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Prove that if $f$ satisfies $|f(x,y)| \leq (x^2 + y^2)$ then $f$ is differentiable at $(0,0)$.

I understand how to prove this: one can deduce that $f(0,0)=0$, and then we can assume that $L_{(0,0)} (x,y)=0$, then prove that $\lim_{(x,y) \to (0,0)} \frac{R_{1,(0,0)}(x,y)}{\sqrt{x^2+y^2}} =0$.

However, how do we deduce that $f_x(0,0)$ and $f_y(0,0)$ both equal zero? Sorry if this seems like a silly question.

sequence
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2 Answers2

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A couple of conceptual points first:

We can't assume that the derivative at the origin is the zero map, as you imply. Rather if you can show the derivative $L_{(0,0)}$ is the zero map, then you have calculated the derivative. It is certainly true however that $f(0,0) = 0$, as $|f(0,0)| \leq 0^2 + 0^2$.

Also, having shown the derivative is the zero map, it follows immediately that the partial derivatives at the origin are also zero, because if the derivative exists at a point, then the partial derivatives are equal to the entries of the matrix that defines the derivative as a linear map.

So to show that $L_{(0,0)} = \bf 0$, by the definition of the derivative, you need to show that

$$\lim_{(x,y)\to(0,0)} \frac{\left|\,f(x,y) - f(0,0) - L_{(0,0)}((x,y)-(0,0))\right|}{||(x,y) - (0,0)||} = 0$$

As $f(0,0) = 0$ and by hypothesis $L_{(0,0)} = \bf 0$, establishing this limit is equivalent to showing

$$\lim_{(x,y)\to(0,0)} \frac{|f(x,y)|}{||(x,y)||} = 0$$

This last result is true. Can you take it from here?

Simon S
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  • Yes, I've just edited my question - I meant "... differentiable at (0,0)". – sequence Jan 30 '15 at 00:45
  • I think I have a gap in knowledge. I'm not sure what the zero map is, and I haven't seen the matrix that defines derivatives as a linear map. – sequence Jan 30 '15 at 00:49
  • The derivative of a function $\mathbb R^2 \to \mathbb R$ at a point $p$ is a linear map from $\mathbb R^2 \to \mathbb R$ considered as vector spaces. If the derivative of a function $f : \mathbb R^2 \to \mathbb R$ exists at some point $p$, write it as $L_p$. Then in the standard basis, $L_p = (f_x \ f_y)$ by which I mean applying $L_p$ to a vector $v = (a,b) \in \mathbb R^2, L_p(v) = af_x + bf_y$. In other words, the derivative gives the partial derivatives. In any case, the way to find the derivative here isn't to find the partial derivatives, but to show $L$ – Simon S Jan 30 '15 at 00:54
  • ...has the property I gave above. That's the definition of the derivative. If $L_{(0,0)}$ is the zero map, then $L_{(0,0)}(x,y) = \bf 0$ for all points $(x,y)$. That's why in the second limit that term disappears. – Simon S Jan 30 '15 at 00:54
  • But derivatives at $(0,0)$ can be other than $0$. I think I'm misunderstanding something. – sequence Jan 30 '15 at 03:02
  • In general, yes, the derivative could be any linear function $\mathbb R^2 \to \mathbb R$. But here the derivative is zero. – Simon S Jan 30 '15 at 04:12
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Hint: (based on one of the answers below): $0\leq \dfrac{|f(x,y)|}{||(x,y)||} \leq \sqrt{x^2+y^2}$

DeepSea
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