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 !

  • $\begingroup$ 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

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).


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