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. Full Proof Verification

Question

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 = $\|f\|_{2,\infty, [0,1]} = sup_{x \in[0,1]}|f(x)| + sup_{x\in[0,1]}|f'(x)| + sup_{x \in [0,1]}|f''(x)|$

I had originally posted this question trying to seek clarification on a particular part, but as I've gone over that solution it didn't make sense to me so I've instead just attempted to reproduce a full solution and get feedback on this from start to finish.

Attempt:

To start I have to examine what is precisely meant by the three conditions given.

Condition 1: $f(x) > 0 \forall \ x \in [0,1]$:

By the Extreme value theorem, there exists a value $x_{1} \in [0,1]$ s.t. $f(x_{1}) = minimum$ over $[0,1]$. By condition $f(x) > 0$, this means there exists a $\delta_{1} > 0 $ s.t $f(x) > f(x_{1}) > \delta_{1} > 0$. In particular this means $f(x_{1}) - \delta_{1} > 0$.

Condition 2: $\|f'(x)\|_{\infty} < 1$:

By the EVT again, there exists an $x_{2} \in [0,1]$ s.t. $f'(x_{2}) = maximum$ over $[0,1]$. By condition $\|f'(x)\|_{\infty} < 1$, there exists a $\delta_{2} >0$ s.t. $\|f'(x)\|_{\infty} = \sup_{x \in [0,1]} |f'(x)| = f'(x_{2}) < 1 - \delta_{2} < 1$.

Condition 3: $|f''(0)| > 2$:

This condition means there exists a $\delta_{3} > 0 $ s.t. $f''(0) > \delta_{3} > 2$. Equivalently one can say: $f''(0) - \delta_{3} > 2$. Also $\sup_{x \in [0,1]}f''(x) - \delta_{3} > f''(0) - \delta_{3} > 2$. (Not sure if this last observations means much.)

Therefore in order to show that $D$ is open (i.e $B_{\delta}(f) \subset D$), I have to establish that $g(x) \in B_{\delta}(f)$. This means $g(x)$ has to satisfy the three observations above.

Proof:

Let $y \in [0,1]$, let $\frac{\delta_{g}}{3} = \frac{min(\delta_{1}, \delta_{2}, \delta_{3})}{3}$

1) Consider $\|g(y) - f(y)\|_{\infty} < \frac{\delta_{g}}{3}$

$\Rightarrow \sup_{y \in [0,1]}|g(y) - f(y)| < \frac{\delta_{g}}{3} \\ \Rightarrow -\frac{\delta_{g}}{3} < g(y) - f(y) < \frac{\delta_{g}}{3} \\ \Rightarrow g(y) \geq f(y) - \frac{\delta_{g}}{3} > f(x_{1}) - \frac{\delta_{g}}{3} \geq f(x_{1}) - \frac{\delta_{1}}{3} > 0 \\ \therefore g(y) > 0$

2) Consider $\|g'(y) - f'(y) \|_{\infty} < \frac{\delta_{g}}{3}$

$\Rightarrow \sup_{y \in [0,1]}|g'(y) - f'(y)| < \frac{\delta_{g}}{3} \\ \Rightarrow -\frac{\delta_{g}}{3} < g'(y) - f'(y) < \frac{\delta_{g}}{3} \\ \Rightarrow g'(y) \leq f'(y) + \frac{\delta_{g}}{3} < f'(x_{2}) + \frac{\delta_{g}}{3} < 1 - \delta_{2} + \frac{\delta_{g}}{3} \leq 1 - \delta_{2} + \frac{\delta_{2}}{3} < 1 - \frac{\delta_{2}}{3} < 1 \\ \therefore g'(y) < 1$

3) Consider $\|g''(0) - f''(0)\|_{\infty} < \frac{\delta_{g}}{3} $

$\Rightarrow \sup_{y \in [0,1]} |g''(0) - f''(0)\| < \frac{\delta_{g}}{3} \\ \Rightarrow -\frac{\delta_{g}}{3} < g''(0) - f''(0) < \frac{\delta_{g}}{3} \\ \Rightarrow g''(0) > f''(0) - \frac{\delta_{g}}{3}$ [Really stuck here]

Comments:

My first concern is whether the first two parts were interpreted in the correct way or am I missing things from my reasoning. As for the third condition, I am utterly loss......Help would definitely be appreciated.

Answer 1

You wrote "This condition means there exists a $\delta_{3} > 0 $ s.t. $f''(0) > \delta_{3} > 2$." You're missing the negative case, e.g., $f''(0)=-3$

I would let $\delta_3 = |f''(0)| - 2.$ Then $|g''(0)-f''(0)|<\delta_3$ implies $ ||g''(0)|-|f''(0)||<\delta_3.$  Thus $|g''(0)| > |f''(0)| - \delta_3 =2.$

Answer 2

Alternatively, notice that $\phi, \psi : C^2[0,1] \to \mathbb{R}$ given by $\phi(f) = \|f'\|_\infty$ and $\psi(f) = |f''(0)|$ are continuous functions w.r.t the given norm on $C^2[0,1]$.

You already more-less established that $\{f \in C^2[0,1] : f(x) > 0, \forall x\in [0,1]\}$ is open so we have \begin{align} D &= \{f \in C^2[0,1] : f(x) > 0, \forall x\in [0,1]\} \cap \{f \in C^2[0,1] : \|f'\|_\infty < 1\} \cap \{f \in C^2[0,1] : |f''(0)| > 2\}\\ &= \{f \in C^2[0,1] : f(x) > 0, \forall x\in [0,1]\} \cap \phi^{-1}\big((-\infty, 1)\big) \cap \psi^{-1}\big((2, \infty)\big) \end{align} which is open as an intersection of three open sets.