22.6. A MEAN VALUE INEQUALITY 717
Proof: Let v =∑nk=1 akvk. Then
f(x+v)− f(x) = f
(x+
n
∑k=1
akvk
)− f(x) .
Then letting ∑0k=1 ≡ 0, f(x+v)− f(x) is given by
n
∑k=1
[f
(x+
k
∑j=1
a jv j
)− f
(x+
k−1
∑j=1
a jv j
)]
=n
∑k=1
[f(x+akvk)− f(x)]+
n
∑k=1
[(f
(x+
k
∑j=1
a jv j
)− f(x+akvk)
)−
(f
(x+
k−1
∑j=1
a jv j
)− f(x)
)](22.6.13)
Consider the kth term in 22.6.13. Let
h(t)≡ f
(x+
k−1
∑j=1
a jv j + takvk
)− f(x+ takvk)
for t ∈ [0,1] . Then
h′ (t) = ak limh→0
1akh
(f
(x+
k−1
∑j=1
a jv j +(t +h)akvk
)− f(x+(t +h)akvk)
−
(f
(x+
k−1
∑j=1
a jv j + takvk
)− f(x+ takvk)
))
and this equals (Dvk f
(x+
k−1
∑j=1
a jv j + takvk
)−Dvk f(x+ takvk)
)ak (22.6.14)
Now without loss of generality, it can be assumed that the norm on X is given by
||v|| ≡max
{|ak| : v =
n
∑j=1
akvk
}
because this is a finite dimensional space, all norms on X are equivalent. Therefore, from22.6.14 and the assumption that the Gateaux derivatives are continuous,
∣∣∣∣h′ (t)∣∣∣∣ =
∥∥∥∥∥(
Dvk f
(x+
k−1
∑j=1
a jv j + takvk
)−Dvk f(x+ takvk)
)ak
∥∥∥∥∥≤ ε |ak| ≤ ε ||v||