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Consider $(\mathcal C^1([0,1]),\|\cdot \|)$ where $$\|f \|=\|f\|_\infty +\|f'\|_\infty $$ where $\|g\|_\infty :=\sup_{[0,1]}|g|.$ Let $$F=\{g\in \mathcal C^1([0,1])\mid \exists x\in [0,1], (f(x),f'(x))=(0,0)\}.$$

1) Is $F$ open or close in $\mathcal C^1([0,1])$ ?

2) Find the interior or $F$.

I'm sorry, but I have no idea how to solve this exercice. I really have problem to visualize such spaces. I know that $$B_\varepsilon(f)=\{g\in\mathcal C^1([0,1])\mid \|f-g\|<\varepsilon\},$$ So for 2) I have to finde all $f\in\mathcal C^1([0,1])$ s.t. there is an $\varepsilon>0$ s.t. $B_\varepsilon(f)\subset D$, but how ?

For 1) I have no idea.

idm
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  • You should probably unify the use of $f,g$ in the definition of $F$. – PhoemueX Oct 25 '15 at 12:46
  • for 1) and the question of whether $F$ is closed consider $f_n\in F$ converging to $f\in C^1$. Then try to show $f\in F$. If this is true, $F$ is closed. If you know how to answer 2) you will probably also know whether $F$ is open... – Thomas Oct 25 '15 at 12:46
  • To expand on the comment of Thomas: Since $f_n \in F$, you get $x_n \in [0,1]$ with $f_n (x_n) = 0 = f_n ' (x_n)$. Now you are looking for a point $x$ with $f(x) = 0=f'(x)$. How could you get such a point? – PhoemueX Oct 25 '15 at 12:47
  • Note that $f(x) =0$ (the constant function) is in $F$, but $f(x) = c$ for $c\ne 0$ is not. This shows that $f= 0$ is not an interior point...then consider $f$ such that $f(0) < 0 , f(1)>0$ and the same inequalities hold for $f^\prime$. By the mean value theorem, this function will be in $F$. Clearly also any small perturbation will satisfy these inequalities, so such an $F$ is an interior point. Enough hints.... – Thomas Oct 25 '15 at 12:49

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