128 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES
that∣∣ fk( vm
m
)∣∣ ≥ 1. Since f is continuous, and vm/m→ 0, it follows that f( vm
m
)→ 0 in V.
However, by Lemma 4.4.7, fk( vm
m
)→ 0, a contradiction.
⇐= If each coordinate function is continuous, then
∥ f (v)− f (v̂)∥W =
∥∥∥∥∥ n
∑k=1
fk (v)wk−n
∑k=1
fk (v̂)wk
∥∥∥∥∥≤ n
∑k=1| fk (v)− fk (v̂)|∥wk∥W
Since each fk is continuous, this shows that f is also. ■
5.3 ComparisonsHere are some useful lemmas about comparisons. Here |·| will be the usual norm but onecould generalize.
Lemma 5.3.1 Suppose S,T are linear, defined on a finite dimensional normed linearspace, S−1 exists, and let δ ∈ (0,1). Then whenever ∥S−T∥ is small enough, it followsthat
|Tv||Sv|
∈ (1−δ ,1+δ ) (5.1)
for all v ̸= 0. Similarly if T−1 exists and ∥S−T∥ is small enough,
|Tv||Sv|
∈ (1−δ ,1+δ )
Proof: Say S−1 exists. Then v → |Sv| is a norm. Then by equivalence of norms,Theorem 4.4.9, there exists η > 0 such that for all v, |Sv| ≥ η |v| . Say ∥T −S∥< r < δη
|Tv||Sv|
=|Sv− (S−T )v|
|Sv|≥ |Sv|−∥T −S∥|v|
|Sv|≥ |Sv|−δη |v|
|Sv|≥ |Sv|−δ |Sv|
|Sv|= 1−δ
|Tv||Sv|
=|Sv+(T −S)v|
|Sv|≤ |Sv|+∥T −S∥|v|
|Sv|≤ |Sv|+δη |v|
|Sv|≤ |Sv|+δ |Sv|
|Sv|= 1+δ
Next suppose that T−1 exists. Then, letting δ̂ be small enough,(
1− δ̂ ,1+ δ̂
)⊆(
11+δ
, 11−δ
). From what was just shown, if ∥S−T∥ is small enough,
|Sv||Tv|
∈(
1− δ̂ ,1+ δ̂
)⊆(
11+δ
,1
1−δ
)so|Tv||Sv|
∈ (1−δ ,1+δ ) . ■
In short, the above lemma says that if one of S,T is invertible and the other is close to it,then it is also invertible and the quotient of |Sv| and |Tv| is close to 1. Then the followinglemma is fairly obvious.
Lemma 5.3.2 Let S,T be n×n matrices which are invertible. Then
o(Tv) = o(Sv) = o(v)
and if L is a continuous linear transformation such that for a < b,
supv ̸=0
|Lv||Sv|
< b, infv ̸=0
|Lv||Sv|
> a
If ∥S−T∥ is small enough, it follows that the same inequalities hold with S replaced withT . Here ∥·∥ denotes the operator norm.