Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
    
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 ;)

share|cite|improve this answer

Hint:

  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.
share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.