Let $X$ be a compact Hausdorff space and $E$ be a Banach space, and consider the Banach space $C(X,E)$ with the sup norm with respect to the norm in $E$. We denote the dual of a Banach space $A$ by $A'$. It is known that the dual of $C(X,E)$ is the set of $E'$- valued regular Borel measures on $X$ with the total variation norm given by
$\|F\|=\text{sup}\{\sum_{i}\|F(X_{i})\|:\{X_{i}\} \text{ partition of }X\}$.
Fix $x\in X$ and consider the map $P:C(X,E)'\rightarrow C(X,E)'$ given by
$P(F)=F\restriction{\{x\}}=\delta(x)\otimes F(\{x\})$, i.e.
$P(F)f=F(\{x\})f(x)$.
Then $P$ is a norm-one projection on $C(X,E)'$.
The following are to be shown:
$\|F\|=\|F\restriction{\{x\}}\|+\|F-F\restriction{\{x\}}\| \, \forall F\in C(X,E)'$ and
$\|\Lambda\|=\text{max }\{\|P^{*}(\Lambda)\|, \|\Lambda-P^{*}(\Lambda)\|\} \, \forall \Lambda\in C(X,E)''$.
I'd be grateful for a hint on how to show this.
Does the first equality follow simply because the total variation is a measure?
In the second equality, it is clear that $\|P^{*}(\Lambda)\|\leq \|\Lambda\|$ and $\|\Lambda-P^{*}(\Lambda)\|\leq \|\Lambda\|$, but how do we show that $\|\Lambda\|$ is in fact the maximum of the two?
Thanks in advance.