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Suppose a function $f$ is defined as follows:

$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$

I want to determine whether or not the function is Fréchet differentiable at $(0,0)$? I don't think that it is.

Here's my reasoning: if $f$ is Fréchet differentiable at $(0,0)$, then from the definition of Fréchet differentiability, the Fréchet derivative must be f itself. But this is not a linear map ($f(1,0)=0$ and $f(0,1)=0$ but $f(1,1) \neq 0$). By contradiction, f is not Fréchet differentiable at $(0,0)$.

Is this correct?

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    $\begingroup$ You don't seem to have the correct definition of Fréchet differentiable in mind, else you wouldn't make such a statement as the Fréchet derivative must be $f$ itself. Go check it out again $\endgroup$ – b00n heT Jun 12 '16 at 16:14
  • $\begingroup$ You are working with a problem in finite dimensional space, what is the difference between Frechet differentiability and the normal derivative in this case? Does the standard derivative exist in this case? $\endgroup$ – David Jun 12 '16 at 16:15
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    $\begingroup$ No. There's no reason that differentiability at the origin should imply that the derivative at the origin must be $f$. The function $f(x,y)=xy$ is not linear but it's certainly differentiable at the origin. $\endgroup$ – David C. Ullrich Jun 12 '16 at 16:15
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    $\begingroup$ Hint for a correct solution: What is $f(t,t)$? $\endgroup$ – David C. Ullrich Jun 12 '16 at 16:17
  • $\begingroup$ I think my mistake is in thinking that if we have Fréchet differentiability then the numerator in the definition of Fréchet differentiability must tend to zero. But I don't see why I'm wrong in thinking that. The denominator in the definition of Fréchet differentiability tends to zero because we are dealing with a norm and since the overall limit of the fraction is zero, surely the numerator must tend to zero too. $\endgroup$ – Pierre Jun 12 '16 at 16:42
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You got all you need at an answer to your earlier question... put in polar,

$$\frac{x^3y^2}{x^4+y^4}= r\,\, \left(\frac{\cos^3(\theta)\sin^2(\theta)}{\cos^4(\theta)+\sin^4(\theta)}\right)$$

This says that your function is continuous, because the quantity in the parentheses has some maximum absolute value which you could find with calculus.

It also says that the function is Gateaux differentiable at the origin, and tells you the directional derivative for any (unit length) direction.

Finally, it says the function cannot be Frechet differentiable, because the directional derivatives along the x,y axes are zero; these are the partial derivatives. If Frechet differentiable, linerarity would demand that the directional derivative in every direction be zero, but that is not the case, just plug in $\theta = \pi / 4$ to get something nonzero. Notice that David C. Ulrich suggested you look at $(t,t),$ same thing.

Extra credit: even if the directional derivatives of some complicated function are linear, for example all zero at some point, that still does not guarantee Frechet differentiability. Explicit bounds are needed for the linear approximation. There are many answers about this distinction, here is mine: Directional derivatives in any direction are all equal with example

Take $$ f(0,0) = 0, \; \mbox{otherwise} \; \; f(x,y) = \frac{x^{12/5} \; y^{6/5}}{x^4 + y^2} $$ If you are uncomfortable with the fractional exponents for numbers that may be negative, switch to $$ f(0,0) = 0, \; \mbox{otherwise} \; \; f(x,y) = \frac{|x|^{12/5} \; |y|^{6/5}}{x^4 + y^2} $$ which is the same.

EDIT:

found one with integer exponents,

Take $$ g(0,0) = 0, \; \mbox{otherwise} \; \; g(x,y) = \frac{x^5 \; y^5}{x^{12} + y^8} $$ Everything looks promising in polar coordinates, but then $$ \frac{|g(t^2, t^3)|}{\sqrt{t^4 + t^6}} \geq \frac{1}{4|t|} $$ when $t \neq 0$ and $|t| \leq \sqrt 3.$

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