0
$\begingroup$

I have two question concerning continuity and differentiability, because something is unclear to me.

Let’s say I have a function : $f:\mathbb{R} \rightarrow \mathbb{R}$ (this function is not necessarily continuous), and now let’s defined the function $F$ as follow :

$$\forall (r,t) \in \mathbb{R}^+\times \mathbb{R} F(r,t)=(r \cos f(t), r \sin f(t))$$

Now my questions are :

Is $F$ continuous on a neighborhood of $(0,0)$ even-if $f$ is not continuous ?

According to me it’s most likely true because : $$\lim_(r,t \rightarrow (0,0)) F(r,t) = 0$$

If it wasn’t continuous on a neighborhood of $(0,0)$ we can’t talk about the limit of such function at the point $(0,0)$, so the function is continuous at $(0,0)$.

Now I would like to know why I can’t say that $F$ is différentiable at the point $(0,0)$ using the following argument :

If $F$ is differentiable at the point $(0,0)$ then there is a function $u$ such that :

$$F(h,h) = F(0,0) + u(0)(h) + o(||h||)$$

So by taking the limit as $h \rightarrow 0$ we have : $$u(0)(h) = 0 $$

And hence $F$ is differentiable at the point $(0,0)$, and $\mathrm{d}F(0) = x \mapsto 0$.

Yet this argument is actually incorrect, but why ?

Thank you, for taking your time !

$\endgroup$
  • $\begingroup$ You haven't proved that such an $u$ exists. And the condition you get for $u$ is false. Taking the limit you get the identity $F(0,0)=F(0,0)$. By the way, $h$ can have different components $h_1$ and $h_2$. $\endgroup$ – Dog_69 Jun 18 '18 at 23:50
0
$\begingroup$

We have that $F$ is continuous, because $\sin$ and $\cos$ are bounded and $r\to 0$. The function $f$ has no say in what happens here. But if ${\rm D}F(0,0)$ is the zero map, we would have to check that $$\lim_{(r,t)\to (0,0)} \frac{F(r,t)}{\sqrt{r^2+t^2}} =(0,0),$$since $F(0,0)=(0,0)$. This is equivalent to $$\lim_{(r,t)\to (0,0)} \frac{r}{\sqrt{r^2+t^2}}\cos f(t) =\lim_{(r,t)\to (0,0)} \frac{r}{\sqrt{r^2+t^2}}\sin f(t) =0. $$ Although $r/\sqrt{r^2+t^2}$ is bounded, the terms $\cos f(t)$ and $\sin f(t)$ need not go to zero (take $f=1$ or whatever).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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