  1. Closed: In any Hausdorff space (and hence any metric space), compact sets are closed.
  2. Bounded: Fix$x_0 \in K$. The open balls$B(x_0, m) = \{x \in X : \|x - x_0\| < m\}$for$m \in \mathbb{N}$form an open cover of$K$. Since$K$is compact, there exists a finite subcover, meaning$K \subset B(x_0, N)$for some$N \in \mathbb{N}$. Thus,$K$is bounded.
• $(\impliedby)$Let$\{e_1, \dots, e_n\}$be a basis of$X$. Suppose$K$is closed and bounded. Let$(y_k)_{k=1}^\infty$be an arbitrary sequence in$K$. We write:
  $$y_k = \sum_{i=1}^n \alpha_i^{(k)} e_i.$$
  Since$K$is bounded, there exists$M > 0$such that$\|y_k\| \le M$for all$k \ge 1$. By the Linear Independence Lemma, there is a constant$c > 0$such that:
  $$M \ge \|y_k\| \ge c \sum_{i=1}^n \left| \alpha_i^{(k)} \right| \ge c \left| \alpha_i^{(k)} \right| \quad \text{for each } i = 1, \dots, n.$$
  Thus, for all$k \ge 1$and all$i = 1, \dots, n$:
  $$\left| \alpha_i^{(k)} \right| \le \frac{M}{c}.$$
  The coordinate sequences$(\alpha_i^{(k)})_{k=1}^\infty$are bounded sequences of scalars. By the Bolzano-Weierstrass theorem, we can extract a subsequence$(y_{k_j})$of$(y_k)$(using the same diagonal subsequence extraction technique as in the lemma) such that all its coordinates converge:
  $$\alpha_i^{(k_j)} \to \alpha_i \quad \text{as } j \to \infty, \quad \text{for } i = 1, \dots, n.$$
  Let$y = \sum_{i=1}^n \alpha_i e_i$. Since:
  $$\|y_{k_j} - y\| \le \sum_{i=1}^n \left| \alpha_i^{(k_j)} - \alpha_i \right| \|e_i\| \to 0 \quad \text{as } j \to \infty,$$
  the subsequence$(y_{k_j})$converges to$y$. Since$K$is closed and all$y_{k_j} \in K$, we have$y \in K$. Hence, every sequence in$K$has a subsequence that converges in$K$. Thus,$K$is compact.$\blacksquare$

Counterexample in Infinite Dimensions

In an infinite-dimensional normed space, a closed and bounded set is not necessarily compact. Let$X = l^2$(the space of square-summable sequences) with the norm$\|x\|_2 = \left( \sum |\xi_i|^2 \right)^{1/2}$. Consider the set of standard basis vectors$K = \{e_1, e_2, e_3, \dots\}$, where$e_k$has$1$at the$k$-th position and$0$elsewhere.

1. Bounded:$\|e_k\|_2 = 1$for all$k$, so$K$is bounded.
2. Closed: For any$j \ne k$, the distance is:
   $$\|e_j - e_k\|_2 = \sqrt{1^2 + 1^2} = \sqrt{2}.$$
   Since the distance between any two distinct points in$K$is$\sqrt{2}$,$K$has no limit points. Thus, it contains all its limit points vacuously and is closed.
3. Non-compact: The sequence$(e_k)_{k=1}^\infty$in$K$cannot have any convergent subsequence because for any subsequence, the distance between any two distinct terms is$\sqrt{2}$, which prevents it from being Cauchy. Thus,$K$is a closed and bounded set that is not compact.

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5. Riesz's Lemma

Riesz's Lemma allows us to find a unit vector that is "almost orthogonal" to any proper closed subspace.

Theorem (Riesz's Lemma)

Let$X$be a normed space, and let$Y$be a proper closed subspace of$X$. Then for any$\theta \in (0, 1)$, there exists a vector$x_\theta \in X$such that:

1. $\|x_\theta\| = 1$,
2. $\|x_\theta - y\| \ge \theta$for all$y \in Y$.

Proof

Since$Y$is a proper closed subspace, choose$x_0 \in X \setminus Y$. The distance$d = \inf_{y \in Y} \|x_0 - y\| > 0$because$Y$is closed and$x_0 \notin Y$. For a given$\theta \in (0, 1)$, we have$1/\theta > 1$, which implies$d/\theta > d$. Thus, there exists some$\epsilon > 0$such that:

$$d + \epsilon = \frac{d}{\theta}.$$

By the definition of the infimum, there must exist$y_0 \in Y$such that:

$$\|x_0 - y_0\| < d + \epsilon = \frac{d}{\theta}.$$

Define:

$$x_\theta = \frac{x_0 - y_0}{\|x_0 - y_0\|}.$$

Clearly,$\|x_\theta\| = 1$. For any$y \in Y$, we compute:

$$\|x_\theta - y\| = \left\| \frac{x_0 - y_0}{\|x_0 - y_0\|} - y \right\| = \frac{1}{\|x_0 - y_0\|} \| x_0 - (y_0 + \|x_0 - y_0\| y) \|.$$

Since$Y$is a subspace and$y_0, y \in Y$, the vector$y' = y_0 + \|x_0 - y_0\| y$is in$Y$. By the definition of$d$, we have$\|x_0 - y'\| \ge d$. Using this bound, we obtain:

$$\|x_\theta - y\| \ge \frac{d}{\|x_0 - y_0\|} > \frac{d}{d/\theta} = \theta. \quad \blacksquare$$

Why is$\theta \in (0, 1)$needed?

Can we choose$\theta = 1$? In general infinite-dimensional normed spaces, no. We cannot always find a unit vector that is at distance exactly$1$from a closed subspace.

Counterexample for$\theta = 1$

Let$X = \{ f \in C([0,1]) : f(0) = 0 \}$with the supremum norm$\|f\|_\infty = \sup_{t\in[0,1]} |f(t)|$. Let$Y$be the proper closed subspace of$X$defined by:

$$Y = \left\{ f \in X : \int_0^1 f(t) dt = 0 \right\}.$$

Consider the vector$x_0(t) = t \in X \setminus Y$. Let's compute the distance$d(x_0, Y) = \inf_{g \in Y} \|x_0 - g\|_\infty$. For any$g \in Y$, we have:

$$\frac{1}{2} = \left| \int_0^1 (t - g(t)) dt \right| \le \int_0^1 |t - g(t)| dt.$$

Since$t - g(t)$is continuous, vanishes at$0$, and has a non-zero integral, it cannot be a constant function. Thus, we have a strict inequality:

$$\int_0^1 |t - g(t)| dt < \|t - g\|_\infty.$$

This implies that for all$g \in Y$:

$$\|x_0 - g\|_\infty > \frac{1}{2} \implies d(x_0, Y) \ge \frac{1}{2}.$$

By defining a sequence of functions$f_n(t) = t - g_n(t)$that rise sharply from$0$to$1/2$on$[0, 1/n]$and then stay slightly above$1/2$to make their integral exactly$1/2$, we can show that$\inf_{g \in Y} \|x_0 - g\|_\infty = 1/2$. However, as shown above, this infimum is never attained.

Now, suppose there existed a unit vector$x \in X$(so$\|x\|_\infty = 1$) such that$d(x, Y) = 1$. Since$Y$has codimension$1$in$X$, any vector$x \in X$can be written as$x = \lambda x_0 + g_0$for some$\lambda \in \mathbb{K}$and$g_0 \in Y$. Since$d(x, Y) = 1$:

$$1 = d(\lambda x_0 + g_0, Y) = |\lambda| d(x_0, Y) = \frac{|\lambda|}{2} \implies |\lambda| = 2.$$

However, we also have$\|x\|_\infty = 1$:

$$1 = \left\| \lambda x_0 + g_0 \right\|_\infty = 2 \left\| x_0 + \frac{g_0}{\lambda} \right\|_\infty.$$

Since$-\frac{g_0}{\lambda} \in Y$, by the fact that the infimum is never attained, we must have:

$$\left\| x_0 - \left(-\frac{g_0}{\lambda}\right) \right\|_\infty > d(x_0, Y) = \frac{1}{2}.$$

This implies:

$$1 = 2 \left\| x_0 + \frac{g_0}{\lambda} \right\|_\infty > 2 \cdot \frac{1}{2} = 1,$$

which is a contradiction ($1 > 1$). Thus, there does not exist any$x \in X$with$\|x\|_\infty = 1$such that$d(x, Y) = 1$.

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6. Riesz's Characterization of Finite-Dimensional Spaces

The compactness of the closed unit ball uniquely distinguishes finite-dimensional spaces from infinite-dimensional ones.

Theorem (Compact Unit Ball)

Let$X$be a normed space. The closed unit ball$B_X = \{ x \in X : \|x\| \le 1 \}$is compact if and only if$X$is finite-dimensional.

Proof

• $(\impliedby)$If$X$is finite-dimensional,$B_X$is closed and bounded. By the Heine-Borel theorem (Section 4),$B_X$is compact.

• $(\implies)$Suppose$B_X$is compact, but$X$is infinite-dimensional.

  1. Pick$x_1 \in X$with$\|x_1\| = 1$. Let$Y_1 = \text{span}\{x_1\}$. Since$Y_1$is finite-dimensional, it is closed (Section 3 Corollary) and proper (since$X$is infinite-dimensional).
  2. By Riesz's Lemma with$\theta = 1/2$, there exists$x_2 \in X$with$\|x_2\| = 1$such that:
     $$d(x_2, Y_1) \ge \frac{1}{2} \implies \|x_2 - x_1\| \ge \frac{1}{2}.$$
  3. Let$Y_2 = \text{span}\{x_1, x_2\}$. As a finite-dimensional subspace,$Y_2$is closed and proper. By Riesz's Lemma with$\theta = 1/2$, there exists$x_3 \in X$with$\|x_3\| = 1$such that:
     $$d(x_3, Y_2) \ge \frac{1}{2} \implies \|x_3 - x_1\| \ge \frac{1}{2} \text{ and } \|x_3 - x_2\| \ge \frac{1}{2}.$$
  4. Inductively, suppose we have chosen$x_1, \dots, x_k$of norm$1$such that$\|x_i - x_j\| \ge 1/2$for all$1 \le j < i \le k$. We define the finite-dimensional subspace:
     $$Y_k = \text{span}\{x_1, \dots, x_k\}.$$
     Since$Y_k$is finite-dimensional, it is closed and proper. By Riesz's Lemma with$\theta = 1/2$, there exists$x_{k+1} \in X$with$\|x_{k+1}\| = 1$such that:
     $$d(x_{k+1}, Y_k) \ge \frac{1}{2} \implies \|x_{k+1} - x_j\| \ge \frac{1}{2} \quad \text{for all } j = 1, \dots, k.$$

  Through this inductive construction, we obtain an infinite sequence$(x_n)_{n=1}^\infty$in$B_X$such that:

  $$\|x_n\| = 1 \text{ for all } n \in \mathbb{N}, \quad \text{and} \quad \|x_n - x_m\| \ge \frac{1}{2} \text{ for all } n \ne m.$$

  Since the distance between any two distinct terms is at least$1/2$,$(x_n)$cannot have any Cauchy subsequence, and thus no convergent subsequence. This contradicts the hypothesis that the closed unit ball$B_X$is compact.

  Thus,$X$must be finite-dimensional.$\blacksquare$

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7. Exercises

Here are some essential exercise questions to test your understanding of finite-dimensional normed spaces, along with hints for solving them.

Exercise 1 (Comparison of Norms on$C([0,1])$)

Let$X = C([0,1])$be the vector space of all continuous functions on$[0,1]$. Define the supremum norm$\|f\|_\infty = \sup_{t \in [0,1]} |f(t)|$and the integral norm$\|f\|_1 = \int_0^1 |f(t)| dt$.

1. Prove that$\|f\|_1 \le \|f\|_\infty$for all$f \in X$.
2. Prove that these two norms are not equivalent.
3. Why does this not contradict the theorem that all norms on a finite-dimensional space are equivalent?

Hint: For part (2), construct a sequence of functions$f_n(t) = t^n \in X$for$n \in \mathbb{N}$. Show that$\|f_n\|_\infty = 1$for all$n$, but$\|f_n\|_1 = \frac{1}{n+1} \to 0$as$n \to \infty$. If the norms were equivalent, we would have$\|f_n\|_\infty \le C \|f_n\|_1$, which fails as$n \to \infty$. For part (3), note that$C([0,1])$is infinite-dimensional.

Exercise 2 (Non-Equivalence of Norms in Infinite Dimensions)

Show that on any infinite-dimensional vector space$X$, one can always construct two norms that are not equivalent.

Hint: Let$\{e_i\}_{i \in I}$be a Hamel basis of$X$. Since$X$is infinite-dimensional, we can choose a countably infinite subset$\{e_1, e_2, e_3, \dots\} \subset \{e_i\}_{i \in I}$. Define two norms:

$$\|x\|_a = \sum |c_i| \quad \text{and} \quad \|x\|_b = \sum i |c_i|,$$

where$x = \sum c_i e_i$(note that all but finitely many$c_i$are zero). Consider the sequence of vectors$x_n = e_n$, and show that no constant$K > 0$can satisfy$\|x_n\|_b \le K \|x_n\|_a$for all$n \in \mathbb{N}$.

Exercise 3 (Equivalence of Norms and Topology Preservation)

Let$\|\cdot\|_a$and$\|\cdot\|_b$be equivalent norms on a vector space$X$. Prove directly from the definitions that:

1. A subset$K \subset X$is compact in$(X, \|\cdot\|_a)$if and only if it is compact in$(X, \|\cdot\|_b)$.
2. The space$(X, \|\cdot\|_a)$is complete (a Banach space) if and only if$(X, \|\cdot\|_b)$is complete.

Hint: For part (1), let$(y_n)$be a sequence in$K$. Compactness in$(X, \|\cdot\|_a)$implies$(y_n)$has a subsequence converging in$K$under$\|\cdot\|_a$. Use the convergence equivalence proposition to conclude that this same subsequence converges to the same limit under$\|\cdot\|_b$. For part (2), use the Cauchy Sequence Equivalence corollary.

Exercise 4 (Distance to Subspaces)

Let$Y$be a finite-dimensional subspace of a normed space$X$. For any$x_0 \in X \setminus Y$, prove that the distance$d(x_0, Y)$is achieved. That is, show there exists a vector$y_0 \in Y$such that:

$$\|x_0 - y_0\| = d(x_0, Y).$$

Hint: Let$d = d(x_0, Y) = \inf_{y \in Y} \|x_0 - y\|$. By definition of the infimum, there is a sequence$(y_n)$in$Y$such that$\|x_0 - y_n\| \to d$. Show that$(y_n)$is a bounded sequence in$Y$. Since$Y$is finite-dimensional, apply the Heine-Borel theorem to extract a convergent subsequence$y_{n_k} \to y_0 \in Y$, and show$\|x_0 - y_0\| = d$using the continuity of the norm.

Exercise 5 (Detailed Calculations for Riesz's Lemma Counterexample)

Carry out the full calculations for the counterexample in Section 5. Specifically, show that for$x_0(t) = t$and$Y = \{ f \in C([0,1]) : f(0) = 0, \int_0^1 f(t) dt = 0 \}$, we have:

1. $Y$is a proper closed subspace of codimension 1.
2. $d(x_0, Y) = 1/2$.
3. For any$g \in Y$,$\|x_0 - g\|_\infty > 1/2$.
4. No unit vector$x \in X$satisfies$d(x, Y) = 1$.

Hint: Follow the step-by-step reasoning in Section 5. For part (3), show that if$\|x_0 - g\|_\infty = 1/2$, then$t - g(t)$must be identically equal to either$1/2$or$-1/2$almost everywhere to satisfy the integral condition, which contradicts$g(0) = 0$. For part (4), write$x = \lambda x_0 + g_0$and prove that$\|x\|_\infty = 1 \implies |\lambda| < 2 \implies d(x, Y) < 1$.