Show $D = \{f \in C^{2}[0,1]): f(x) > 0, \ \forall x \in [0,1], \|f'\|_{\infty}<1, |f''(0)| > 2\}$ is open w.r.t Sup Norm.
Sup Norm = $\|f\|_{2,\infty, [0,1]} = sup_{x \in[0,1]}|f(x)| + sup_{x\in[0,1]}|f'(x)| + sup_{x \in [0,1]}|f''(x)|$
The only condition I was having problems showing was the third one: $|f''(0)| > 2$. How to specifically manipulate the objects.
Attempt:
To start I said let $\delta_{g} = \frac{min(\delta_{1}, \delta_{2}, \delta_{3})}{2}$ Where each $\delta_{i}$ just corresponded to each condition of the set.
To show the condition means I have to show $\forall g \in B_{\delta}(f), \ |g''(0)| > 2$
Given that $[0,1]$ is a compact set, by Intermediate Value THeorem, there exists a $\delta_{3} > 0$ s.t $f''(0)> 2 + \delta_{3} > 2$
Consider $\|g'' - f'' \|_{\infty} < \delta_{g}$
$\Rightarrow |g''(0) - f''(0)| < \delta_{g} \\ \Rightarrow g''(0) > f''(0) - \delta_{g}$
And this is where I'm stuck..... What I would like since I feel it would complete this part would be to somehow show: $g''(0) >\ something\ > 2 + \delta_{g} > 2$
Am I on the right path?
You can complete the proof by noting the following: $|g''(0)| \geq |f''(0)|-|g''(0)-f''(0)|>2$ provided $|g''(0)-f''(0)| < |f''(0)|-2$.
There is a better way of answering this question: the map $f \to (f,\|f'\|_{\infty},f'')$ is a continuous map from $C^{2}[0,1]$ into $\mathbb R^{3}$ and $D$ is the inverse image of the open set $\{(a,b,c) \in \mathbb R^{3}: a>0 ,b<1,c>2\}$.