C*-algebra inequality closed under pertubation?

Question

Let $A$ be a C*-algebra. Let $a \in A_+$, the positive cone of $A$. Let $f,g \geq 0$ be real functions with compact supports contained in $(0,\infty)$. Further, assume that $g(x) = f(x) + 1$ for all $x$ in the support of $f$. So, we have something like this:

Now, from functional calculus, $f(a) \leq g(a)$ and in fact any positive $b$ in the C*-algebra generated by $a$ that is close to $g(a)$ will be greater than $f(a)$. What I'm unable to figure out is whether this holds, more generally, in all of $A$.

Question: Is there an $\epsilon > 0$ such that $b \in A_+$, and $\|b-g(a)\| < \epsilon$ implies $f(a) \leq b$?

Added: note the crucial assumption that $b \geq 0$

Roughly does the controlled gap between $f$ and $g$ imply that $g(a)$ is "substantially" bigger than $f(a)$ in the above precise sense.

Answer 1

This is not true. Take $f$ such that $f(1) = 1$. For $t \geq 0$ let $p_t$ denote the rank 1 projection onto the span of $e_1 + t e_2 \in \mathbb C^2$, then $f(p_0) = p_0$, $g(p_0) = 2 p_0$, and $\lim_{t \to 0} \| 2 p_t - 2 p_0 \| = 0$. However, for all $t > 0$, $p_0 \not\leq 2p_t$ since $\ker(2p_t) \not\subset \ker(p_0)$.

Answer 2

Yes.

Let $\textbf{1}$ be the identity, adjoin one if necesarry. You norm condition is equivalent to $-\epsilon\textbf{1}< (b-g(a))$

$-\epsilon\textbf{1}< (b-g(a))\Rightarrow g(a)-\epsilon\textbf{1}<b$ So choose an $\epsilon$ such that $f(a)< g(a)-\epsilon\textbf{1}$.