There are theorems which can be used to get differentiability of a function based on existence
and continuity of the partial derivatives. A generalization of this was given above. Here a
function defined on a product space is considered. It is very much like what was presented
above and could be obtained as a special case but to reinforce the ideas, I will do it from
scratch because certain aspects of it are important in the statement of the implicit function
theorem.
The following is an important abstract generalization of the concept of partial derivative
presented above. Insead of taking the derivative with respect to one variable, it is taken with
respect to several but not with respect to others. This vague notion is made precise in the
following definition. First here is a lemma.
Lemma 5.9.1Suppose U is an open set in X × Y. Then the set, U_{y}definedby
U ≡ {x ∈ X : (x,y) ∈ U }
y
is an open set in X. Here X ×Y is a finite dimensional vector space in which the vector spaceoperations are defined componentwise. Thus for a,b ∈ F,
a (x1,y1) +b (x2,y2) = (ax1 + bx2,ay1 + by2)
and the norm can be taken to be
||(x,y )|| ≡ max (||x||,||y||)
Proof: Recall by Theorem 3.7.4 it does not matter how this norm is defined and
the definition above is convenient. It obviously satisfies most axioms of a norm.
The only one which is not obvious is the triangle inequality. I will show this now.
||(x,y) + (x ,y )|| ≡ ||(x + x ,y + y )|| ≡ max (||x +x ||,||y+ y ||)
1 1 1 1 1 1
≤ max (||x||+||x1 ||,||y||+ ||y1||)
≤ max (∥x∥,∥y∥)+ max (∥x1∥,∥y1∥)
≡ ||(x,y)||+ ||(x1,y1)||
Let x ∈ U_{y}. Then
(x,y)
∈ U and so there exists r > 0 such that
B ((x,y),r) ∈ U.
This says that if
(u,v)
∈ X × Y such that
||(u,v)− (x,y)||
< r, then
(u,v)
∈ U. Thus
if
||(u,y)− (x,y)|| = ||u− x|| < r,
then
(u,y)
∈ U. This has just said that B
(x,r)
, the ball taken in X is contained in U_{y}. This
proves the lemma. ■
Or course one could also consider
Ux ≡ {y :(x,y) ∈ U }
in the same way and conclude this set is open in Y . Also, the generalization to many factors
yields the same conclusion. In this case, for x ∈∏_{i=1}^{n}X_{i}, let
( )
||x|| ≡ max ||xi||Xi : x = (x1,⋅⋅⋅,xn)
Then a similar argument to the above shows this is a norm on ∏_{i=1}^{n}X_{i}. Consider the
triangle inequality.
∥(x1,⋅⋅⋅,xn) + (y1,⋅⋅⋅,yn)∥ = max (||xi + yi|| ) ≤ max (∥xi∥ + ∥yi∥ )
i Xi i Xi Xi
( ) ( )
≤ maxi ||xi||Xi + maix ||yi||Xi
Corollary 5.9.2Let U ⊆∏_{i=1}^{n}X_{i}be an open set and let
{ ( ) }
U(x1,⋅⋅⋅,xi−1,xi+1,⋅⋅⋅,xn) ≡ x ∈ Fri :x1,⋅⋅⋅,xi− 1,x,xi+1,⋅⋅⋅,xn ∈ U .
Then U_{}
(x1,⋅⋅⋅,xi−1,xi+1,⋅⋅⋅,xn)
is an open set in F^{ri}.
Proof: Let z ∈ U_{}
(x1,⋅⋅⋅,xi−1,xi+1,⋅⋅⋅,xn)
. Then
(x1,⋅⋅⋅,xi−1,z,xi+1,⋅⋅⋅,xn)
≡ x ∈ U by
definition. Therefore, since U is open, there exists r > 0 such that B
(x,r)
⊆ U. It follows that
for B
(z,r)
_{Xi} denoting the ball in X_{i}, it follows that B
(z,r)
_{Xi}⊆ U_{(x1,⋅⋅⋅,xi−1,xi+1,⋅⋅⋅,xn)
}
because to say that
Next is a generalization of the partial derivative.
Definition 5.9.3Let g : U ⊆∏_{i=1}^{n}X_{i}→ Y , where U is an open set. Then the map
( )
z → g x1,⋅⋅⋅,xi−1,z,xi+1,⋅⋅⋅,xn
is a function from the open set in X_{i},
{ ( ) }
z : x = x1,⋅⋅⋅,xi− 1,z,xi+1,⋅⋅⋅,xn ∈ U
to Y . When this map is differentiable, its derivative is denoted by D_{i}g
(x)
. To aid in thenotation, for v ∈ X_{i}, let θ_{i}v ∈∏_{i=1}^{n}X_{i}be the vector
(0,⋅⋅⋅,v,⋅⋅⋅,0)
where the v is in thei^{th}slot and forv ∈∏_{i=1}^{n}X_{i}, let v_{i}denote the entry in the i^{th}slot of v. Thus, bysaying
( )
z → g x1,⋅⋅⋅,xi−1,z,xi+1,⋅⋅⋅,xn
is differentiable is meant that for v ∈ X_{i}sufficiently small,
g (x+ θiv)− g(x) = Dig (x)v + o(v).
Note D_{i}g
(x)
∈ℒ
(Xi,Y)
.
Definition 5.9.4Let U ⊆ X be an open set. Then f : U → Y is C^{1}
(U)
if f isdifferentiable and the mapping
x → Df (x),
is continuous as a function from U to ℒ
(X,Y )
.
With this definition of partial derivatives, here is the major theorem. Note the
resemblance with the matrix of the derivative of a function having values in ℝ^{m} in terms of
the partial derivatives.
Theorem 5.9.5Let g,U,∏_{i=1}^{n}X_{i}, be given as in Definition 5.9.3. Then g isC^{1}
(U)
if and only if D_{i}g exists and is continuous on U for each i. In this case, g isdifferentiable and
∑
Dg (x)(v) = Dkg (x) vk (5.15)
k
(5.15)
where v =
(v ,⋅⋅⋅,v )
1 n
.
Proof: Suppose then that D_{i}g exists and is continuous for each i. Note that
k
∑ θ v = (v ,⋅⋅⋅,v ,0,⋅⋅⋅,0).
j=1 j j 1 k
Thus ∑_{j=1}^{n}θ_{j}v_{j} = v and define ∑_{j=1}^{0}θ_{j}v_{j}≡ 0. Therefore,
exists and equals the formula given in 5.15. Also x →Dg
(x)
is
continuous since each of the D_{k}g
(x)
are.
Next suppose g is C^{1}. I need to verify that D_{k}g
(x)
exists and is continuous. Let v ∈ X_{k}
sufficiently small. Then
g(x+ θkv)− g (x ) = Dg (x )θkv+ o(θkv)
= Dg (x )θ v+ o(v)
k
since
||θkv||
=
||v||
. Then D_{k}g
(x )
exists and equals
Dg (x)∘ θk
Now x → Dg
(x)
is continuous. Since θ_{k} is linear, it follows from Lemma 3.7.1 that
θ_{k} : X_{k}→∏_{i=1}^{n}X_{i} is also continuous. ■
Note that the above argument also works at a single point x. That is, continuity at x of
the partials implies Dg
(x)
exists and is continuous at x.
The way this is usually used is in the following corollary which has already been obtained.
Remember the matrix of Df
(x)
. Recall that if a function is C^{1} in the sense that x → Df
(x )
is continuous then all the partial derivatives exist and are continuous. The next corollary says
that if the partial derivatives do exist and are continuous, then the function is differentiable
and has continuous derivative.
Corollary 5.9.6Let U be an open subset of F^{n}and letf :U → F^{m}be C^{1}in the sensethat all the partial derivatives of f exist and are continuous. Then f is differentiableand
∑n ∂f--
f (x + v) = f (x) + ∂xk (x)vk + o(v).
k=1
Similarly, if the partial derivatives up to order k exist and are continuous, then the function isC^{k}in the sense that the first k derivatives exist and are continuous.