352
P. Xiang and Y. A. Wang
The Gâteaux differential is homogeneous in s in the sense that
dT(x; 𝛼s) = 𝛼dT(x; s)
(106)
but is in general neither linear nor continuous in s. Nor does the existence of the
Gâteaux differential at x ensure continuity of T at x. For example,
f (𝜉 1 , 𝜉 2 ) =
{ 𝜉
3
1
𝜉 2
(𝜉 1 , 𝜉 2 ≠ 0)
0 (𝜉 1 = 𝜉 2 = 0)
.
(107)
At point (0, 0), it can be easily shown that the Gâteaux differential exists and it is
zero. Clearly, the Gâteaux differential is a continuous linear operator. However, f is
not continuous at (0,0). Therefore, we cannot relate the Gâteaux differentiability of
T to the continuity of T.
Let us go forward on the basis that  is also a normed vector space. Suppose
dT(x; s) is linear and continuous in s for some x ∈ , then we may write
dT(x; s) = lim
𝜆→0
T(x + 𝜆s) − T(x)
𝜆
= T
′
G (x)s .
(108)
The operator T
′
G
is by definition, a mapping  →  and is linear and continuous:
we may conclude that
T
′
G (x) ∈ (,  ) .
(109)
This operator is called the Gâteaux or weak derivative of T at x. It is very important
to note that when speaking of the linearity and continuity of T
′
G
(x), we means those
properties in the operator sense with respect to a fixed s. T
′
G
itself may be a function of
x, but its continuity and linearity with respect to the variable x are complete different
things from the continuity and linearity we discussed here.
When T
′
G
(x) exists, it is certainly true that
T(x + 𝜆s) − T(x) = T
′
G (x)𝜆s + 𝜖(x, s, 𝜆) ,
(110)
where 𝜖∕𝜆 → 0 as 𝜆 → 0 with x and s fixed. However, the convergence may not be
uniform with respect to s and in that case T cannot be approximated by a linear operator with uniform accuracy in the neighborhood of x. If we further demand uniform
convergence then we arrive at the strong derivative.
Definition 12 Let  and  be normed vector spaces. An operator T ∶  →  is
Fréchet differentiable at x ∈ Dom(T) ⊂  if there exists a continuous linear operator T
′
F
(x) ∈ (,  ) such that, for all s ∈ ,
T(x + s) − T(x) = T
′
F (x)s + 𝜖(x; s)
(111)