Is this function Gateaux differentiable?

Question

Consider the real-valued function $f:\mathbb{R}\rightarrow\mathbb{R}$ of two real variables defined by $$f(x,y)=\begin{cases} \frac{x^3}{x^2+y^2} & \mbox{ if } (x, y)\ne (0, 0)\\ 0 & \mbox{ if } (x, y)=(0, 0). \end{cases}$$ Is this function Gateux differentiable and why? Thanks

Answer 1

This $f$ is $C^1$ on all of $\dot{\mathbb R}^2$, hence a fortiori Gateaux differentiable in this domain. It remains to investigate what happens at the origin. One computes $$f_x(0,0)=\lim_{x\to 0}{f(x,0)-f(0,0)\over x}=1,\quad f_y(0,0)=\lim_{y\to 0}{f(0,y)-f(0,0)\over y}=0\ .$$ If $f$ were Fréchet differentiable, or just: differentiable at $(0,0)$ it would therefore have to be the case that $$\lim_{(x,y)\to(0,0)}{f(x,y)-f(0,0)-x\over\sqrt{x^2+y^2}}=0\ .$$ But $${f(x,y)-f(0,0)-x\over\sqrt{x^2+y^2}}={r\cos^3\phi-r\cos\phi\over r}=-\sin^2\phi\cos\phi$$ does not converge to $0$ when $r\to0$. It follows that $f$ is not differentiable at $(0,0)$.

For Gateaux differentiability we have to choose a fixed tangent vector $(u,v)$ at $(0,0)$ and to consider the limit $$d_{(u,v)}f(0,0):=\lim_{t\to0}{f(tu,tv)-f(0,0)\over t}=\lim_{t\to0}{t^3u^3\over t(t^2u^2+t^2v^2)}={u^3\over u^2+v^2}$$ if $(u,v)\ne(0,0)$, and $=0$ otherwise. Since the limit exists for each $(u,v)$ the function $f$ is indeed Gateaux differentiable at $(0,0)$.