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 17.8.1Suppose U is an open set in X × Y. Then the set, U_{y}defined by
Uy ≡ {x ∈ X : (x,y) ∈ U}
is an open set in X. Here X ×Y is a finite dimensional vector space in which the vector space operationsare 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 11.5.4 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||X ) ≤ max (∥xi∥X + ∥yi∥X )
i i i i i
( ) ( )
≤ max ||xi||Xi + max ||yi||Xi
i i
Corollary 17.8.2Let 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 17.8.3Let 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 the notation, forv ∈ X_{i}, let θ_{i}v ∈∏_{i=1}^{n}X_{i}be the vector
(0,⋅⋅⋅,v,⋅⋅⋅,0)
where the v is in the i^{th}slot and forv ∈∏_{i=1}^{n}X_{i}, let v_{i}denote the entry in the i^{th}slot of v. Thus, by saying
( )
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 17.8.4Let U ⊆ X be an open set. Then f : U → Y is C^{1}
(U)
if f is differentiable and themapping
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 17.8.5Let g,U,∏_{i=1}^{n}X_{i}, be given as in Definition 17.8.3. Then g is C^{1}
(U)
ifand only if D_{i}g exists and is continuous on U for each i. In this case, g is differentiable and
Dg (x) (v) = ∑ D g (x )v (17.8)
k k k
(17.8)
where v =
(v1,⋅⋅⋅,vn)
.
Proof: Suppose then that D_{i}g exists and is continuous for each i. Note that
∑k
θjvj = (v1,⋅⋅⋅,vk,0,⋅⋅⋅,0).
j=1
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 17.8. 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)θkv+ o (v )
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 Theorem 11.6.3 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 17.8.6Let U be an open subset of F^{n}and letf :U → F^{m}be C^{1}in the sense that all the partialderivatives of f exist and are continuous. Then f is differentiable and
∑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 is C^{k}in thesense that the first k derivatives exist and are continuous.