Prove that the functin is differentiable at (0, ,0).

[michonamona]

Homework Statement

Let r>0, and let f be a function from B_{r}(\textbf{0}) \rightarrow \textbf{R}, and suppose that there exists an \alpha > 1 such that |f(\textbf{x})| \leq ||\textbf{x}||^{\alpha} for all \textbf{x} \in B_{r}(\textbf{0}). Prove that f is differentiable at 0.

What happens to this result when \alpha = 1?

Homework Equations

Note:

The characters in bold are vectors.

B_{r}(\textbf{0}) is an open ball in R^{n} about 0

The symbol ||.|| represents the norm of the vector.

Some definitions of differentiability of a function

A function f from R \rightarrow R^{n} is differentiable if

lim_{\textbf{h} \rightarrow \textbf{0}} \frac{f(\textbf{0} + \textbf{h}) - f(\textbf{0}) - Df(\textbf{a})(\textbf{h})}{||\textbf{h}||} = \textbf{0}.

Where Df(\textbf{a}) is the matrix of first order partials. In our case, since the function is from R to R^n, this would be the gradient of f.

The Attempt at a Solution

The started this problem by computing that the first order partials exists. But that by itself is not sufficient to concluded that f is differentiable at 0. The next step would be to show that the first-order partials of f are continuous at 0. This is where I got stuck.

 

[LeonhardEuler]

Can you explain what you are thinking about this problem?

 

[michonamona]

Can you explain what you are thinking about this problem?

Hi Euler,

Thank you for your reply.

There's a theorem in my textbook that says that if the first-order partial derivatives of a vector-valued function f exists at a, and if these first-order partial derivatives are continuous at a, then f is differentiable at a.

I know how to show the existence of the first-order partial derivatives of f at 0, but I'm having trouble proving that they are continuous at 0. Any insights?

 

[LeonhardEuler]

I see how you are stuck. Have you thought about avoiding this theorem, and working directly from the definition of the derivative?

 

[michonamona]

I see how you are stuck. Have you thought about avoiding this theorem, and working directly from the definition of the derivative?

You mean using:

lim_{\textbf{h} \rightarrow \textbf{0}} \frac{f(\textbf{0} + \textbf{h}) - f(\textbf{0}) - Df(\textbf{a})(\textbf{h})}{||\textbf{h}||} = \textbf{0}.

directly?

 

[LeonhardEuler]

yes.