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I am looking for confirmation so that I can be sure I understand what is being asked here. I need to show that the following function $f(x,y)$ is not differentiable at $(0,0)$ but that $g(x,y)=yf(x,y)$ is:

$$ f(x,y) = \left \{ \begin{array}{ll} {\frac {x^2 y} {x^2 + y^2} } & \mbox{ if $x,y\neq 0$ } \\ 0 & \mbox{if $x=y=0$} \end{array} \right. $$

If the differential exists at $(x,y)$, then there is a linear mapping $f'(x,y):\mathbb{R}^2 \rightarrow \mathbb{R}$ such that

$$f(x+tu,y+tv) - f(x,y) = f'(x,y)(tu,tv) + r(tu,tv)$$

where $r(tu,tv)/t \rightarrow 0$ as $t \rightarrow 0$. At $(x,y)=(0,0)$, this implies that

$$ \frac{u^2 v}{u^2 + v^2} = f(tu,tv) = f'(0,0)(tu,tv) + r(tu,tv) $$

since $f$ is nonlinear, $f'(0,0)$ cannot exist.

However, following the same line of reasoning for $g(x,y)$ gives:

$$ \frac{u^2 v^2 t}{u^2 + v^2} = g'(0,0)(tu,tv) + r(tu,tv) $$

Thus, as $t \rightarrow 0$, we have that $g'(0,0)(tu,tv) \rightarrow 0$, which means that $g'(0,0)$ is simply the linear operator defined by $g'(0,0)(x,y)=0$.

AnonSubmitter85
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1 Answers1

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Since $f = 0$ on both coordinate axis, the partial derivative of $f$ (if they exist) must be $0$. Hence, if $f$ is differentiable at $(0,0)$ it would have to be the case that

$$ f(x,y) - f(0,0) = \frac{x^2y}{x^2+y^2} = r(x,y), $$

where $r(x,y)/\sqrt{x^2+y^2} \to 0$ as $(x,y) \to (0,0)$. Analyzing this limit, by passing to polar coordinates, we get $$ \frac{r(x,y)}{\sqrt{x^2+y^2}} = \frac{x^2y}{(x^2+y^2)^{3/2}} = \frac{r^3 \cos^2\theta \sin\theta}{r^3} = \cos^2\theta\sin\theta $$ and it's clear that this does not tend to $0$ (for most values of $\theta$) as $r \to 0$.

I'll leave the corresponding analysis of $yf(x,y)$ to you.

mrf
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