Analysis Problem (proving differentiability at a point)

[AKG]

Need some hints on how to go about doing this:

f(x, y)=\left\{\begin{array}{cc}\frac{x^4 + y^4}{x^2 + y^2},&amp;\mbox{ if }<br /> (x, y)\neq (0,0)\\0, &amp; \mbox{ if } (x, y) = 0\end{array}\right.

Show that f is differentiabile at (0, 0).

I've tried a number of things, too ugly and not worth writing down here (all got me nowhere). As far as I can tell, I want to show that some linear transformation \lambda \ :\ \mathbb{R}^2 \rightarrow \mathbb{R} satisfies the equation:

\lim _{h \rightarrow 0} \frac{|f(0 + h) - f(0) - \lambda (h)|}{|h|} = 0

\lim _{h \rightarrow 0} \frac{|f(h) - \lambda (h)|}{|h|} = 0

\lim _{h \rightarrow 0} \left ( \frac{h_1^4 + h_2^4}{|h|^3} - \frac{\lambda (h)}{|h|} \right ) = 0

Note that h = (h_1, h_2) \in \mathbb{R}^2

Now, if we let \mu = -\lambda, then we have, and h_2 = 0, then we have:

\lim _{h \rightarrow 0} \left ||h_1| + \frac{\mu (h)}{|h_1|} \right |

\leq \lim _{h \rightarrow 0} |h_1| + \left | \frac{\mu (h)}{|h_1|} \right |

We know that there exists some real M > 0 such that |\mu (v)| \leq M|v|\ \forall \ v \in \mathbb{R}^2, so:

\leq \lim _{h \rightarrow 0} |h_1| + M = M

So, if the function is differentiable at 0, then M = 0, so the linear transformation \mu is the zero transformation, so the derivative at 0 is the zero transformation. Tell me if I've made a mistake so far, because, if not, I think I can prove that it's also not the zero transformation. If I try to picture the graph, I think it should be zero. But I want to show that there exists some transformation such that :

\lim _{h \rightarrow 0} \frac{|f(0 + h) - f(0) - \lambda (h)|}{|h|} = 0

holds, and when I try \lambda = 0, I just can't seem to evaluate the limit right (or rather, prove that it will evaluate to zero). However, I do seem to be able to show that it will be greater than zero, meaning that, if there is a derivative, the zero transformation is not it (contrary to what I just showed above with M = 0). So, clearly, I'm stuck. Any help would be appreciated.

EDIT: Actually, I think I can do the proof, here's what I have.

Assuming what I've done is right so far, and \lambda = \mu = 0 (the zero transformation), then I need to prove:

L = \lim _{h \rightarrow 0} \frac{|f(h)|}{|h|} = 0

L = \lim _{h \rightarrow 0} \frac{|h_1^4 + h_2^4|}{(h_1^2 + h_2^2)^{3/2}}

= \lim _{h \rightarrow 0} \frac{|(h_1^2 + h_2^2)^2 - 2h_1^2h_2^2|}{(h_1^2 + h_2^2)^{3/2}}

= \lim _{h \rightarrow 0} \left ||h| - \frac{2h_1^2h_2^2}{(h_1^2 + h_2^2)^{3/2}} \right |

= 2\lim _{h \rightarrow 0} \frac{h_1^2h_2^2}{(h_1^2 + h_2^2)^{3/2}}

= 2\lim _{h \rightarrow 0} \left (|h|\frac{h_1^2h_2^2}{(h_1^2 + h_2^2)^2} \right )

= 2\lim _{h \rightarrow 0} |h| \left (\frac{h_1h_2}{h_1^2 + h_2^2} \right )^2

Now, consider the function g(z) = z + \frac{1}{z} for positive z \in \mathbb{R}. Simple analysis shows that g reaches a minimum at 2, so:

z + \frac{1}{z} \geq 2

Now, let \frac{|h_1|}{|h_2|} = z. Now, if either component of h is zero, we could have proven that the L = 0 long ago, so for the case where neither is zero, we can assign z as we have above. Now, we have:

\frac{|h_1|}{|h_2|} + \frac{|h_2|}{|h_1|} \geq 2

h_1^2 + h_2^2 \geq 2|h_1||h_2|

\frac{1}{2} \geq \frac{|h_1||h_2|}{h_1^2 + h_2^2}

\frac{1}{4} \geq \left ( \frac{h_1h_2}{h_1^2 + h_2^2} \right )^2

2|h|\frac{1}{4} \geq 2|h|\left ( \frac{h_1h_2}{h_1^2 + h_2^2} \right )^2

So:

L \leq \frac{1}{2}\lim _{h \rightarrow 0} |h| = 0

And so the proof is done. Did I do it right?

 

[AKG]

Originally, I had trouble with this question (which is why I posted it here). However, I think I got it (see edit in original post) so if anyone could check and make sure I did it right, that would be great. Thanks.

 

[wais]

Your proof looks correct. Here are some hints on how to approach this problem:

1. First, try to understand the function f(x,y) and what it looks like. This will help you in visualizing the problem and understanding what you need to prove.

2. The key to proving differentiability at a point is to find a linear transformation that approximates the function well enough at that point. In this case, you need to find a linear transformation that approximates f(x,y) at (0,0).

3. Use the definition of differentiability to set up the limit equation as you have done. This will give you an idea of what the linear transformation should look like.

4. Use algebraic manipulation and inequalities to simplify and prove the limit equation. This will involve using properties of limits and manipulating the numerator and denominator.

5. Remember that the linear transformation you are looking for is not unique, so there may be multiple ways to prove differentiability at a point. Keep experimenting and trying different approaches until you find one that works.

Overall, the key to proving differentiability at a point is to have a good understanding of the function and to think creatively about how to approximate it with a linear transformation. Good luck!