4.4. EQUIVALENCE OF NORMS 103
Proof: These numbers exist thanks to Theorem 4.4.5. It cannot be that δ = 0 because ifit were, you would have |α|= 1 but ∑
nj=1 αkv j = 0 which is impossible since {v1, · · · ,vn}
is linearly independent. The first of the above inequalities follows from δ ≤∥∥∥θ−1 α|α|
∥∥∥ =f(
α|α|
)≤ ∆. The second follows from observing that θ
−1α is a generic vector v in V . ■Note that these inequalities yield the fact that convergence of the coordinates with re-
spect to a given basis is equivalent to convergence of the vectors. More precisely, to saythat limk→∞vk = v is the same as saying that limk→∞ θvk = θv. Indeed, δ |θvn−θv| ≤∥vn−v∥ ≤ ∆ |θvn−θv|.
Now we can draw several conclusions about (V,∥·∥) for V finite dimensional.
Theorem 4.4.8 Let (V,∥·∥) be a finite dimensional normed linear space. Then thecompact sets are exactly those which are closed and bounded. Also (V,∥·∥) is complete. IfK is a closed and bounded set in (V,∥·∥) and f : K→R, then f achieves its maximum andminimum on K.
Proof: First note that the inequalities 4.21 and 4.22 show that both θ−1 and θ are
continuous. Thus these take convergent sequences to convergent sequences.Let {wk}∞
k=1 be a Cauchy sequence. Then from 4.22, {θwk}∞
k=1 is a Cauchy sequence.Thanks to Theorem 4.4.5, it converges to some β ∈ Fn. It follows that limk→∞ θ
−1θwk =
limk→∞wk = θ−1β ∈V . This shows completeness.
Next let K be a closed and bounded set. Let {wk} ⊆ K. Then {θwk} ⊆ θK whichis also a closed and bounded set thanks to the inequalities 4.21 and 4.22. Thus there is asubsequence still denoted with k such that θwk→ β ∈ Fn. Then as just done, wk→ θ
−1β.Since K is closed, it follows that θ
−1β ∈ K.
This has just shown that a closed and bounded set in V is sequentially compact hencecompact.
Finally, why are the only compact sets those which are closed and bounded? Let K becompact. If it is not bounded, then there is a sequence of points of K,{km}∞
m=1 such that∥km∥ ≥
∥∥km−1∥∥+1. It follows that it cannot have a convergent subsequence because thepoints are further apart from each other than 1/2. Indeed,∥∥km−km+1∥∥≥ ∥∥km+1∥∥−∥km∥ ≥ 1 > 1/2
Hence K is not sequentially compact and consequently it is not compact. It followsthat K is bounded. If K is not closed, then there exists a limit point k which is not in K.(Recall that closed means it has all its limit points.) By Theorem 3.1.8, there is a sequenceof distinct points having no repeats and none equal to k denoted as {km}∞
m=1 such thatkm→ k. Then this sequence {km} fails to have a subsequence which converges to a pointof K. Hence K is not sequentially compact. Thus, if K is compact then it is closed andbounded.
The last part is the extreme value theorem, Theorem 3.7.2. ■Next is the theorem which states that any two norms on a finite dimensional vector
space are equivalent.
Theorem 4.4.9 Let ∥·∥ ,∥·∥1 be two norms on V a finite dimensional vector space.Then they are equivalent, which means there are constants 0 < a < b such that for all v,
a∥v∥ ≤ ∥v∥1 ≤ b∥v∥