2. Lemma If two functions f,g e M coincide on a neighborhood U of p e M
-------------------------
and D is a derivation at p then D(f) = D(g).
Proof.
-------------------------
Sublemma. If x e R
n
and B(x,e) is a fixed open ball about it, then there
n
exists a smooth function h:R
R such that h(y) is equal to 1 if y e
----------------------------------------
----------L
B(x,e/2), positive if y e B(x,e) and zero if y m B(x,e).
Define the function h
of the real variable t by the formula
&e-(1-t2)-1if t e (-1,1)
h (t) = {
0
70
otherwise.
It is a good exercise to prove that h is a smooth function on R. Set
0
h (y) := h (4NyN /e), let c denote the characteristic function of the ball
1
0
B(x,3e/4), and define the function h as follows
2
0
i
h (y) :=
2
jnc(z)h1(y-z)dz.
R
If we put h(y) = h (y)/h (x) then we get a desirable function.
2
2
Now let us prove the lemma. Using the construction above we can define a
smooth function h on M which is zero outside U and such that h(p) = 1. In
this case h(f-g) is the constant function 0 on M. Thus we have
0 = D(0) = D(h(f-g)) = D(h) (f(p)-g(p)) + h(p) D(f-g) = D(f) - D(g).
44444
Remarks.
-----------------------------------
i) The sublemma shows that the mapping i:M
----------L
F*(M)
above is indeed an
inclusion. If p $ q are distinct points of M, then there is a smooth function
h on M such that
[i(p)](h) = h(p) = 1 $ [i(q)](h) = h(q) = 0.
ii) We can extend a derivation D at a point p on functions f defined only
in a neighborhood U of p by taking a smooth function h on M such that h is
zero outside U and constant 1 in a neighborhood of p and putting D(f) :=
~
D(f), where
(
~
f(x)h(x)
f(x) = {
0
9
for
for
xeU
x m U.
By lemma 2 this extension of D is correctly defined.
Lemma 3. Let f:B
----------L
-------------------------
R be a smooth function defined on an open ball B C Rn
around the origin. Then there exist smooth functions g 1< i<n on B such that
i
n
f(x) = f(0) + S x g (x)
for x = (x ,...,x ) e B
ii
1
n
i=1
and
df
g (0) =
(0).
i
dx
i
--------------------
4