What condition on $f$ to be differentiable on $(0,0)$?

Question

Let $a > 0$ , $b > 0$ and $f_{a,b} :\mathbb{R}^2 \rightarrow \mathbb{R} $ defined as:

$$f_{a,b} (x,y) = x +y + |x|^a |y|^b$$

Give a necessary and sufficient condition on $ (a,b) $ for $f_{a,b}$ to be differentiable at (0,0).

Using the definition of the differentiation on the point $(0,0)$, I need to get :

$$\lim_{(x,y) \to (0,0)} \frac{|f_{a,b}(x,y) - f_{a,b}(0,0)|}{\sqrt{x^2 + y^2}} = 0$$

We have:

$$ \frac{|f_{a,b}(x,y) - f_{a,b}(0,0)|}{\sqrt{x^2 + y^2}} = \frac{|x + y + |x|^a|y|^b|}{\sqrt{x^2 + y^2}} \leq \frac{|x|^a|y|^b}{\sqrt{x^2 + y^2}} $$

As a sufficient condition, the last term needs to tend to zero to get the differentiability, for that to happen, I do not how to derive the condition on $a$ and $b$.

I don't know how to get the necessary condition.

Thank you.

Answer 1

Here's a useful theorem:

If $f$ is differentiable at $(0,0)$ then the directional derivative $$\partial_{\vec u} f (0,0) = \lim_{h \to 0} \frac{f(t\vec u)}{t} $$ exists for every vector $\vec u$. In particular, the two partial derivatives exist, namely $$\frac{\partial f}{\partial x}(0,0) = \partial_{\langle 1,0 \rangle} f(0,0) $$ and $$\frac{\partial f}{\partial y}(0,0) = \partial_{\langle 0,1 \rangle} f(0,0) $$ Furthermore, for any vector $\vec u = \langle s,t \rangle$ we have $$\partial_{\vec u} f(0,0) = s \cdot \frac{\partial f}{\partial x}(0,0) + t \cdot \frac{\partial f}{\partial y}(0,0) $$

So, fixing values of $a,b$, you can determine whether the directional derivatives exist for every $\vec u = \langle s,t \rangle$. And if they do, then you should be able to get formulas for those directional derivatives expressed using $a$ and $b$, and then you should be able to use those formulas to determine whether the last equation is true for every $\vec u = \langle s,t \rangle$. This procedure should give you specific conditions on $a$ and $b$ which will help answer the question.

Answer 2

First not that the $x+y$ has no effect on differentiability. So drop those terms. The partial derivatives are both $0$. Hence, if the function is differentiable then (the dertivative is necessarily $0$ and) $\frac {|x|^{a}|y|^{b}} {\sqrt {x^{2}+y^{2}}} $ must tend to $0$ as $(x,y) \to 0$. In other words, $\frac {x^{2a} y^{2b}} {x^{2}+y^{2}} $ must converge as $(x,y) \to 0$. Put $x=y$ to see that $a+b >1$ is a necessary condition. This condition is also sufficent: use polar coordinates to prove this.