There is a way to get the differentiability of a function from the existence and continuity of
the Gateaux derivatives. This is very convenient because these Gateaux derivatives are
taken with respect to a one dimensional variable. The following theorem is the main
result.
Theorem 5.6.1Let X be a normed vector space having basis
{v1,⋅⋅⋅,vn}
andlet Y be another normed vector space having basis
{w1,⋅⋅⋅,wm }
. Let U be anopen set in X and let f : U → Y have the property that the Gateaux derivatives,
which proves the continuity of Df because of the assumption the Gateaux derivatives are
continuous. ■
This motivates the following definition of what it means for a function to be
C^{1}.
Definition 5.6.2Let U be an open subset of a normed finite dimensional vectorspace, X and let f : U → Y another finite dimensional normed vector space. Then fis said to be C^{1}if there exists a basis for X,
{v1,⋅⋅⋅,vn }
such that the Gateauxderivatives,
Dvk f (x)
exist on U and are continuous.
Note that as a special case where X = ℝ^{n}, you could let the v_{k} = e_{k} and the condition
would reduce to nothing more than a statement that the partial derivatives
-∂f
∂xi
are all continuous.
Here is another definition of what it means for a function to be C^{1}.
Definition 5.6.3Let U be an open subset of a normed finite dimensional vectorspace, X and let f : U → Y another finite dimensional normed vector space. Then f issaid to be C^{1}if f is differentiable andx →Df
(x)
is continuous as a map from U toℒ
(X,Y )
.
Now the following major theorem states these two definitions are equivalent. This is
obviously so in the special case where X = ℝ^{n} and the special basis is the usual one because,
as observed earlier, the matrix of Df
(x)
is just the one which has for its columns the partial
derivatives which are given to be continuous.
Theorem 5.6.4Let U be an open subset of a normed finite dimensional vectorspace, X and let f : U → Y another finitedimensional normed vector space. Then thetwo definitions above are equivalent.
Proof: It was shown in Theorem 5.6.1, the one about the continuity of the Gateaux
derivatives yielding differentiability, that Definition 5.6.2 implies 5.6.3. Suppose then that
Definition 5.6.3 holds. Then if v is any vector,
f-(x-+-tv)−-f (x) Df (x)tv-+-o(tv-)
ltim→0 t = ltim→0 t
o-(tv)
= Df (x)v+ lt→im0 t = Df (x )v
Thus D_{v}f
(x)
exists and equals Df
(x )
v. By continuity of x → Df
(x)
, this establishes
continuity of x → D_{v}f
(x)
and proves the theorem. ■
Note that the proof of the theorem also implies the following corollary.
Corollary 5.6.5Let U be an open subset of a normed finite dimensional vectorspace, X and let f : U → Y another finite dimensional normed vector space. Then ifthere is a basis of X,
{v1,⋅⋅⋅,vn}
such that the Gateaux derivatives, D_{vk}f
(x)
exist andare continuous. Then all Gateaux derivatives, D_{v}f
(x)
exist and are continuous for allv ∈ X.
From now on, whichever definition is more convenient will be used.