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Let $f : \mathbb{R^2} \rightarrow \mathbb{R}$ be a bounded function and $F(x,y) = f(x,y)\sin(x^2 + y^2)$. How can we demonstrate that $F$ is differentiable at $(0,0)$?

I don't how to do this. I already tried, but I don't get it.

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Can you prove that $(x^2+y^2)f(x,y)$ is differentiable at the origin? – Lubin Jan 7 '13 at 17:58
up vote 1 down vote accepted

Let's try approaching along multiples of $v$, i.e. "calculate $dF_0(v)$" :

$$\lim_{t \rightarrow 0} \frac{f(tv)\sin(t^2\|v\|^2)}{t}$$

Now if $v$ is 0, we get $0$ (as always!), if not writing $\| v \|^2$ as just some $c > 0$, we have $$\lim_{t \rightarrow 0} \frac{\sin(ct^2)}{t} = \lim_{t \rightarrow 0} \frac{ \cos(ct^2) 2tc}{1} = 0$$

And now since $f(tv)$ was bounded our original limit is zero too!

To see we're done, well, the 0 map from $\mathbb{R}^2 \rightarrow \mathbb{R}$ is certainly linear ;)

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  1. Show that $F(0,0)=0$.
  2. Show that for all real number $t$, $|\sin t|\leqslant |t|$.
  3. Use this to get that $\partial_xF(0,0)=\partial_yF(0,0)=0$.
  4. Conclude, using the definition of differentiability.
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