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I am trying to show that $g(x,y)=y f(x,y)$ with

$$f: \mathbb R^2 \rightarrow \mathbb R $$ $$ f(x,y) = \begin{cases} \dfrac {2xy^2}{x^2+y^4} & (x,y)\ne (0,0) \\\\ 0 & (x,y)=(0,0) ~ \end{cases} $$

is not totally differentiable in $(0,0)$.

What I already have found out about the functions is that $f$ is not continuous in $(0,0)$, the function $g$ has all directional derivatives and they are equal to $0$.

The criterion I use for total differentials is that $f$ is totally differentiable in $x$, if there exists a linear map $L$ such that

$$\lim_{h \to 0}\frac{g(x+h)-g(x)-Lh}{|h|}=0$$

For $(x,y)=(0,0)$ we get

$$\lim_{h \to 0}\frac{g(h)-g(0)-Lh}{|h|}=\lim_{h \to 0}\frac{h_2f(h)-0f(0)-Lh}{|h|}=\lim_{h \to 0}h_2\frac{f(h)-f(0)}{|h|}-\frac{Lh}{|h|}=0 $$

Is it legitimate to deduce from that that $g$ is not totally differentiable in $(0,0)$ because $$lim_{h \to 0}\frac{f(h)-f(0)}{|h|}$$ does not exist, as $f$ is not continous in $(0,0)$?

B.Swan
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2 Answers2

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Its not because $f$ is discontinuous at $(0,0)$, but because only that the limit does not exist as you can choose two paths $(x,y) = (t,t), (t,2t^2)$ and you got different limits. Try some new paths as this is a hint of idea.

DeepSea
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  • By the path you mean the way in which $h$ approaches $0$? – B.Swan Jun 13 '17 at 16:06
  • I do not know which limit you adress by "the limit", do you mean the last one? Or the one in the criterion for total differentials? – B.Swan Jun 13 '17 at 16:32
  • Yes, the way $h =(x,y) \to (0,0)$. – DeepSea Jun 13 '17 at 16:40
  • When I plug those in, I get $\frac{\frac{2}{t^{-2}+1}-Lh}{\sqrt{2}t}$ for $(t,t)$ and $\frac{\frac{2}{16^{-1}t^{-2}+1}-Lh}{\sqrt{5}t}$ for $(t,2t)$, which seem to have the same limit for me. Do you care to clarify closer what you mean? – B.Swan Jun 13 '17 at 16:54
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Edited from the original: We have

$$g(x,y) = 2\frac{xy^3}{x^2 + y^4}, (x,y)\ne (0,0).$$

Suppose $Dg(0,0)$ exists. Then $Dg(0,0)(x,y) = ax + by,$ where $a = \partial g/\partial x (0,0), b=\partial g/\partial y(0,0).$ Because $g$ vanishes on the axes, $a,b = 0.$ Thus $Dg(0,0)$ is the zero map. It follows that

$$\frac{g(x,y)}{(x^2+y^2)^{1/2}} \to 0.$$

Check the paths $(x,x),(x,\sqrt x)$ to see this doesn't happen.

zhw.
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