The following is from Bruckner's Real Analysis :
In Exercise 12:9.3. :
Part (a) : If we want to make sure that $F(f)$ be common only on the value $1- \ln 2$ and because the lower bound of integration starts at $ \ln 2$ wouldn't that the minimum value for $a$ be $1$, no less?
Part (b) : If to consider $g=t - b$ then $a$ perhaps can be chosen less than $1$ but how the rigorous connection happens, I don't know?
Let $X=\mathcal{C}[0,1]$. Let $h(t)=ae^t$ where $a\geq 0$.
Recall that $A=\{f \in X : \|f\|\leq 1\}$ and $B=\{f \in X : \|f-h\|\leq 1\}$.
Item a. Note that for all $f\in A$, we have $F(f)\leq 1 - \ln 2$. We also have that the constant function $1$ belongs to $A$ and $F(1)=1 - \ln 2$. So, $\sup \{F(f): f\in A\}= 1- \ln 2$.
So, in order that $F$ separates $A$ and $B$, we must have, for all $f \in B$, $F(f) \geq 1- \ln 2$. It is easy to see that this is true if and only if $F(ae^t-1) \geq 1- \ln 2$. So, \begin{align*} F(ae^t-1) & = \int_0^1(ae^t-1) dg = \int_{\ln 2}^1(ae^t-1) dg = \\ & =[ae^t-1]_{\ln 2}^1= (e-2)a-(1 -\ln 2) \geq 1- \ln 2 \end{align*} So, $$ a \geq \frac{2(1 -\ln 2)}{e-2}$$ So $F$ separates $A$ and $B$ if and only if $ a \geq \frac{2(1 -\ln 2)}{e-2}$. So, the smallest value of $a$ for which $F$ separates $A$ and $B$ is $ a = \frac{2(1 -\ln 2)}{e-2}$.
item b: It is easy to see that \begin{align*} a > \frac{2}{e} & \implies ae-1 >1 \implies \|ae^t-1\| >1 \implies \\ & \implies \text{ there is a functional that separates $A$ and $B$. } \end{align*}
In fact, if $a > \frac{2}{e}$, we have $(\ln2 -\ln a) <1$. Define $g_a$ by $g_a(t)= 0$, if $0 \leq t \leq (\ln2 -\ln a)$ and $g_a(t)= t-(\ln 2 - \ln a) $, if $ (\ln2 -\ln a) \leq t \leq 1$. As in example 12.45, we define, for all $f \in X$, $$ F_a(f)=\int_0^1 f dg_a$$ So, we have for all $f \in A$, $$ F_a(f) \leq \int_0^1 1 dg_a= \int_{\ln2 -\ln a}^1 1 dg_a = 1 -(\ln2 -\ln a)$$ Now, if $a > \frac{2}{e}$, we have $(\ln2 -\ln a) <1$ and we also have $ae^t-1 \geq 1$, for $(\ln2 -\ln a) \leq t \leq 1$. So, for all $f \in B$. \begin{align*} F_a(f) & \geq \int_0^1 (ae^t-1)dg_a=\int_{\ln2 -\ln a}^1 (ae^t-1) dg_a \geq \\ & \geq\int_{\ln2 -\ln a}^1 1 dg_a = 1 -(\ln2 -\ln a) \end{align*}
Now note that is $a \leq \frac{2}{e}$, then $ae^t - 1 <1$ for all $t\in [0,1)$ and $ae^t - 1 \leq 1$ for $t=1$. So, $$A \cap B= \{f \in X : \forall t, (ae^t - 1) \leq f(t) \leq 1 \} \ne \emptyset $$
Any functional $F$ that would separate $A$ and $B$ must assign the same value all $f \in A \cap B$. But any functional $F$ is of the form $$ F(f)=\int_0^1 f dg$$ for some $g \in \text{NBV}[0,1]$. So, since $ae^t - 1 <1$ for all $t\in [0,1)$ and $F$ must assign the same value all $f \in A \cap B$, we can easily conclude that $g(t)=0$ for all $t\in [0,1)$ (see Remark below). So $F$ is identically null and does not separate $A$ and $B$.
So we have shown that:
If $a > \frac{2}{e}$, there is a functional that separates $A$ and $B$.
If $a \leq \frac{2}{e}$, there is no functional that separates $A$ and $B$.
So, there is no smallest value of $a$ for which some linear functional separates $A$ and $B$. The infimum of the $a$ for which there are such linear functional is $\frac{2}{e}$.
Item c. If the question were asked for open balls, the answer for item b would not change.
In fact, if the question were asked for open balls, we would have $A_1=\{f \in X : \|f\|<1\}$ and $B_1=\{f \in X : \|f-h\|< 1\}$. It is easy to see that a linear functional $F$ separates open balls $A_1$ and $B_1$ if and only if it separates the corresponding closed balls $A$ and $B$ (the closure of the open balls). So the answer for item b would not change.
It is interesting to consider what happens for $a=\frac{2}{e}$. In this case, $A_1 \cap B_1 =\emptyset$ (because if $f\in A_1$, then $f(1) <1$ and if $f\in B_1$, $f(1)> ae-1= 2-1=1$ ). We could then think that there is a functional that separates $A_1$ and $B_1$. However, there is no such functional, because, as said above, a linear functional $F$ separates open balls $A_1$ and $B_1$ if and only if it separates the corresponding closed balls $A$ and $B$ (the closure of the open balls).
$\square$
Remark: Let us prove in detail that if $F$ must assign the same value all $f \in A \cap B$, then $g(t)=0$ for all $t\in [0,1)$.
Note that, by Riesz Representation Theorem, $g \in \text{NBV}[0,1]$, so $g$ is a bounded variation function, $g$ is right continuous and $g(0)=0$.
Now, for a contradiction, suppose there is a $t_0 \in (0,1)]$ such that $g(t_0) \ne 0$. Without loss of generality, suppose $g(t_0) > 0$. Since $g$ is right continuous, there is $t_1 >t_0$ such that $g(t)>0$ for $t \in (t_0,t_1)$. Let $\varepsilon <\frac{t_1-t_0}{2}$ and let $f_0 \in \mathcal{C}[0,1]$ such that
$f_0(t) \in [0, 2-ae^{t_1} ]$, for all $t\in [0,1]$
$f_0(t) = 0$ for $t \in [0,t_0] \cup [t_1, 1]$
$f_0(t)=2-ae^{t_1}$ for $t\in [t_0+\varepsilon , t_1-\varepsilon]$
Clearly $$F(f_0)= \int_0^1 f_0(t)dg > 0$$ But, it is easy to see that, for all $t \in [0, 1]$, $(ae^t - 1) \leq 1- f_0(t) \leq 1$. So, $1- f_0 \in A \cap B$. Since the constant function $1$ is also in $A\cap B$, we have $F( 1- f_0(t))= F(1)$. So, $$ F(f_0)= F(1 - (1- f_0(t))) = F(1) - F( 1- f_0(t)) =0$$ Contradiction. So we have $g(t)=0$ for all $t\in [0,1)$.