3
$\begingroup$

I have trouble answering the following question:

"Let $f: [0,+ \infty) \to \mathbb{R}$ be a function and consider the function $g: \mathbb{R}^2 \to \mathbb{R}: (x,y) \mapsto f(\left\lVert (x,y) \right\rVert)$. Give a necessary and sufficient condition on $f$ such that $g$ is totally differentiable."

I have tried to solve this and thought a good condition could be that $f$ is differentiable (I don't know if it is correct, though).

  • For the sufficiency of this condition, I have the following. Since $\mathbb{R}^2 \to \mathbb{R}: (x,y) \mapsto \left\lVert (x,y) \right\rVert$ is totally differentiable on $\mathbb{R}^2 \backslash \left\{(0,0)\right\}$, the chain rule gives us that $g$ is totally differentiable on $\mathbb{R}^2 \backslash \left\{(0,0)\right\}$. But then I don't know why $g$ would also be totally differentiable in $(0,0)$...

  • For the necessity of the condition, I don't know why total differentiability of $g$ would imply differentiability of $f$...

Thanks in advance for any help.

  • 0
    Nicely stated question. – 2017-01-02

1 Answers 1

2

Suppose $g$ is totally differentiable on $\mathbb R^2.$ Let $x>0.$ Then for $|h|$ small and nonzero,

$$ \frac{f(x+h)-f(x)}{h} = \frac{g(x+h,0)-g(x,0)}{h}.$$

Now $g$ totally differentiable certainly implies $g$ has partial derivatives at each point. Thus the limit as $h\to 0$ of the right side above exists, which implies the limit of the left side exists. I.e., $f'(x)$ exists.

At $x=0,$ verify that $f'(0)$ exists (from the right) as above. But notice that

$$\frac{f(|h|)-f(0)}{h} = \frac{g(h,0)-g(0,0)}{h}.$$

The limit on the right exists as $h\to 0,$ hence the limit on the left exists. But check that the the limit on the left exists iff $f'(0)=0.$

Summary: If $g$ is totally differentiable on $\mathbb R^2,$ then $f$ is differentiable on $[0,\infty)$ with $f'(0) = 0.$

I'm pretty sure the converse is true. Give it a try!