# Analysis Problem (proving differentiability at a point)

1. Oct 26, 2004

### AKG

Need some hints on how to go about doing this:

$$f(x, y)=\left\{\begin{array}{cc}\frac{x^4 + y^4}{x^2 + y^2},&\mbox{ if } (x, y)\neq (0,0)\\0, & \mbox{ if } (x, y) = 0\end{array}\right.$$

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?

Last edited: Oct 26, 2004
2. Oct 26, 2004

### 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.