Open sets in a particular norm. Let $f \in C^2[0,1]$, define $||f||_{2,\infty} := \underset{x \in [0,1]} {\text{sup}} |f(x)|+ \underset{x \in [0,1]} {\text{sup}}|f'(x)|+ \underset{x \in [0,1]} {\text{sup}}|f''(x)|$. Then I wish to check whether the following sets are open w.r.t. this norm.
(a) $A = \{f \in C^2[0,1]\ :\ f(1)<0, f'(1) = 0, f''(1)>0  \}.$ 
(b) $B = \{f \in C^2[0,1]\ :\ f(x)\cdot f'(x) > 0\ \forall\ x \in [0,1]\}.$
What I did was to use the definition of an open set in a metric space to see that if these sets are open but somehow I cannot derive the conditions using the definition of open sets for part (a). So, I guess that the set $A$ is not open. However, I cannot find some $C^2[0,1]$ functions such that $\epsilon$ neighbourhood of it lies outside $A$. I also tried to use Intermediate and extreme value theorems but can conclude only that $f(1)$ is a local minimum for $f \in A$. 
Any hints would be nice. Thanks.  
 A: The $A$ is not open. If $f \in A$ then $f(x)+\epsilon x$ is not in this set for any $\epsilon >0$. Since $f(x)+\epsilon x \to f$ in the given  norm as $\epsilon \to 0$ it follow that no open ball around $f$ can be contained in $A$.
$B$ is open because $f \to \inf \{f(x)f'(x): 0 \leq x \leq 1\}$ is a continuous function and $B$ is the inverse of $(0,\infty)$ under this function. [I am using the fact that the infimum here is attained at some point. I have also used the fact that if $g_n \to g$ uniformly then $\inf g_n \to \inf g$]. 
A: $B$ is open. Assume $f \in B$. Notice that $f'f > 0$ is continuous on $[0,1]$ so it attains a minimum $m > 0$. We claim that for $\|g-f\| < \varepsilon$ we have $g \in B$ where $\varepsilon = \sqrt{\|f\|^2+m}-\|f\|$.
We have $\|g\| < \|f\|+\varepsilon$ so
$$|g'(x)g(x)-f'(x)f(x)| \le |g'(x)-f'(x)||g(x)|+|g(x) - f(x)||f'(x)| < \|g-f\|(2\|f\|+\varepsilon) < \varepsilon(2\|f\|+\varepsilon)$$
so $$|g'(x)g(x)| > |f'(x)f(x)|-\varepsilon(2\|f\|+\varepsilon) \ge m-\left(\sqrt{\|f\|^2+m}-\|f\|\right)\left(\sqrt{\|f\|^2+m}+\|f\|\right) = 0$$
