92 CHAPTER 4. THE DERIVATIVE

Definition 4.1.2 A function g is o(v) if

lim∥v∥→0

g(v)∥v∥

= 0 (4.1)

A function f : U → Y is differentiable at x ∈U if there exists a linear transformation L ∈L (X ,Y ) such that

f(x+v) = f(x)+Lv+o(v)

This linear transformation L is the definition of Df(x). This derivative is often called theFrechet derivative.

Note that from Theorem 2.7.4 the question whether a given function is differentiable isindependent of the norm used on the finite dimensional vector space. That is, a function isdifferentiable with one norm if and only if it is differentiable with another norm. In infinitedimensions, this is not clearly so and in this case, simply regard the norm as part of thedefinition of the normed linear space which incidentally will also typically be assumed tobe a complete normed linear space.

The definition 4.1 means the error, f(x+v)− f(x)−Lv converges to 0 faster than ∥v∥.Thus the above definition is equivalent to saying

lim∥v∥→0

∥f(x+v)− (f(x)+Lv)∥∥v∥

= 0 (4.2)

or equivalently,

limy→x

∥f(y)− (f(x)+Df(x)(y−x))∥∥y−x∥

= 0. (4.3)

The symbol, o(v) should be thought of as an adjective. Thus, if t and k are constants,

o(v) = o(v)+o(v) , o(tv) = o(v) , ko(v) = o(v)

and other similar observations hold.

Theorem 4.1.3 The derivative is well defined.

Proof: First note that for a fixed nonzero vector v, o(tv) = o(t). This is because

limt→0

o(tv)|t|

= limt→0∥v∥ o(tv)∥tv∥

= 0

Now suppose both L1 and L2 work in the above definition. Then let v be any vector and lett be a real scalar which is chosen small enough that tv+x ∈U . Then

f(x+ tv) = f(x)+L1tv+o(tv) , f(x+ tv) = f(x)+L2tv+o(tv) .

Therefore, subtracting these two yields (L2−L1)(tv) = o(tv) = o(t). Therefore, dividingby t yields (L2−L1)(v) = o(t)

t . Now let t → 0 to conclude that (L2−L1)(v) = 0. Sincethis is true for all v, it follows L2 = L1. ■

Lemma 4.1.4 Let f be differentiable at x. Then f is continuous at x and in fact, thereexists K > 0 such that whenever ∥v∥ is small enough,∥f(x+v)− f(x)∥ ≤ K ∥v∥ . Also if fis differentiable at x, then

o(∥f(x+v)− f(x)∥) = o(v)