718 CHAPTER 22. THE DERIVATIVE

provided ||v|| is sufficiently small. Since ε is arbitrary, it follows from Lemma 22.6.1 theexpression in 22.6.13 is o(v) because this expression equals a finite sum of terms of theform h(1)−h(0) where ||h′ (t)|| ≤ ε ||v|| whenever ∥v∥ is small enough. Thus

f(x+v)− f(x) =n

∑k=1

[f(x+akvk)− f(x)]+o(v)

=n

∑k=1

Dvk f(x)ak +n

∑k=1

[f(x+akvk)− f(x)−Dvk f(x)ak

]+o(v) .

Consider the kth term in the second sum.

f(x+akvk)− f(x)−Dvk f(x)ak = ak

(f(x+akvk)− f(x)

ak−Dvk f(x)

)where the expression in the parentheses converges to 0 as ak → 0. Thus whenever ||v|| issufficiently small, ∣∣∣∣f(x+akvk)− f(x)−Dvk f(x)ak

∣∣∣∣≤ ε |ak| ≤ ε ||v||

which shows the second sum is also o(v). Therefore,

f(x+v)− f(x) =n

∑k=1

Dvk f(x)ak +o(v) .

Defining

Df(x)v≡n

∑k=1

Dvk f(x)ak

where v = ∑k akvk, it follows Df(x) ∈L (X ,Y ) and is given by the above formula.It remains to verify x→ Df(x) is continuous.

||(Df(x)−Df(y))v||

≤n

∑k=1

∣∣∣∣(Dvk f(x)−Dvk f(y))

ak∣∣∣∣

≤ max{|ak| ,k = 1, · · · ,n}n

∑k=1

∣∣∣∣Dvk f(x)−Dvk f(y)∣∣∣∣

= ||v||n

∑k=1

∣∣∣∣Dvk f(x)−Dvk f(y)∣∣∣∣

(Note that ∥v∥ ≡max{|ak| ,k = 1, · · · ,n} where v = ∑k akvk) and so

||Df(x)−Df(y)|| ≤n

∑k=1

∣∣∣∣Dvk f(x)−Dvk f(y)∣∣∣∣

which proves the continuity of Df because of the assumption the Gateaux derivatives arecontinuous.

In particular, if Dvk f(x) exist and are continuous functions of x, this shows that f isGateaux differentiable and in fact the Gateaux derivatives are continuous. The followinggives the corresponding result for functions defined on infinite dimensional spaces.