# Function from $\mathbb {R}$ to $\mathbb{R}^2$ - real analysis problem

Function $$f(x)$$ maps $$(-a,a)$$ to $$\mathbb{R}^2$$ and $$f \in C^1$$ (continously differentiable). Is it possible that image of every open interval $$(-b,b)$$ (for $$b of course) contains neighborhood of $$f(0)$$?

I've tried to figure out this problem, but I have no idea how to do this. I know that inverse function theorem or implicit function theorem can be helpful, anyway I don't see any way in which I could apply these theorems in this problem.

• Let $r$ be the slope of the tangent at $x=0$. Take $\epsilon>0$ so small that the angle between the slopes $r\pm \epsilon$ is less than $\pi/2$. Then, by definition of differentiable there is $c>0$, such that for all $0\leq b<c$ the curve on $(-b,b)$ is between the lines passing through $f(0)$ with slopes $r\pm\epsilon$. The cone covered by these two lines doesn't cover any disc with center $f(0)$. – conditionalMethod Nov 8 '19 at 7:39
• yes it is possible : take $f$ any constant function. – Olivier Roche Nov 8 '19 at 13:16
• @Olivier a constant function from $\Bbb R$ to $\Bbb R ^2$ have a nowhere density image (a singleton), so certainly it cannot contain any open set – Masacroso Nov 8 '19 at 13:33
• @Masacroso Indeed, thanks – Olivier Roche Nov 8 '19 at 14:48

The answer is no. One way to show it is showing that a compact rectifiable curve in $$\Bbb R ^2$$ cannot contain an open disc.
Now note that if $$f|_{[-b,b]}$$ is continuously differentiable then necessarily it image is a rectifiable compact curve in $$\Bbb R ^2$$.