1012 CHAPTER 29. THE AREA FORMULA

and so

1−δ ≤ 1− r |v|η |v|

≤ |Sv|−∥T −S∥|v||Sv|

≤ |T v||Sv|

≤ |Sv|+∥T −S∥|v||Sv|

≤ 1+r |v|η |v|

= 1+δ

The last assertion follows by noting that if T−1 is given to exist and S is close to T then

|Sv||T v|

∈ (1−δ ,1+δ ) so|T v||Sv|∈(

11+δ

,1

1−δ

)⊆(

1− δ̂ ,1+ δ̂

)By choosing δ appropriately, one can achieve the last inclusion for given δ̂ .

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 |T v| is close to 1. Then the followinglemma is fairly obvious.

Lemma 29.2.2 Let S,T be n×n matrices which are invertible. Then

o(T v) = o(Sv) = o(v)

and if L is a continuous linear transformation such that for a < b,

supv̸=0

|Lv||Sv|

< b, infv̸=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.

Proof: Consider the first claim. For

|o(T v)||v|

=|o(T v)||T v|

|T v||v|≤ |o(T v)||T v|

∥T∥

Thus o(T v) = o(v) . It is similar for T replaced with S.Consider the second claim. Pick δ sufficiently small. Then by Lemma 29.2.1

supv̸=0

|Lv||T v|

= supv̸=0

|Lv||Sv||Sv||T v|

≤ (1+δ )supv̸=0

|Lv||Sv|

< b

if δ is small enough. The other inequality is shown exactly similar.

29.3 A DecompositionThis follows [47] which is where I encountered this material. Assume the following:

Dh(x) exists at a.e.x ∈ G say at all x ∈ A⊆ G (29.3.4)