( ( k−1 ) )
(Dv f(x+ ∑ ajvj + takvk) − Dv f (x + takvk)) ak (20.6.14)
k j=1 k
(20.6.14)
Now without loss of generality, it can be assumed that the norm on X is given
by
( )
{ ∑n }
||v|| ≡ max |ak| : v = akvk
( j=1 )
because this is a finite dimensional space, all norms on X are equivalent. Therefore,
from 20.6.14 and the assumption that the Gateaux derivatives are continuous,
which proves the continuity of Df because of the assumption the Gateaux derivatives are
continuous. ■
In particular, if D_{vk}f
(x)
exist and are continuous functions of x, this shows that f is
Gateaux differentiable and in fact the Gateaux derivatives are continuous. The
following gives the corresponding result for functions defined on infinite dimensional
spaces.
Theorem 20.6.4Suppose f : U → Y where U is an open set in X, a normedlinear space. Suppose that f is Gateaux differentiable on U and that the Gateauxderivative is continuous on an open set containing x. Then f is Frechet differentiableat x.
-1--∥f (x + v)− f (x)− G (x)v∥ =-1-- sup |〈y ∗,f (x + v)− f (x)− G (x)v〉|
∥v∥ ∥v∥ ∥y∗∥≤1
-1-- ∗
= ∥v∥ ∥syu∗∥p≤1|〈y ,G(x + tv)v − G (x)v〉| ≤ s|tu|≤p1 ∥G (x + tv) − G (x)∥ℒ(X,Y)
which converges to 0 as
∥v∥
→ 0 thanks to continuity of G at x. This proves the
following.
Theorem 20.6.6Suppose f : U → Y where U is an open set in X, a normed linearspace. Suppose that f is Gateaux differentiable on U and that the Gateaux derivativeis continuous at x. Then f is Frechet differentiable at x and Df
(x)
v = D_{v}f
(x)
.
Example 20.6.7Let X be C_{0}^{2}
( )
¯Ω
where Ω is a bounded open set in ℝ^{n}consisting ofthose functions which are twice continuously differentiable and vanish near ∂Ω. The normwill be
{ }
∥u∥X ≡ ∥u∥∞ + max {∥u,i∥∞ ,i}+ max ∥u,ij∥∞i,j
Then let f : X → ℝ be defined by
1 ∫
f (u) ≡- ∇u ⋅∇udx
2 Ω
Show f is differentiable at u ∈ X.
Consider the Gateaux differentiability.
f (u +tv)− f (u ) t∫ ∇u ⋅∇vdx 1 ∫
tli→m0 -------t-------= ltim→0--Ω---t------+ t2 ∇v ⋅∇v
Ω
so it converges to
∫ ∫
∇u ⋅∇vdx = − Δuvdx
Ω Ω
the last step comes from the divergence theorem. Clearly v →−∫_{Ω}Δuvdx is linear and
ℝ valued.
so it seems that this is a
differentiable function and
∫
Df (u )(v) = − Δuvdx
Ω
Definition 20.6.8Let f : U → Y where U is an open set in X. Then f is calledC^{1}
(U)
if it Gateaux differentiable and the Gateaux derivative is continuous on U.
As shown, this implies f is differentiable and the Gateaux derivative is the Frechet
derivative. It is good to keep in mind the following simple example or variations of
it.
Example 20.6.9Define
{ x2sin(1) x ⁄= 0
f (x ) ≡ 0 if x =x0
This function has the property that it is differentiable everywhere but is not C^{1}
(ℝ)
.
In fact the derivative fails to be continuous at 0.