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Show that $f(x,y)=\dfrac{xy^2}{x^2+y^2}$ (with $(x,y)\not=(0,0)$ and $f(0,0)=0$) is continuous but not differentiable at $(0,0)$.

I tried to show continuity with an $\epsilon -\delta$ argument but I don't know how to factorize the expression so that I can have something useful.

For differentiability I think I should show that the partials are not continuous at $(0,0)$. But finding the partials is also painful.

Xena
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4 Answers4

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For continuity, a common trick is to express $f(x,y)=g(x,y)h(x,y)$ where $g$ has limit $0$ at $(0,0)$ and $h$ is bounded in a punctured neighborhood of $(0,0)$. This is easy here: $$ g(x,y)=x,\qquad h(x,y)=\frac{y^2}{x^2+y^2} $$ because it's obvious that $0\le h(x,y)\le 1$ for all $(x,y)\ne(0,0)$.

Differentiability doesn't imply continuity of the partial derivatives; in some sense it's the other way round.

This function has partial derivatives at $(0,0)$ and both are zero: $$ \lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}= \lim_{h\to0}\frac{1}{h}\frac{h\cdot0^2}{h^2+0^2} =0 $$ and $$ \lim_{h\to0}\frac{f(0,0+h)-f(0,0)}{h}= \lim_{h\to0}\frac{1}{h}\frac{0\cdot h^2}{h^2+0^2} =0 $$ So, if the function is differentiable at $(0,0)$, its differential must be zero. Can you go on?

egreg
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  • No :( Could you give me the whole answer, what happens if its differential must be zero? – Xena Jul 16 '13 at 17:46
  • @Bedi Apply the definition: the limit at $(0,0)$ of $|f(x,y)|/\sqrt{x^2+y^2}$ should be zero. Is it? – egreg Jul 16 '13 at 18:12
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By the inequality

$$|f(x,y)|=\dfrac{|x|y^2}{x^2+y^2}\leq \dfrac{|x|(x^2+y^2)}{x^2+y^2}=|x|$$ we see that $f$ is continuous at $(0,0)$.

For the differentiability, your thinking is correct so do not hesitate and do not be afraid of calculus

  • Out of curiosity, what is this inequality called? – Sujaan Kunalan Jul 16 '13 at 16:56
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    This inequality has no name unless you want call it Sami's inequality:) –  Jul 16 '13 at 16:59
  • Great! But I have doubts about my thinking of the differentiability part: we know that continuous partials imply differentiability but that doesn't necessarily mean that if the partials are discontinous the function is not differentiable, here is an example: http://mathinsight.org/differentiable_function_discontinuous_partial_derivatives – Xena Jul 16 '13 at 17:33
  • @Bedi Yes but to prove that $f$ isn't differentiable at $(0,0)$ it suffices to prove that its partial derivatives aren't continuous at $(0,0)$. –  Jul 16 '13 at 18:26
  • @SamiBenRomdhane "it suffices to prove that its partial derivatives aren't continuous at (0,0)"... No, no, and no. – Did Jul 17 '13 at 07:25
  • @Did Why? I extracted this paragraph; "Cependant une fonction possédant des dérivées partielles continues sur $E$ est différentiable sur $E$ et est dite de classe $C^1$" from this –  Jul 17 '13 at 11:24
  • Continuous partial derivatives imply differentiability. Not the other way round. – Did Jul 18 '13 at 08:20
  • To add something interesting, there's another inequality that I like to use all the time in these problems: x*y/(x^2+y^2) <= 1/2 – Darsen Dec 12 '19 at 18:26
  • I don't know if it has a name, but it comes from (x+y)^2 >= 0. (And sorry about the typewriting, I'm still learning LaTex). – Darsen Dec 12 '19 at 18:27
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We guess the limit is $0$ and so should verify that: $$\forall\epsilon>0~\exists\delta>0,~\forall(x,y)\left(0<||(x,y)-(0,0)||<\delta\Longrightarrow\big|\frac{xy^2}{x^2+y^2}-0\big|<\epsilon\right)$$

When $0<||(x,y)-(0,0)||<\delta$ so $\sqrt{x^2+y^2}<\delta$ and from this we get $$|x|<\delta,~~|y|<\delta$$ Now if we set $z=\text{max}(|x|,|y|)$ then $z<\delta$ and also $$|\frac{xy^2}{x^2+y^2}-0|=\frac{|x||y|^2}{|x|^2+|y|^2}<\frac{z^3}{z^2+0}=z<\delta$$ So, it is enough to choose $\epsilon=\delta$. This shows that the function is continuous at the origin.

Mikasa
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For proving continuity you may put $x = r\cos\theta$ and $y= r \sin\theta$. Change the function to polar coordinates. As $(x,y) \rightarrow (0,0)$, $r \rightarrow 0$. This may be a method for showing continuity.

For your second question related to differentiability see the following.

For any $v \in \mathbb{R}^2$, and $t \in \mathbb{R}$ $f(tv) = t f(v)$.

From $\epsilon , \delta$ definition you may easily show $D_{(1,0)} f(0,0) = D_{(0,1)} f(0,0) = 0$ where $D_{(1,0)}$ and $D_{(0,1)}$ are partial derivative along $x$ and $y$ axis, respectively.

But $D_{(1,1)}f(0,0) = \frac{1}{2}$.

So the function is not differentiable at $(0,0)$.

Did
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Supriyo
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