Kenneth Kuttler

150 CHAPTER 6. THE DERIVATIVE

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)) (6.13)

Using Theorem 6.5.2,∥∥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 6.13 is o(v). Now

(f (x+akvk)−f (x))−Dvkf (x)ak =

f (x+akvk)−(f (x)+Dvkf (x)ak

)= ak

(f (x+akvk)−f (x)

ak−Dvkf (x)

)= o(v)

because ∥∥∥∥ak

(f (x+akvk)−f (x)

ak−Dvkf (x)

)∥∥∥∥≤ ∥v∥

∥∥∥∥(f (x+akvk)−f (x)

ak−Dvkf (x)

)∥∥∥∥ .Collecting terms in 6.13,

f (x+v)−f (x) = o(v)+n

∑k=1

(f (x+akvk)−f (x)) = o(v)+n

∑k=1

Dvkf (x)ak

which shows that Df (x)(v) = ∑nk=1 Dvkf (x)ak where v = ∑

nk=1 akvk. This formula also

shows that x→ Df (x) is continuous because of the continuity of these Dvkf . ■Note how if X =Rp and the basis vectors are the ek, then the a are just the components

of the vector v taken with respect to the usual basis vectors. Thus this gives the above resultabout the matrix of Df (x).

This motivates the following definition of what it means for a function to be C1.

Definition 6.6.2 Let 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 C1 if there exists a basis for X ,{v1, · · · ,vn} such that the Gateaux deriva-tives,Dvkf (x) exist on U and are continuous functions of x.

Note that as a special case where X = Rn, you could let the vk = ek and the conditionwould reduce to nothing more than a statement that the partial derivatives ∂f

∂xiare all con-

tinuous. If X =R, this is not a very interesting condition. It would say the derivative existsif the derivative exists and is continuous.

Here is another definition of what it means for a function to be C1.

150 CHAPTER 6. THE DERIVATIVEThe mapping v — (a,...,@,) is an isomorphism of V and F” and we can define a normas )y lael which is equivalent to the norm on V thanks to Theorem 5.2.4. Let hy (x) =f (@+z) “ f (x). Then collecting the terms,nf(x+v)- =) ( hy (x + agv,) — + Lis (a+agv,)—f(x)) (6.13)k=1Using Theorem 6.5.2,|| Daw, x (w+ taxv,)|| = \]axDv, hx (a + tav,)]||k-1ak (pf [++ Yan sume) - Def (2 +10) |j=lIACllvlleprovided ||v]|| is sufficiently small, thanks to the assumption that the D,, f are continuous.It follows, since € is arbitrary that the first sum on the right in 6.13 is o(v). Now(f (a+ avy) — f (@)) — Do, f (@) a =f (w@+ayux) — f (x)akf (20401) ~ (F (@) + Daf (@) ax) =a ( Dy, (@)) =0(0)becausea (fetes — f (x) ~Dy.f («)) |ak< jo | (Fees )_p, Fle \)).Collecting terms in 6.13,flet+v)— f(a vedo (a +ayvg) — f (@)) = 0(v) +. Dag f (a) a4fo k=lwhich shows that Df (a) (v) = Yf_; Du, f (aw) ax where v = Yf_, a;v¢. This formula alsoshows that « — Df (a) is continuous because of the continuity of these D,, f.Note how if X = R? and the basis vectors are the e;, then the @ are just the componentsof the vector v taken with respect to the usual basis vectors. Thus this gives the above resultabout the matrix of Df (x).This motivates the following definition of what it means for a function to be C!.Definition 6.6.2 Let 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' if there exists a basis for X,{v1,-++,Un} such that the Gateaux deriva-tives, Dy, f (a) exist on U and are continuous functions of «.Note that as a special case where X = R”, you could let the vz; = e, and the conditionwould reduce to nothing more than a statement that the partial derivatives of are all con-tinuous. If X = R, this is not a very interesting condition. It would say the derivative existsif the derivative exists and is continuous.Here is another definition of what it means for a function to be C!.