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 Incoming links: Fréchet differentiability

Fix

(x, y) \in U

and a sufficiently small rectangle inside

U

. Its rectangular increment

Δ = f (x + h, y + k) - f (x + h, y) - f (x, y + k) + f (x, y)

Δ = \int_{x}^{x + h} \int_{y}^{y + k} D_{2} D_{1} f (s, t) d t d s = \int_{y}^{y + k} \int_{x}^{x + h} D_{1} D_{2} f (s, t) d s d t .

Divide by

hk

and let

h, k \to 0

. Continuity of the mixed partial derivatives makes the two limits

D_{2} D_{1} f (x, y)

and

D_{1} D_{2} f (x, y)

. They are equal everywhere on $U$ .

Repeated interchange of adjacent partial derivatives reduces every order-

m

D_{1}^{j} D_{2}^{m - j} f

0 \leq j \leq m

. Hence there are at most

m + 1

distinct functions. This maximum is attained by

f (x, y) = e^{x + 2 y}

, whose listed partial derivatives are the distinct functions

2^{m - j} e^{x + 2 y}

. The answer is $m + 1$ , rather than the

2^{m}

possible written orders.

For the supplied

f

f (t, t) = 1/2

for

t \neq = 0

, whereas

f (0, 0) = 0

. It is not even continuous at the origin, so $f$ is neither differentiable nor infinitely differentiable there.

For

g

∣ g (x, y) ∣ \leq ∣ x y ∣ \leq (x^{2} + y^{2}) /2

. Thus

g (x, y) = o (x^{2} + y^{2})

, proving Fréchet differentiability at the origin with derivative zero. However,

D_{1} g (0, y) = - y, D_{2} g (x, 0) = x,

including zero at the origin. Therefore

D_{2} D_{1} g (0, 0) = - 1

and

D_{1} D_{2} g (0, 0) = 1

. $g$ is differentiable but not infinitely differentiable at the origin: it cannot have continuous second partial derivatives near that point.