# For which values of the parameter, $f(x, y)$ is differentiable at $(0, 0)$?

For which values of $$\alpha$$, the following function is differentiable at the origin?

$$f(x, y) = \begin{cases} \frac{x^2y}{(x^4+y^2)^{\alpha}} & (x, y) \neq (0, 0) \\\\ 0 & (x, y) = (0, 0)\end{cases}$$

attempts

I know that continuity is not a sufficient condition for a function to be differentiable at a point, yet it's necessary because if $$f$$ is not continuous then it's not differentiable (at a point).

Then I at least need the continuity, when restricting to the lines though the origin $$y = mx$$ I get

$$\lim_{x\to 0} f(x, mx) = \lim_{x\to 0} \frac{mx^3}{(x^{2\alpha}(x^2+m^2)} = 0 \quad \iff \alpha < \frac{3}{2}$$

Now, I know that for some $$f$$, if $$\exists$$ the partial derivatives at a point of $$f$$ and they are continuous too, then $$f$$ is differentiable at that point.

• Is there a rapider way to check for the differentiability when $$f$$ depends on a parameter, or should I check for existence and continuity of $$\partial f$$?

Thank you.

• Writing a complete and slightly more general answer, request you to wait Mar 10, 2023 at 8:55
• @SarveshRavichandranIyer Thank you so much, I appreciate that. No rush! Mar 10, 2023 at 12:15
• @Numb3rs I need to tender an apology at this point. While the technique I mentioned for continuity works very nicely, it breaks down horribly for differentiability, to the extent that I have been unable to fix the answer with consistent effort. I would like you to remove the acceptance and upvote so that I can delete the answer, but I promise that I'll be back because this is a question which deserves a very good answer. Once again I'm really sorry about this. Having said that, I still believe that $\alpha < \frac 34$ is the differentiability threshold, although I need to check this. Mar 16, 2023 at 12:59
• +1 for this nice question Mar 16, 2023 at 23:45
• @SarveshRavichandranIyer I have to thank you again for this! By the way, I got an idea which I believe to actually be wrong, or incomplete. But I don't know if I will be able to develop here completely, in a comment. I could try to shrink it, because it's actually not a big deal in the end. Would you take a look at it and straightly tell me how wrong is it? :D Mar 18, 2023 at 19:09

$$f(x,y)=\frac{x^{2}y}{(x^{4}+y^{2})^{a}}.$$ Since $$\partial_{x}f(0,0)=\partial_{y}f(0,0)=0$$, we have \begin{align*} & \frac{f(x,y)-\partial_{x}f(0,0)(x-0)-\partial_{y}f(0,0)}{\sqrt{x^{2}+y^{2}} }\\ & =\frac{x^{2}y}{\sqrt{x^{2}+y^{2}}(x^{4}+y^{2})^{a}}. \end{align*} We have \begin{align*} \frac{x^{2}|y|}{\sqrt{x^{2}+y^{2}}(x^{4}+y^{2})^{a}} & =\frac{\sqrt{x^{2} }(x^{4})^{1/4}(y^{2})^{1/2}}{\sqrt{x^{2}+y^{2}}(x^{4}+y^{2})^{a}}\leq \frac{\sqrt{x^{2}+y^{2}}(x^{4}+y^{2})^{1/4+1/2}}{\sqrt{x^{2}+y^{2}} (x^{4}+y^{2})^{a}}\\ & =\frac{(x^{4}+y^{2})^{1/4+1/2}}{(x^{4}+y^{2})^{a}}=(x^{4}+y^{2} )^{3/4-a}\rightarrow0 \end{align*} if $$\frac{3}{4}>a$$. If $$a\ge \frac{3}{4}$$, $$x^{2}=y$$ with $$x>0$$ gives $$\frac{x^{4}}{\sqrt{x^{2}+x^{4}}(x^{4}+x^{4})^{a}}=\frac{x^{4}}{2^{a} \sqrt{1+x^{2}}x^{1+4a}}=\frac{x^{3-4a}}{2^{a}\sqrt{1+x^{2}}}\rightarrow \left\{ \begin{array} [c]{cc} \frac{1}{2^{a}} & \text{if }a=\frac{3}{4}.\\ \infty & \text{if }a>\frac{3}{4}. \end{array} \right.$$
• the said quantity does not go to infinity when $\alpha=3/4$. the denominator goes to a non zero constant for every positive $a$. i agree with the first part,though. Mar 19, 2023 at 12:05
• Thank you very much.i would like you to add some details in the inequality you used above for $a<3/4$, it looks like cauchy Schwarz or holder but I am not sure Mar 19, 2023 at 15:33
• @SarveshRavichandranIyer I think he just used some majorizations, like when you for example write $x^2 \leq x^2+y^2$ or when you majorize with the $\ell^2$ norm like $|x| + |y| \leq C\sqrt{x^2+y^2}$ and so on. In the very first passage he uses a trick though, for $(x^4)^{1/4}(y^2)^{1/2}$ is just $xy$. Smooth! Mar 19, 2023 at 17:08