(1)EQUIVALENCE OF NORMS 1

EQUIVALENCE OF NORMS

1. Equivalence of Normed Spaces If A ∈ M_mn(R), we have seen that

kAk_∞≤ kAk_op≤ kAk₂.

This inequality was used to prove the completeness of Mmn(R). It is not difficult to show that kAk₂ ≤√

mnkAk∞. We obtain

kAk_∞≤ kAk_op≤√

mnkAk∞.

This leads to the definition of equivalence of norms on a real vector space.

Definition 1.1. Let V be a real vector space. Two norms k · k1 and k · k2 on V are said to be equivalent if there exists C₁, C₂> 0 such that

C₁kvk₂ ≤ kvk₁ ≤ C₂kvk₂ for any v ∈ V.

Let x ∈ Rⁿ. The p-norm (1 ≤ p < ∞) of x is defined to be kxk_p = (|x1|^p+ · · · + |xn|^p)^1/p and the ∞-norm of x is defined to be

kxk∞= max{|x₁|, · · · , |x_p|}.

For any x ∈ Rⁿ,

kxk∞≤ kxk_p≤ √^p

nkxk∞. In fact, we can prove the following

Theorem 1.1. Any two norms on Rⁿ are equivalent.

Proof. We only need to show that any norm on Rⁿis equivalent to the Euclidean norm k · k2

on Rⁿ.

Let k · k be a norm on Rⁿ. Denote φ : Rⁿ→ R by φ(v) = kvk for v ∈ V. Let {ei} be the standard basis for Rⁿ. For v ∈ Rⁿ, we write v = a1e1+ · · · + anen. By triangle inequality and the Schwarz inequality,

φ(v) ≤

n

X

i=1

|a_i|kφ(e_i)k ≤

n

X

i=1

kφ(e_i)k²

!1/2

kvk₂.

Let C2 = Pn

i=1kφ(e_i)k²
1/2

. By triangle inequality,

|φ(v) − φ(w)| ≤ kv − wk = φ(v − w) ≤ C₂kv − wk₂.

This shows that φ : Rⁿ→ R is continuous on (Rⁿ, k · k2). Let Sⁿ⁻¹ = {x ∈ Rⁿ: kxk2 = 1}.

Then Sⁿ⁻¹ is sequentially compact in (Rⁿ, k · k₂). By the Extremum Value Theorem, φ attains its minimum on Sⁿ⁻¹. There exists x0 ∈ Sⁿ⁻¹ such that φ(x0) = C1 where C1 =

1

2 EQUIVALENCE OF NORMS

min{φ(x) : x ∈ Sⁿ⁻¹}. Hence φ(x) ≥ C₁ for any x ∈ Sⁿ⁻¹. Let v be any nonzero vector in Rⁿ. Denote x = v/kvk₂. Then φ(x) ≥ C₁. On the other hand,

φ(x) =

v kvk₂

= kvk kvk₂ ≥ C₁.

This shows that kvk ≥ C₁kvk₂ for any v 6= 0. Of course, this inequality is true when v = 0.

We prove that kvk ≥ C1kvk₂ for any v ∈ Rⁿ. We obtain that C1kvk₂≤ kvk ≤ C₂kvk₂ for any v ∈ Rⁿ.

Now we need to show that C1> 0. If C1 = 0, φ(x0) = 0. Hence kx0k = 0. This implies that x0 = 0. This leads to the contradiction to kx0k₂= 1. We prove our assertion. 

Using this, we would like to prove the following result.

Corollary 1.1. If k · k is a norm on Rⁿ, then (Rⁿ, k · k) is a real Banach space.

In fact, we can prove a more general fact.

Proposition 1.1. Let k · k₁ and k · k₂ be equivalent norms on a real vector space V. Then (V, k · k1) is a Banach space if and only if (V, k · k2) is a Banach space.

To prove this result, we use the following lemma.

Lemma 1.1. Let V be a real vector space. Suppose that k · k₁ and k · k₂ are norms on V such that kvk1 ≤ Ckvk₂ for any v ∈ V, where C is a positive real number. Let (vn) be a sequence in V.

(1) If (v_n) is convergent in (V, k · k₂), then it is convergent in (V, k · k₁);

(2) If (vn) is a Cauchy sequence in (V, k · k2), then it is a Cauchy sequence in (V, k · k1).

Proof. This is left to the reader as an exercise. 

Example 1.1. Let V = C[0, 1] be the space of all real valued continuous functions on [a, b].

For f ∈ V, we define

kf k_∞= sup

x∈[0,1]

|f (x)|,

kf k₁ = Z ₁

0

|f (x)|dx,

kf k₂ =

Z 1 0

|f (x)|²dx

1/2

.

Then k · k∞ and k · k1 and k · k2 are norms on V. Furthermore, for any f ∈ V, kf k₂ ≤ kf k₁ ≤ kf k∞.

We have already shown that (V, k · k∞). Are these norms equivalent? Since (V, k · k∞) is complete, if these norms are equivalent, then V are complete with respect to one of these norms. Unfortunately, these norms are not equivalent.

Theorem 1.2. Any finite dimensional real normed space is a Banach space.

Proof. Let {v₁, · · · , v_n} be a basis for V. For each v ∈ V, we can write v = a₁v₁+ · · · + a_nv_n for a unique choice of (a1, · · · , an) ∈ Rⁿ. We obtain a linear isomorphism T : Rⁿ → V

EQUIVALENCE OF NORMS 3

sending a = (a1, · · · , an) to Pn

i=1aivi. Moreover by triangle inequality and the Schwarz inequality,

kT (a)k ≤

n

X

i=1

|a_i|kT (e_i)k ≤ C₂kak₂ where C2 = pPn

i=1kT (e_i)k². This proves that T is continuous on Rⁿ. Using a similar technique as above, we can find C1 > 0 such that kT (a)k ≥ C1kak₂ for any a ∈ Rⁿ. We obtain that

C1kak₂≤ kT (a)k ≤ C₁kak₂.

Let (xk) be a Cauchy sequence in (V, k · k). Then for any  > 0, there exists K ∈ N such that

kx_k− x_lk < C₁ whenever k, l ≥ K_.

Let a_k = T⁻¹(x_k) for k ≥ 1. Then x_k = T (a_k) for any k ≥ 1. When k, l ≥ K_. ka_k− a_lk₂ ≤ 1

C1

kx_k− v_lk < .

This shows that (ak) is a Cauchy sequence in Rⁿ. By completeness of Rⁿ, (ak) is convergent to some a ∈ Rⁿ. By continuity of T,

k→∞lim x_k = lim

k→∞T (a_k) = T (a).

This shows that (x_k) is convergent to T (a) in (V, k · k). We prove that (V, k · k) is a Banach

space.