Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let's concider two definitions of twice differentiability:

Definition 1. $f(x,y)$ is twice differentiable at $(x_0,y_0)$ iff a)$f^\prime_x, f^\prime_y$ are differentiable functions of two variables at $(x_0,y_0)$, b) $f$ is differentiable in a neighbourhood of $(x_0,y_0)$.

Definition 2. $f(x,y)$ is twice differentiable at $(x_0,y_0)$ iff a)$f^\prime_x, f^\prime_y$ are differentiable functions of two variables at $(x_0,y_0)$.

It's easy to show that Definition 1 is equivalent to (Fréchet) differentiability of $f^\prime:R^2\to L(R^2;R)$ at $(x_0,y_0)$ and this definition seems quite natural. But usually Definition 2 is used for functions of several variables and it is enough to prove some important theorems (Young' theorem, Taylor's theorem with the Peano form of the remainder). My question: is anybody aware of an example of a function, which satisfies Definition 2 but doesn't satisfy Definition 1?? If such a function exists, there must be a special term for those functions that satisfy Definition 2, but not twice Fréchet differentiable. Constructing such an example won't be very trivial because a) implies that function $f$ is uniformly continuous in a n-hood of $(x_0,y_0)$. Anyway, I believe that this question must have been studied carefully, though you can really find different definitions of n times differentiable function of several variables in different books.

share|improve this question
    
In case anyone else shares my momentary confusion upon reading the question: (a) implies that $f$ is differentiable at $(x_0,y_0)$, since its partial derivatives are continuous there (since they are differentiable there); the additional assumption of (b) is that it is differentiable in a neighborhood of $(x_0,y_0)$. –  Mike Shulman May 12 at 23:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.