Differentiability through paths

Question

Let $f:\mathbb{R}^{2} \to \mathbb{R}$ defined by $$f(x) = \begin{cases}\displaystyle\frac{x^{3}}{x^{2}+y^{2}},& (x,y)\neq (0,0)\\ 0,& (x,y) = (0,0)\end{cases}.$$ Show that $f$ is not differentiable in $(0,0)$, however, show that for every differentiable path $\lambda: (0,1) \to \mathbb{R}^{2}$, passing through origin, $f \circ \lambda$ is differentiable.

My problem is in the second part. If $\lambda: \mathbb{R} \to \mathbb{R}^{2}$ is a line pararallel to $e_{i}$ passing through $0$, $\lambda(t) = (0,0) + te_{i}$, I can show that $(f\circ \lambda)'(0) = \frac{\partial}{\partial x_{i}}f(0,0)$. I've tried to generalize this to an arbitrary path with domain $(0,1)$, but I couldnt. Can someone help me?

Answer 1

Denote $\lambda(t) = (\lambda_1(t), \lambda_2(t))$. Since $\lambda$ goes through the origin, we have that $\lambda(c) = (0,0)$ for some $c\in(0,1)$. We can assume that $\lambda'(c) \ne 0$ so that $\lambda$ isn't constant on a neighbourhood of the origin.

Then $(f\circ \lambda)(c) = 0$ so by definition:

$$(f\circ \lambda)'(c) = \lim_{t\to 0} \frac{(f\circ \lambda)(c+t)- (f\circ \lambda)(c)}t = \lim_{t \to 0} \frac{(f\circ \lambda)(c+t)}t= \lim_{t \to 0} \frac{\lambda_1(c+t)^3}{t(\lambda_1(c+t)^2 + \lambda_2(c+t)^2)}$$

Notice that $$\lim_{t\to 0} \frac{\lambda_1(c+t)}{t} = \lim_{t\to 0} \frac{\lambda_1(c+t)-\lambda_1(c)}{t}= \lambda_1'(c)$$ and similarly for $\lambda_2$.

Therefore $$(f\circ \lambda)'(c) = \lim_{t\to 0} \frac{\lambda_1(c+t)^3}{t^3} \cdot \frac{t^2}{\lambda_1(c+t)^2 + \lambda_2(c+t)^2}=\lim_{t\to 0} \left(\frac{\lambda_1(c+t)}{t}\right)^3 \cdot \frac1{\left(\frac{\lambda_1(c+t)}{t}\right)^2 + \left(\frac{\lambda_2(c+t)}{t}\right)^2}$$

Since $\lambda'(c) \ne 0$, the product rule for limits yields $$(f\circ \lambda)'(c) = \frac{\lambda_1'(c)^3}{\lambda_1'(c)^2 + \lambda_2'(c)^2}$$

Answer 2

A broad hint, but (deliberately) not a complete solution

Suppose that $\alpha$ and $\beta$ are two differentiable paths with $\alpha(0) = \beta(0) = (0,0)$ and $\alpha'(0) = \beta'(0) \ne 0$. In short form, '$\alpha$ and $\beta$ agree to first order at $0$'.

You can prove that if $f \circ \alpha$ is differentiable at $0$, then so is $f \circ \beta$, and the derivatives agree. (I think I'd use an epsilon-delta argument to show this. The proof should not depend on $f$, by the way, but the assumption that the derivative $\alpha'(0)$ is nonzero will be essential.)

Once you know that, you need only consider a few possible curves through the origin, namely those of the form $$ \alpha(t) = t \mathbf{v} $$ where $\mathbf{v} = (a, b) = r (\cos \theta, \sin \theta) $ is some nonzero vector, i.e., $r \ne 0$. For each such vector, it's easy to write down $f \circ \alpha(t)$ and observe that it's differentiable, and then (with the help of the second paragraph) you're done.