# Prove that if $|f(x,y)|\leq x^2+y^2$ for all $(x,y) \in \mathbb{R}^2$ then $f$ is differentiable at $(0,0)$

My attempt:

We want

$$\lim\limits_{(x,y)\to(0,0)} \frac{|f(x,y)-L_{(0,0)}(x,y)|}{\sqrt{x^2+y^2}}=0 \tag1$$

Setting up for a squeeze/sandwich theorem we have

$$\left|\frac{|f(x,y)-L_{(0,0)}(x,y)|}{\sqrt{x^2+y^2}}-0\right|$$

By triangle inequality:

$$\leq \frac{|f(x,y)|+ |L_{(0,0)}(x,y)|}{\sqrt{x^2+y^2}}$$

Also from our assumption $f(0,0)\leq0^2+0^2=0$

$$\leq \frac{0+|L_{(0,0)}(x,y)|}{\sqrt{x^2+y^2}}$$

Expanding linearization and using our assumption again:

$$=\frac{|f(0,0)+f_x(0,0)x+f_y(0,0)y|}{\sqrt{x^2+y^2}}\leq \frac{|0+ f_x(0,0)x+f_y(0,0)y|}{\sqrt{x^2+y^2}}$$

This is where I'm stuck, how do I proceed further to get a limit $\to$ 0?
Or is there a better approach to this problem?

EDIT: tried to find partial derivatives using definition
(since the limit above would hold true if partials at $(0,0)$ is $0$)

Claim:

$$f_x(0,0) = \lim\limits_{h\to0} \frac{f(0+h,0)-f(0,0)}{h} = 0$$

Proof: $$\left|\frac{f(0+h,0)-f(0,0)}{h}-0\right|\leq\frac{|f(0+h,0)|+|f(0,0)|}{|h|}\leq\frac{|h^2|+|0|}{|h|}=|h|\to0$$

Hence by squeeze (and symmetry) both $f_x(0,0) = f_y(0,0) = 0$

Therefore, equation the limit $(1)$ goes to $0$ and thus $f$ is differentiable at $(0,0)$

Would this be a complete answer?

This is all very laboured. A function $f:\Bbb R^2\to\Bbb R$ is differentiable at $(0,0)$ with derivative $a$ if $f(v)-f(0,0)=a\cdot v+o(|v|)$ as $v\to0$ where $v$ is a vector, $a\cdot v$ is the dot product of $a$ and $v$, and $|v|$ is the Euclidean length of $v$. Here $$f(v)=(0,0)\cdot v+O(|v|^2)$$ so $f$ if differentiable with derivative $(0,0)$ at the origin.

• Sorry but I'm not familiar with $O(|v|^2)$ what does it mean? – bolt997 Oct 15 '17 at 6:41
• – Angina Seng Oct 15 '17 at 6:47

$|f_x(0,0)|= \left|\displaystyle \lim_{h \to 0}\dfrac{f(h,0) - f(0,0)}{h}\right|=\displaystyle \lim_{ h \to 0}\left|\dfrac{f(h,0)}{h}\right|\le\displaystyle \lim_{h \to 0} \dfrac{h^2+0^2}{|h|}=0$,and the other partial at $(0,0)$ is also $0$, thus the numerator is $0$, hence the fraction is $0$.

Note: I simply continue what you left off, and to complete the proof you would use all the work you started.

• so? the existence of partials is not enough – qbert Oct 15 '17 at 6:33
• Thanks!! It took a while for me to type my edit but it seems I reached the same conclusion as you did. – bolt997 Oct 15 '17 at 6:39

As you showed in your question, $f(0,0)=0$, and $$\lim_{(x,y)\rightarrow(0,0)} \frac{f(x,y) - f(0,0)}{\sqrt{x^2 +y^2}}=\lim_{(x,y)\rightarrow(0,0)} \frac{f(x,y) - 0}{\sqrt{x^2 +y^2}} = \lim_{(x,y)\rightarrow(0,0)} \frac{f(x,y)}{\sqrt{x^2 +y^2}}.$$

So it suffices to show the limit on the right exists.

Claim: $\lim_{(x,y)\rightarrow(0,0)} \frac{f(x,y)}{\sqrt{x^2 +y^2}}= 0$.

Proof: Observe that

$$\left|\frac{f(x,y)}{\sqrt{x^2 +y^2}}\right| \leq \frac{x^2+y^2}{\sqrt{x^2 +y^2}} = \sqrt{x^2+y^2}.$$

Hence $\frac{f(x,y)}{\sqrt{x^2 +y^2}}\rightarrow 0$ as $(x,y)\rightarrow 0$.

• Oh right, I could've deleted the linearization instead of the function... – bolt997 Oct 15 '17 at 6:42