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Let $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be defined as $f(0,0)=0$ and $f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}$ for $(x,y)\neq (0,0)$

Then question asks to prove that $f$ is differentiable.

Hint that is given is :

Show that $D_1f$ equals product of $y$ and a bounded function and $D_2f$ equals product of $x$ and a bounded function.

I calculated $D_1f=y\frac{x^4+4x^2y^2-y^4}{(x^2+y^2)^2}$ and clearly $\left|\frac{x^4+4x^2y^2-y^4}{(x^2+y^2)^2}\right|\leq 3$ So, we have $D_1f$ as product of $y$ and a bounded function.. Similarly $D_2f$ is product of $x$ with a bounded function..

I know that if $D_1f$ and $D_2f$ are bounded then $f$ is continuous..

But here it is product of a bounded function with $y$..

I do not know how to proceed..

Please help me...

P.S : I am supposed to prove that it is differentiable.

I think i should use condition that if partial derivatives are continuous then $f$ is differentiable...

kk lm
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  • Why calculate partial derivatives to show continuity? (Btw proving continuity is pretty easy here.) – zhw. Jan 17 '16 at 18:07
  • questions were in that order.. find partial derivatives and then show that $f$ is continuous and it was asking to use the hint that partial derivaives has that property that i have stated above @zhw. – kk lm Jan 17 '16 at 18:08
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    @kklm You're not forced to use the hint. Recall that $$\forall x,y\in \mathbb R\left(2|xy|\leq x^2+y^2\right).$$ Hence $$\forall x,y\in \mathbb R\left(|f(x,y)|\leq \text{Something pretty}\right).$$ – Git Gud Jan 17 '16 at 18:12
  • What questions? I see one question only, and as it stands it is a bit bizarre. – zhw. Jan 17 '16 at 18:12
  • @zhw. : I am sorry for stating it wrongly.. I wanted to prove that $f$ is differentiable.. I can say that if i know that partial derivatives are continuous... Now, i want to see if there is any better way to prove that partial derivatives are continuous... – kk lm Jan 17 '16 at 18:16
  • @GitGud : Thanks.. For proving continuity your idea works pretty well... :) – kk lm Jan 17 '16 at 18:20
  • @kklm Basically the problem is asking you to use the fact that if $\lim \limits_{(x,y)\to(0,0)}(D_1f)(x,y)=0=\lim \limits_{(x,y)\to(0,0)}(D_2f)(x,y)$, then $f$ is $C^1$. – Git Gud Jan 17 '16 at 18:21
  • @kklm No problem, but zhw.'s way of proving continuity is much more obvious. – Git Gud Jan 17 '16 at 18:21
  • @GitGud : I have got one more question.. I could prove that partial derivatives are continuous at $(0,0)$ but i have no idea how to prove that they are continuous every where – kk lm Jan 17 '16 at 18:25
  • The origin is the only problematic problem, for the other points you just need to note that $f$ is a rational function where the denominator is never null. – Git Gud Jan 17 '16 at 18:26
  • @GitGud : Oh yes yes... I dont know how i forgot that :D Thank you :) – kk lm Jan 17 '16 at 18:28
  • @kklm You're welcome. – Git Gud Jan 17 '16 at 18:29

2 Answers2

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Differentiability (using continuity of partial derivatives): Check that both partial derivatives of $f$ at $(0,0)$ are zero. So you want to show $D_1f(x,y) \to 0$ as $(x,y) \to (0,0).$ But you've shown $D_1f(x,y) = y\cdot g(x,y),$ where $g$ is bounded. That implies what you want, right? Same for $D_2f(x,y).$

We can also show $Df(0,0)$ exists more directly. We know both partial derivatives equal $0$ at $(0,0).$ So if $Df(0,0)$ exists, it is the zero linear transformation. So we want to show

$$f(x,y) = f(0,0) + 0\cdot x + 0\cdot y + o(\sqrt {x^2+y^2}) = o(\sqrt {x^2+y^2}).$$

The estimate $|f(x,y)| \le |xy|$ (from below) gives this.

Previous answer on continuity:

Just use

$$|f(x,y)| \le |xy|\frac{x^2 + y^2}{x^2+y^2} = |xy|.$$

We know $xy\to 0$ as $(x,y) \to (0,0)$ and that gives continuity at $(0,0).$

zhw.
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  • Ok... This does not answer correct question but i see this as a hint for my correct question... To prove partial derivatives are continuous... $|D_1f|\leq 3|y|$ so first partial derivative is continuous at $(0,0)$ and similarly second partial derivative is continuous at $(0,0)$ and thus $f$ is differentiable at $(0,0)$.. – kk lm Jan 17 '16 at 18:18
  • On your edit.. Yes Yes.. I got it :) Thanks.. – kk lm Jan 17 '16 at 18:29
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For $p=(x,y)$ and $r>0$ let $B(p,r)=\{(x',y'):(x'-x)^2+(y'-y)^2<r^2\}.$ Suppose $g_1:S_1\to R$ and $g_2:S_2\to R$ are differentiable, where $S_1\subset R^2\supset S_2,$ and $S_1\cap S_2\supset B(p,r)$. Then $\bullet$(1): $g_1,g_2, g_1+g_2,$ and $g_1 g_2$ are differentiable on $B(p,r).$ $\bullet$(2): If $g_1 \neq 0$ on $B(p,r)$ then $g_2/g_1$ is differentiable on $B(p,r).$..... In your Q, with $S_1=S_2=R^2\backslash \{(0,0)\}$ and $p=(x,y)\ne (0,0),$ apply (1) and (2) repeatedly with $r=\sqrt {x^2+y^2}.$

DanielWainfleet
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