[proofplan] We treat the four parts in order. (i) is a real-imaginary splitting via the conjugate functional $\bar\varphi(f) := \overline{\varphi(\bar f)}$. (ii) reduces the complex norm of a real functional to its real norm by an explicit polar argument: pick a unit-modulus rotation $\lambda$ that sends $\varphi(f)$ to the positive real axis and exploit that $\varphi$ kills imaginary parts of real-rotated test functions. (iii) characterises positivity via the norm-attaining condition at the unit element; the converse uses an imaginary perturbation $f + it\mathbb{1}_K$ that detects any non-real value of $\varphi$. (iv) constructs the positive part by the lattice-style Riesz-type sup formula $\varphi^+(f) := \sup\{\varphi(g) : 0 \le g \le f\}$ on $C_+(K)$, verifies positive homogeneity and additivity via a splitting lemma using the lattice operation $g \wedge f_1$ in $C(K)$, and extends to all of $C(K)$ by linearity. [/proofplan]

[step:Construct the conjugate functional and split into real and imaginary parts] Define \begin{align*} \bar\varphi: C(K) &\to \mathbb{C} \\ f &\mapsto \overline{\varphi(\bar f)}. \end{align*} Linearity: for $f, g \in C(K)$ and $\alpha, \beta \in \mathbb{C}$, \begin{align*} \bar\varphi(\alpha f + \beta g) = \overline{\varphi(\bar\alpha \bar f + \bar\beta \bar g)} = \overline{\bar\alpha \varphi(\bar f) + \bar\beta \varphi(\bar g)} = \alpha\, \overline{\varphi(\bar f)} + \beta\, \overline{\varphi(\bar g)} = \alpha \bar\varphi(f) + \beta \bar\varphi(g). \end{align*} Boundedness: $|\bar\varphi(f)| = |\varphi(\bar f)| \le \|\varphi\| \|\bar f\|_\infty = \|\varphi\| \|f\|_\infty$, and applying the same to $\bar\varphi$ gives $\|\bar\varphi\| = \|\varphi\|$. So $\bar\varphi \in M(K)$. Set \begin{align*} \varphi_1 := \frac{\varphi + \bar\varphi}{2}, \qquad \varphi_2 := \frac{\varphi - \bar\varphi}{2i}. \end{align*} Then $\varphi_1, \varphi_2 \in M(K)$ as linear combinations of bounded linear functionals, and $\varphi = \varphi_1 + i\varphi_2$ by direct computation. To verify $\varphi_j \in M_\mathbb{R}(K)$ (i.e. $\varphi_j(f) \in \mathbb{R}$ for all $f \in C_\mathbb{R}(K)$): if $f = \bar f$, then \begin{align*} \bar\varphi_1(f) = \overline{\varphi_1(\bar f)} = \overline{\varphi_1(f)}, \end{align*} and $\bar\varphi_1 = \tfrac{1}{2}(\bar\varphi + \overline{\bar\varphi}) = \tfrac{1}{2}(\bar\varphi + \varphi) = \varphi_1$ (the conjugate of $\bar\varphi$ at $f$ is $\overline{\bar\varphi(\bar f)} = \overline{\overline{\varphi(f)}} = \varphi(f)$). So $\varphi_1(f) = \overline{\varphi_1(f)}$, hence real. The same argument applies to $\varphi_2$. Uniqueness: suppose $\varphi = \psi_1 + i\psi_2$ with $\psi_j \in M_\mathbb{R}(K)$. Then $\bar\varphi = \psi_1 - i\psi_2$ (since real-valued functionals satisfy $\bar\psi = \psi$, as the same calculation above shows). Solving the linear system gives $\psi_1 = (\varphi + \bar\varphi)/2 = \varphi_1$ and $\psi_2 = (\varphi - \bar\varphi)/(2i) = \varphi_2$. [/step]

[step:Prove the restriction $M_\mathbb{R}(K) \to C_\mathbb{R}(K)^*$ is isometric] Fix $\varphi \in M_\mathbb{R}(K)$. The restriction $\varphi|_{C_\mathbb{R}(K)}$ is real-linear and bounded with $\|\varphi|_{C_\mathbb{R}(K)}\| \le \|\varphi\|$ (the supremum is taken over a smaller set). We prove the reverse inequality. Let $f \in C(K)$ with $\|f\|_\infty \le 1$. Choose $\lambda \in \mathbb{C}$ with $|\lambda| = 1$ and $\lambda \varphi(f) = |\varphi(f)|$ (any unit-modulus complex number that rotates $\varphi(f)$ to the non-negative real axis; if $\varphi(f) = 0$, take $\lambda = 1$). Then $\lambda f \in C(K)$ with $\|\lambda f\|_\infty = \|f\|_\infty \le 1$, and \begin{align*} |\varphi(f)| = \lambda \varphi(f) = \varphi(\lambda f) = \varphi(\operatorname{Re}(\lambda f)) + i \varphi(\operatorname{Im}(\lambda f)), \end{align*} where $\operatorname{Re}(\lambda f), \operatorname{Im}(\lambda f) \in C_\mathbb{R}(K)$. The left-hand side is real and non-negative, and the second term on the right is $i$ times the real number $\varphi(\operatorname{Im}(\lambda f))$ (real because $\varphi \in M_\mathbb{R}(K)$ and $\operatorname{Im}(\lambda f) \in C_\mathbb{R}(K)$). Equating imaginary parts: $\varphi(\operatorname{Im}(\lambda f)) = 0$. Therefore \begin{align*} |\varphi(f)| = \varphi(\operatorname{Re}(\lambda f)) \le \|\varphi|_{C_\mathbb{R}(K)}\| \cdot \|\operatorname{Re}(\lambda f)\|_\infty \le \|\varphi|_{C_\mathbb{R}(K)}\| \cdot \|f\|_\infty, \end{align*} using $|\operatorname{Re}(z)| \le |z|$ pointwise. Taking the supremum over $\|f\|_\infty \le 1$ gives $\|\varphi\| \le \|\varphi|_{C_\mathbb{R}(K)}\|$, hence equality. Surjectivity: given $\psi \in C_\mathbb{R}(K)^*$ (a bounded $\mathbb{R}$-linear functional), define \begin{align*} \varphi: C(K) &\to \mathbb{C} \\ f &\mapsto \psi(\operatorname{Re} f) + i \psi(\operatorname{Im} f). \end{align*} Then $\varphi$ is $\mathbb{C}$-linear: additivity is immediate, and for $\alpha = a + ib \in \mathbb{C}$ and $f \in C(K)$, $\operatorname{Re}(\alpha f) = a\operatorname{Re} f - b \operatorname{Im} f$ and $\operatorname{Im}(\alpha f) = a \operatorname{Im} f + b \operatorname{Re} f$, so \begin{align*} \varphi(\alpha f) &= \psi(a \operatorname{Re} f - b \operatorname{Im} f) + i \psi(a \operatorname{Im} f + b \operatorname{Re} f) \\ &= a \psi(\operatorname{Re} f) - b \psi(\operatorname{Im} f) + i a \psi(\operatorname{Im} f) + i b \psi(\operatorname{Re} f) \\ &= (a + ib)(\psi(\operatorname{Re} f) + i \psi(\operatorname{Im} f)) = \alpha \varphi(f). \end{align*} Boundedness: $|\varphi(f)| \le |\psi(\operatorname{Re} f)| + |\psi(\operatorname{Im} f)| \le 2\|\psi\| \|f\|_\infty$, so $\varphi \in M(K)$. By construction $\varphi(f) = \psi(f)$ for $f \in C_\mathbb{R}(K)$ (then $\operatorname{Im} f = 0$), so $\varphi|_{C_\mathbb{R}(K)} = \psi$. Finally $\varphi \in M_\mathbb{R}(K)$ because $\varphi(f) = \psi(\operatorname{Re} f) \in \mathbb{R}$ for $f \in C_\mathbb{R}(K)$. [/step]

[step:Characterise positivity via the norm-attaining condition at $\mathbb{1}_K$]We prove $M_+(K) = \{\varphi \in M(K) : \|\varphi\| = \varphi(\mathbb{1}_K)\}$. ($\subseteq$). Let $\varphi \in M_+(K)$ (so $\varphi$ is real-linear, positive: $\varphi(f) \ge 0$ for $f \in C(K)$ with $f \ge 0$; in particular $\varphi \in M_\mathbb{R}(K)$). For $f \in C_\mathbb{R}(K)$ with $\|f\|_\infty \le 1$, the inequalities $-\mathbb{1}_K \le f \le \mathbb{1}_K$ hold pointwise, so $\mathbb{1}_K - f \ge 0$ and $\mathbb{1}_K + f \ge 0$. Positivity gives $\varphi(\mathbb{1}_K) - \varphi(f) \ge 0$ and $\varphi(\mathbb{1}_K) + \varphi(f) \ge 0$, hence $|\varphi(f)| \le \varphi(\mathbb{1}_K)$. Combined with the part (ii) isometry, this gives $\|\varphi\| = \|\varphi|_{C_\mathbb{R}(K)}\| \le \varphi(\mathbb{1}_K)$. Since $\|\mathbb{1}_K\|_\infty = 1$ and $\varphi(\mathbb{1}_K) \in \mathbb{R}$ with $\varphi(\mathbb{1}_K) \ge 0$, the supremum is attained: $\|\varphi\| \ge |\varphi(\mathbb{1}_K)| = \varphi(\mathbb{1}_K)$. Hence $\|\varphi\| = \varphi(\mathbb{1}_K)$. ($\supseteq$). Let $\varphi \in M(K)$ with $\|\varphi\| = \varphi(\mathbb{1}_K)$. The right-hand side is real, so $\varphi(\mathbb{1}_K) \in \mathbb{R}$, and $\|\varphi\| \ge 0$ forces $\varphi(\mathbb{1}_K) \ge 0$. We may assume $\|\varphi\| > 0$, else $\varphi \equiv 0 \in M_+(K)$. Normalising, replace $\varphi$ by $\varphi / \varphi(\mathbb{1}_K)$ to assume $\|\varphi\| = \varphi(\mathbb{1}_K) = 1$. We first show $\varphi \in M_\mathbb{R}(K)$. Fix $f \in C_\mathbb{R}(K)$ and write $\varphi(f) = a + ib$ with $a, b \in \mathbb{R}$. We show $b = 0$. For $t \in \mathbb{R}$, the function $f + it\mathbb{1}_K \in C(K)$ has $\|f + it\mathbb{1}_K\|_\infty = \sup_{x \in K} \sqrt{f(x)^2 + t^2} = \sqrt{\|f\|_\infty^2 + t^2}$, and \begin{align*} |\varphi(f + it\mathbb{1}_K)|^2 = |a + ib + it|^2 = a^2 + (b + t)^2. \end{align*} The bound $|\varphi(g)| \le \|\varphi\| \|g\|_\infty = \|g\|_\infty$ gives \begin{align*} a^2 + (b+t)^2 \le \|f\|_\infty^2 + t^2 \quad \text{for all } t \in \mathbb{R}. \end{align*} Expanding: $a^2 + b^2 + 2bt + t^2 \le \|f\|_\infty^2 + t^2$, i.e. $a^2 + b^2 + 2bt \le \|f\|_\infty^2$. If $b \ne 0$, sending $t \to \operatorname{sgn}(b) \cdot \infty$ makes the left-hand side blow up to $+\infty$, contradicting the bound. Hence $b = 0$, i.e. $\varphi(f) \in \mathbb{R}$, so $\varphi \in M_\mathbb{R}(K)$. Next, we show $\varphi$ is positive. Let $f \in C_\mathbb{R}(K)$ with $0 \le f \le \mathbb{1}_K$. Then $\mathbb{1}_K - f \in C_\mathbb{R}(K)$ with $0 \le \mathbb{1}_K - f \le \mathbb{1}_K$, so $\|\mathbb{1}_K - f\|_\infty \le 1$ and $|\varphi(\mathbb{1}_K - f)| \le \|\varphi\| = 1$. Therefore \begin{align*} \varphi(f) = \varphi(\mathbb{1}_K) - \varphi(\mathbb{1}_K - f) = 1 - \varphi(\mathbb{1}_K - f) \ge 1 - |\varphi(\mathbb{1}_K - f)| \ge 1 - 1 = 0. \end{align*} For general $f \ge 0$ in $C_\mathbb{R}(K)$, write $f = \|f\|_\infty \cdot (f / \|f\|_\infty)$ (assuming $f \ne 0$; the case $f = 0$ is immediate); the rescaled function lies in $[0, 1]$, so $\varphi(f / \|f\|_\infty) \ge 0$ by the previous case, and $\varphi(f) = \|f\|_\infty \varphi(f/\|f\|_\infty) \ge 0$. Hence $\varphi \in M_+(K)$. For the inclusion $M_+(K) \subset M(K)$: positive functionals are by definition real-linear and bounded (the bound being $\|\varphi\| = \varphi(\mathbb{1}_K)$ as just shown), and by the surjectivity in Step 2 every real-linear bounded functional on $C_\mathbb{R}(K)$ extends uniquely to a complex-linear bounded functional on $C(K)$.[/step]

[guided]Why does the imaginary perturbation $f + it\mathbb{1}_K$ detect non-real values of $\varphi(f)$? The key identity is $\|f + it\mathbb{1}_K\|_\infty^2 = \|f\|_\infty^2 + t^2$ (for real $f$), which says perturbing by $it\mathbb{1}_K$ enlarges the sup norm only by an additive $t^2$. Meanwhile $|\varphi(f + it\mathbb{1}_K)|^2 = a^2 + (b+t)^2$ has a cross term $2bt$ in $t$. If $b \ne 0$, this linear-in-$t$ growth eventually dominates $t^2$ for large $|t|$ (in the negative direction if $b > 0$ and positive direction if $b < 0$), driving $|\varphi(f + it\mathbb{1}_K)|^2 - \|f + it\mathbb{1}_K\|_\infty^2$ above zero, contradicting the operator norm bound $\|\varphi\| = 1$. For positivity: the trick "$\varphi(f) = 1 - \varphi(\mathbb{1}_K - f)$" exploits that the unit of the algebra $C(K)$ is $\mathbb{1}_K$ and $\|\varphi\|$ is attained at $\mathbb{1}_K$. The perturbation $\mathbb{1}_K - f \in [0,1]$ when $f \in [0,1]$, so its $C$-norm is at most $1$; the bound $|\varphi(\mathbb{1}_K - f)| \le 1$ makes $\varphi(f) \ge 0$ automatic.[/guided]

[step:Define $\varphi^+$ on the positive cone via a sup formula] Fix $\varphi \in M_\mathbb{R}(K)$. Let $C_+(K) := \{f \in C_\mathbb{R}(K) : f \ge 0\}$. Define \begin{align*} \varphi^+: C_+(K) &\to \mathbb{R} \\ f &\mapsto \sup\{\varphi(g) : g \in C_\mathbb{R}(K),\ 0 \le g \le f\}. \end{align*} The supremum is finite: for $0 \le g \le f$, $\|g\|_\infty \le \|f\|_\infty$, so $|\varphi(g)| \le \|\varphi\| \|f\|_\infty$, giving $\varphi^+(f) \le \|\varphi\| \|f\|_\infty < \infty$. Also $\varphi^+(f) \ge \varphi(0) = 0$ (taking $g = 0$). Furthermore $\varphi^+(f) \ge \varphi(f)$ (taking $g = f$). **Positive homogeneity.** For $c \ge 0$ and $f \in C_+(K)$, \begin{align*} \varphi^+(cf) = \sup\{\varphi(g) : 0 \le g \le cf\} = \sup\{\varphi(ch) : 0 \le h \le f\} = c \varphi^+(f), \end{align*} using the bijection $g \leftrightarrow h = g/c$ between $\{g : 0 \le g \le cf\}$ and $\{h : 0 \le h \le f\}$ (for $c > 0$; the case $c = 0$ is direct). **Additivity.** For $f_1, f_2 \in C_+(K)$, we show $\varphi^+(f_1 + f_2) = \varphi^+(f_1) + \varphi^+(f_2)$. ($\ge$). Given $g_j \in C_\mathbb{R}(K)$ with $0 \le g_j \le f_j$, set $g := g_1 + g_2$; then $g \in C_\mathbb{R}(K)$ and $0 \le g \le f_1 + f_2$, so $\varphi(g_1) + \varphi(g_2) = \varphi(g) \le \varphi^+(f_1 + f_2)$. Taking the supremum over $g_1, g_2$ gives $\varphi^+(f_1) + \varphi^+(f_2) \le \varphi^+(f_1 + f_2)$. ($\le$). Given $g \in C_\mathbb{R}(K)$ with $0 \le g \le f_1 + f_2$, define \begin{align*} g_1 &:= \min(g, f_1) = g \wedge f_1, \\ g_2 &:= g - g_1. \end{align*} Both $g_1$ and $g_2$ lie in $C_\mathbb{R}(K)$ (the lattice operation $\wedge$ on $C_\mathbb{R}(K)$ is continuous: $g \wedge f_1 = \tfrac{1}{2}(g + f_1 - |g - f_1|)$, a continuous function of two continuous functions). Pointwise, $0 \le g_1 \le f_1$ (since $0 \le g$ and $g_1 \le f_1$). For $g_2$: $g_2 \ge 0$ since $g_1 \le g$, and $g_2 = g - g \wedge f_1 = (g - f_1)^+ \le ((f_1 + f_2) - f_1)^+ = f_2$ (using $g \le f_1 + f_2$ and $f_2 \ge 0$). So $0 \le g_2 \le f_2$. Therefore \begin{align*} \varphi(g) = \varphi(g_1) + \varphi(g_2) \le \varphi^+(f_1) + \varphi^+(f_2). \end{align*} Taking the supremum over $g$ gives $\varphi^+(f_1 + f_2) \le \varphi^+(f_1) + \varphi^+(f_2)$. [/step]

[step:Extend $\varphi^+$ to a complex-linear functional on $C(K)$] Every $f \in C_\mathbb{R}(K)$ decomposes as $f = f^+ - f^-$ with $f^\pm := \max(\pm f, 0) \in C_+(K)$ (continuous, since $f^+ = \tfrac{1}{2}(f + |f|)$ and similarly for $f^-$). Define \begin{align*} \varphi^+: C_\mathbb{R}(K) &\to \mathbb{R} \\ f &\mapsto \varphi^+(f^+) - \varphi^+(f^-). \end{align*} This is well-defined: if $f = u - v$ with $u, v \in C_+(K)$, then $u + f^- = v + f^+$ in $C_+(K)$, so by additivity (Step 4) $\varphi^+(u) + \varphi^+(f^-) = \varphi^+(v) + \varphi^+(f^+)$, hence $\varphi^+(u) - \varphi^+(v) = \varphi^+(f^+) - \varphi^+(f^-)$, independent of the decomposition. **$\mathbb{R}$-linearity.** Additivity: for $f_1, f_2 \in C_\mathbb{R}(K)$, write $f_1 + f_2 = (f_1^+ + f_2^+) - (f_1^- + f_2^-)$; by additivity on $C_+(K)$, \begin{align*} \varphi^+(f_1 + f_2) = \varphi^+(f_1^+ + f_2^+) - \varphi^+(f_1^- + f_2^-) = \varphi^+(f_1) + \varphi^+(f_2). \end{align*} Homogeneity: for $c \ge 0$, $(cf)^\pm = cf^\pm$, so $\varphi^+(cf) = c\varphi^+(f^+) - c\varphi^+(f^-) = c\varphi^+(f)$ from positive homogeneity. For $c = -1$, $(-f)^\pm = f^\mp$, so $\varphi^+(-f) = \varphi^+(f^-) - \varphi^+(f^+) = -\varphi^+(f)$. Combining, $\varphi^+(cf) = c\varphi^+(f)$ for all $c \in \mathbb{R}$. **Boundedness.** For $f \in C_\mathbb{R}(K)$, $f^\pm \le |f| \le \|f\|_\infty \mathbb{1}_K$, so $\varphi^+(f^\pm) \le \|\varphi\| \|f\|_\infty$ from the bound in Step 4, giving $|\varphi^+(f)| \le 2\|\varphi\|\|f\|_\infty$. So $\varphi^+ \in C_\mathbb{R}(K)^*$. Now extend to $C(K)$ by $\varphi^+(f) := \varphi^+(\operatorname{Re} f) + i \varphi^+(\operatorname{Im} f)$, which gives $\varphi^+ \in M_\mathbb{R}(K)$ by the surjectivity of Step 2. [/step]

[step:Define $\varphi^-$ as $\varphi^+ - \varphi$ and verify positivity] Set $\varphi^- := \varphi^+ - \varphi$, an element of $M_\mathbb{R}(K)$. Then $\varphi = \varphi^+ - \varphi^-$ by construction. We verify $\varphi^\pm \in M_+(K)$. For $f \in C_+(K)$, $\varphi^+(f) = \sup\{\varphi(g) : 0 \le g \le f\} \ge \varphi(0) = 0$, so $\varphi^+ \ge 0$ on $C_+(K)$, hence $\varphi^+ \in M_+(K)$. For $\varphi^-$: take any $f \in C_+(K)$. Then $0 \le f \le f$ shows $f$ itself is admissible in the supremum, so $\varphi^+(f) \ge \varphi(f)$, hence $\varphi^-(f) = \varphi^+(f) - \varphi(f) \ge 0$. So $\varphi^- \ge 0$ on $C_+(K)$, hence $\varphi^- \in M_+(K)$. [/step]

[step:Verify the norm identity $\|\varphi\| = \|\varphi^+\| + \|\varphi^-\|$] By part (iii), $\|\varphi^\pm\| = \varphi^\pm(\mathbb{1}_K)$. Then \begin{align*} \|\varphi^+\| + \|\varphi^-\| = \varphi^+(\mathbb{1}_K) + \varphi^-(\mathbb{1}_K) = \varphi^+(\mathbb{1}_K) + (\varphi^+(\mathbb{1}_K) - \varphi(\mathbb{1}_K)) = 2\varphi^+(\mathbb{1}_K) - \varphi(\mathbb{1}_K). \end{align*} By definition, $\varphi^+(\mathbb{1}_K) = \sup\{\varphi(g) : 0 \le g \le \mathbb{1}_K\}$. We compute this supremum. For the upper bound: for $0 \le g \le \mathbb{1}_K$, write $g = \tfrac{1}{2}((2g - \mathbb{1}_K) + \mathbb{1}_K)$, so $\varphi(g) = \tfrac{1}{2}(\varphi(2g - \mathbb{1}_K) + \varphi(\mathbb{1}_K))$. Since $-\mathbb{1}_K \le 2g - \mathbb{1}_K \le \mathbb{1}_K$, $\|2g - \mathbb{1}_K\|_\infty \le 1$, so $|\varphi(2g - \mathbb{1}_K)| \le \|\varphi\|$, giving $\varphi(g) \le \tfrac{1}{2}(\|\varphi\| + \varphi(\mathbb{1}_K))$. For the lower bound: by definition of $\|\varphi\| = \|\varphi|_{C_\mathbb{R}(K)}\|$ (part ii) and the fact that $\varphi$ is real-valued on $C_\mathbb{R}(K)$, for every $\varepsilon > 0$ there exists $h \in C_\mathbb{R}(K)$ with $\|h\|_\infty \le 1$ and $\varphi(h) > \|\varphi\| - \varepsilon$ (by replacing $h$ with $-h$ if necessary). Set $g := \tfrac{1}{2}(\mathbb{1}_K + h) \in C_\mathbb{R}(K)$; then $0 \le g \le \mathbb{1}_K$ and \begin{align*} \varphi(g) = \tfrac{1}{2}(\varphi(\mathbb{1}_K) + \varphi(h)) > \tfrac{1}{2}(\varphi(\mathbb{1}_K) + \|\varphi\| - \varepsilon). \end{align*} Taking $\varepsilon \to 0$, $\varphi^+(\mathbb{1}_K) \ge \tfrac{1}{2}(\|\varphi\| + \varphi(\mathbb{1}_K))$. Combining, $\varphi^+(\mathbb{1}_K) = \tfrac{1}{2}(\|\varphi\| + \varphi(\mathbb{1}_K))$, hence \begin{align*} \|\varphi^+\| + \|\varphi^-\| = 2 \cdot \tfrac{1}{2}(\|\varphi\| + \varphi(\mathbb{1}_K)) - \varphi(\mathbb{1}_K) = \|\varphi\|. \end{align*} [/step]

[step:Prove uniqueness of the Jordan decomposition] Suppose $\varphi = \psi^+ - \psi^-$ with $\psi^\pm \in M_+(K)$ and $\|\varphi\| = \|\psi^+\| + \|\psi^-\|$. We show $\psi^\pm = \varphi^\pm$. For any $g \in C_\mathbb{R}(K)$ with $0 \le g \le f \in C_+(K)$, positivity of $\psi^-$ gives $\psi^-(g) \ge 0$, so \begin{align*} \varphi(g) = \psi^+(g) - \psi^-(g) \le \psi^+(g) \le \psi^+(f). \end{align*} (The last inequality uses positivity of $\psi^+$ applied to $f - g \ge 0$.) Taking the supremum over $g$, $\varphi^+(f) \le \psi^+(f)$ for all $f \in C_+(K)$. The reverse inequality follows from the norm identity. By part (iii) applied to $\psi^\pm$ and $\varphi^\pm$, \begin{align*} \psi^+(\mathbb{1}_K) + \psi^-(\mathbb{1}_K) = \|\psi^+\| + \|\psi^-\| = \|\varphi\| = \|\varphi^+\| + \|\varphi^-\| = \varphi^+(\mathbb{1}_K) + \varphi^-(\mathbb{1}_K). \end{align*} Subtracting $\varphi(\mathbb{1}_K) = \psi^+(\mathbb{1}_K) - \psi^-(\mathbb{1}_K) = \varphi^+(\mathbb{1}_K) - \varphi^-(\mathbb{1}_K)$ from both sides and dividing by $2$ gives $\psi^-(\mathbb{1}_K) = \varphi^-(\mathbb{1}_K)$, hence $\psi^+(\mathbb{1}_K) = \varphi^+(\mathbb{1}_K)$. Now $\varphi^+(\mathbb{1}_K) \le \psi^+(\mathbb{1}_K)$ becomes equality, and $(\psi^+ - \varphi^+)(\mathbb{1}_K) = 0$. Since $\psi^+ - \varphi^+$ acts as $\psi^+(g) - \varphi^+(g) \ge 0$ for $0 \le g \le \mathbb{1}_K$ (the same argument with the supremum), and equals zero at $\mathbb{1}_K$: for any $f \in C_+(K)$ with $\|f\|_\infty \le 1$, both $f$ and $\mathbb{1}_K - f$ lie in $[0, \mathbb{1}_K]$, so $(\psi^+ - \varphi^+)(f) \ge 0$ and $(\psi^+ - \varphi^+)(\mathbb{1}_K - f) \ge 0$; the latter gives $(\psi^+ - \varphi^+)(f) \le (\psi^+ - \varphi^+)(\mathbb{1}_K) = 0$, hence $(\psi^+ - \varphi^+)(f) = 0$. By scaling and the decomposition $f = f^+ - f^-$, $(\psi^+ - \varphi^+)(f) = 0$ for all $f \in C_\mathbb{R}(K)$, so $\psi^+ = \varphi^+$ on $C_\mathbb{R}(K)$, hence on $C(K)$ (by Step 2). Then $\psi^- = \psi^+ - \varphi = \varphi^+ - \varphi = \varphi^-$. This completes the proof of (iv). [/step]