7.6. EXISTENCE OF THE DERIVATIVE, C1 FUNCTIONS 191

7.6 Existence of the Derivative, C1 FunctionsThere is a way to get the differentiability of a function from the existence and continuity ofone dimensional directional derivatives. The following theorem is the main result. It giveseasy to verify one dimensional conditions for the existence of the derivative. The meaningof ∥·∥ will be determined by context in what follows. This theorem says that if the Gateauxderivatives exist for each vector in a basis and they are also continuous, then the functionis differentiable.

Theorem 7.6.1 Let X be a normed vector space having basis {v1, · · · ,vn} and letY be another normed vector space. Let U be an open set in X and let f : U → Y have theproperty that the one dimensional limits

Dvkf (x)≡ limt→0

f (x+ tvk)−f (x)t

exist and x→ Dvkf (x) are continuous functions of x ∈U as functions with values in Y .Then Df (x) exists and

Df (x) v=n

∑k=1

Dvkf (x)ak

where v= ∑nk=1 akvk. Furthermore, x→ Df (x) is continuous; that is

limy→x∥Df (y)−Df (x)∥= 0.

Proof: Let v= ∑nk=1 akvk where all ak are small enough that for all k ≥ 0,

x+k

∑j=1

a jv j ∈ B(x,r)⊆U,0

∑k=1

akvk ≡ 0.

The mapping v → (a1, ...,an) is an isomorphism of V and Fn and we can define a normas ∑k |ak| which is equivalent to the norm on V thanks to Theorem 5.2.4. Let hk (x) ≡f(x+∑

k−1j=1 a jv j

)−f (x) . Then collecting the terms,

f (x+v)−f (x) =n

∑k=1

(hk (x+akvk)−hk (x))+n

∑k=1

(f (x+akvk)−f (x)) (7.13)

Using Theorem 7.5.2,∥∥Dakvkhk (x+ takvk)∥∥ =

∥∥akDvkhk (x+ takvk)∥∥

=

∥∥∥∥∥ak

(Dvkf

(x+

k−1

∑j=1

a jv j + takvk

)−Dvkf (x+ takvk)

)∥∥∥∥∥≤ C∥v∥ε

provided ∥v∥ is sufficiently small, thanks to the assumption that the Dvkf are continuous.It follows, since ε is arbitrary that the first sum on the right in 7.13 is o(v). Now

(f (x+akvk)−f (x))−Dvkf (x)ak =