# Decomposing a Bounded Linear Functional on Lp as a difference of Positive Bounded Linear functionals

I am learning Measure theory via self study of Bartle "The elements of Integration and Lebesgue Measure". I was stumped by the reasoning in one of the decomposition proofs. The point is to show that a Bounded Linear Functional can be represented as the difference of two positive bounded linear functionals. The proof presented is as follows.

Define $G^+ = sup\{G(f) : g \in L_p : 0 \le g \le f\}$ for all $f \ge 0$.

The next step is to ST $G^+$ is a BLF (Bounded Linear Functional).

It is clear that $G^+(cf) = c G^+ (f)$ for $c \ge 0$, $f \ge 0$ (This part was not a problem). The next step attempts to prove that given

$f_j \ge 0, G^(f_1 + f_2) = G^+(f_1) + G^+(f_2)$

This is where I got confused and did not quite understand the line of reasoning. This is how the proof continues:

If $0 \le g_j \le f_j$ Then $G(g_1) + G(g_2) = G(g_1 + g_2) \le G^+(f_1 + f_2)$

Taking supremum over all $g_j \in L_p$ we claim that

$G^+(f_1) + G^+(f_2) \le G^+(f_1 + f_2)$

I would have thought that it would be te other way around, i.e.

$sup\{G(g_1) : g_1 \in L_p, 0 \le g_1 \le f_1 \}$ + $sup\{G(g_2) : g_2 \in L_p, 0 \le g_2 \le f_2 \}$ >= $sup\{G(g_1 + g_2) : g_1, g_2 \in L_p, 0 \le g_j \le f_j\}$

Which implies $G^+(f_1) + G^+(f_2) \ge G^+(f_1 + f_2)$

The book continues: Conversely if $0 \le h \le f_1 + f_2$ let $g_1 = sup(h - f_2,0)$ and $g_2 = inf(h, f_2)$.

It follows that $g_1 + g_2 = h$ and that $0 \le g_j \le f_j$. Threfore

$G(h) = G(g_1) + G(g_2) \le G^+(f_1) + G^+(f_2)$

Since this is true for all $h \in L_p$ we have $G^+(f_1 + f_2) = G^+(f_1) + G^+(f_2)$

My questions were:

1. Why would $G^+$ be positive?

2. How did the final conclusion $G^+(f_1 + f_2) = G^+(f_1) + G^+(f_2)$ for f_1, f_2 in L_p follow?

I would have thougt that the definition for $G^+$ should have been

$G^+ = sup\{G(f) : g \in L_p : 0 \le g \le f, G(g) \ge 0\}$

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I think that the definition of G^+ should be this: $G^+(f)=\sup\{G(g): g \in L_p,\, 0\leq g\leq f\}$, for $f\geq 0$.
Ad 1. I think it is obvious, If $f\geq 0$, then $\mathbb{0} \in \{g\in L_p: 0\leq g \leq f\}$ and $G(0)=0$, so the supremum can't be less than $0$.
Ad 2. Let $$A(f)=\{g\in L_p: 0\leq g \leq f\},$$ for $f\geq 0$. It is obvious (by definition of $G^+$) that $$G(g_1)+G(g_2)=G(g_1+g_2)\leq G^+(f_1+f_2),$$ for $g_j \in A(f_j)$, $j=1,2$ (becouse $g_1+g_2\in A(f_1+f_2)$). So now,at the first we take supremum over all $g_1$ and we obtain: $$\sup\limits_{g_1\in A(f_1)} G(g_1)+G(g_2)\leq G^+(f_1+f_2),$$ next we take supremum over all $g_2$ and we obtain: $$\sup\limits_{g_1\in A(f_1)}G(g_1)+\sup\limits_{g_2\in A(f_2)}G(g_2)=\sup\limits_{g_2\in A(f_2)}(\sup\limits_{g_1\in A(f_1)} G(g_1)+G(g_2))\leq G^+(f_1+f_2),$$
In the proof of the opposite inequality we just use the fact that taking $h \in A(f_1+f_2)$ we can define $g_1 \in A(f_1)$ and $g_2 \in A(f_2)$ (then $G(g_j)\leq G^+(f_j)$, for $j=1,2$) such that $h=g_1+g_2$ and use the linearity of $G$.