Confusion of the proof of the Jordan decomposition theorem of bounded linear functional in Folland's Book. I am reading Folland Lemma 7.15, the so-called Jordan Decomposition Theorem of bounded linear functional, stated as

For $I\in C([0,1],\mathbb{R})^{*}$, there exists positive linear functional $I^{\pm}\in C_{0}([0,1],\mathbb{R})^{*}$ such that $I=I^{+}-I^{-}$.

Folland used a more general space $X$ that is locally compact Hausdorff, but it then involves some $C_{0}-$space, which is uniform closure of the space of all continuous function with compact support. Let us just use $X=[0,1]$ here for convenience.
Most of the proof is pretty straightforward, but I got stuck in several places, where Folland claims to be obvious.

Firstly, for $f\in C([0,1],[0,\infty)$), he defines $$I^{+}(f)=\sup\{I(g):g\in C([0,1],\mathbb{R}), 0\leq g\leq f\}.$$ I know how to show that $I^{+}(f_{1}+f_{2})=I^{+}(f_{1})+I^{+}(f_{2})$, but I do not see how to show $I^{+}(cf)=cI^{+}(f)$ for $c\geq 0$. The supremum of $I(g)$ overall the choices of $0\leq g\leq cf$ is the same as $c$ multiplying the supremum over all the choices of $0\leq g\leq f$? Why?

Secondly, for $f\in C([0,1], \mathbb{R})$, he then defines $$I^{+}(f):=I^{+}(f^{+})-I^{+}(f^{-}).$$ Again, I understand the closure under addition, but I don't know how to prove $I^{+}(cf)=cI^{+}(f)$ for all $c\in\mathbb{R}$. The case of $c\geq 0$ can follow from the first case, but what happens if we have $c<0$?

Finally, he defines $I^{-}:=I^{+}-I$ and says that by definition of $I^{+}$, it is clear $I^{-}$ is positive. However, I did not see this clearly.. Also, does this definition imply $I^{-}(f)=I^{+}(-f)$? why?
Thank you so much!
 A: For the first point, let
$$S = \{I^+(g) \mid 0 \leq g \leq f\},$$
and
$$T = \{I^+(h) \mid 0 \leq h \leq cf\}.$$
To show that $c \sup S = \sup T,$ I'll show that in fact $c \cdot S = T$ for $c > 0.$ (For $c=0,$ things are easy enough to prove directly...)
Imagine you have a term $I^+(g) \in S,$ corresponding to some $g$ with $0 \leq g \leq f.$ Define $h = cg.$ Then $0 \leq h \leq cf.$ So, $I^+(h) = I^+(cg) = cI^+(g) \in T.$ Thus $c\cdot S \subseteq T.$ Similarly, given $I^+(h) \in T,$ you can recover a corresponding $g$ by dividing by $c$ (here we use $c > 0$).
This is a relatively common trick. If you're comparing objects $0\leq g \leq f$ and $0\leq g \leq cf,$ first give the two different $g$'s different names, and then think of how to relate them. The only obvious relation is multiplication by $c.$
For the second point, work as follows for $c < 0.$ Set $g = cf.$ How do $g^+, g^-$ relate to $f^+, f^-$? Since $c$ is negative, the signs change! So, if $x$ is so that $f(x) > 0,$ then $g(x) < 0$ must follow. Therefore we find that $g^- = -cf^+$ and $g^+ = -cf^-.$ $-c$ is a positive scalar. So, $$I^+(g) = I^+(-cf^-) - I^+(-cf^+) = -cI^+(f^-) + c I^+(f^+) = cI^+(f).$$
Alternatively, you could just do
$$0 =I^+(0) = I^+(cf - cf) = I^+(cf) + I^+(-cf) = I^+(cf) - cI^+(f)$$
since $-c > 0.$
Now, your third point. Suppose $f \geq 0.$ Then (0\leq f \leq f,) so $I^+(f) \geq I(f)$ by definition of supremum. Thus $I^+f - If \geq 0.$
Now, your negation point. This does not need to hold in general. Suppose it did; then always we'd find $I^-(f) = -I^+(f).$ But then $If = I^+f - I^-f = 2I^+f = I^+(2f).$ But this cannot happen for non-negative functions!
A: For the first question, the equality is clear when $c = 0$. Suppose $c\neq 0$, then for $0\leq g\leq cf$, $I(g)\in\{cI(g):0\leq g\leq f\}$. So, $I^+(cf)\leq cI^+(f)$. Conversely, when $0\leq g\leq f, cI(g) = I(cg)\leq I^+(cf)$, so $cI^+(f)\leq I^+(cf)$.
As you have said, for a general $f, I^+(cf) = cI^+(f)$ when $c\geq 0$ from the first part. When $c<0$, we have $cf = (-c)(-f)$ and $(-f)^+ = f^-, (-f)^- = f^+$ therefore,
$$I^+(cf) = I^+((-c)f^-)-I^+((-c)f^+)\\
=(-c)(I^+(f^-)-I^+(f^+)).$$
For your third question, observe that when $f$ is nonnegative, $f\leq f$ so $I(f)\leq I^+(f)$. I don't think $I^-(f) = I^+(-f)$ holds, but I'm not sure about that.
