If $|\Phi(f)|\leq\Lambda(f)$ for all $f\in C_c^+(X)$ then $|\Phi(f)|\leq\Lambda(|f|)$ for all $f\in C_c(X)$. Let $X$  be a locally compact Hausdorff space, and let $C_c(X)$ denote the $\Bbb C$-vector space of all continuous complex-valued functions with compact support. Suppose $\Phi$ is a bounded linear functional on $C_c(X)$ and $\Lambda$ is a positive linear functional on $C_c(X)$. Let $C_c^+(X)$ consists of all $f\in C_c(X)$ such that $f\geq 0$. Suppose we have $|\Phi(f)|\leq\Lambda(f)$ for every $f\in C_c^+(X)$. Using this condition, I am trying to show that $|\Phi(f)|\leq\Lambda(|f|)$ for every $f\in C_c(X)$. If $f$ is real, then it is easily done by using $f=f^+-f^-$, but I am having a hard time with the general case. Any hints?
Note. This questions arises in the proof of Theorem 6.19 of Rudin's Real and Complex Anaylsis. 
 A: Since you mention Rudin's proof, the situation is quite simple. He defines
$$\Lambda(f) := \sup\left\{ |\Phi(h)| : h\in C_c(X), |h|\leq f\right\}$$
for all $f\in C_c^+(X)$. Therefore $\Lambda(|f|) \geq |\Phi(f)|$ is true by definition of $\Lambda$.

Nevertheless, one can answer the question you've asked as well:
Let $f=u+iv$ with real-valued $u,v\in C_c(X)$ and $u=u^+-u^-, v=v^+-v^-$ with non-negative $u^\pm, v^\pm\in C_c^+(X)$, then it is easy to see that
$$|\Phi(f)|\leq \Lambda(|u|)+\Lambda(|v|) \leq 2\Lambda(|f|)$$
Now we can use the tensor power trick to eliminate the constant 2: We apply the same argument to $\Phi^{\otimes N}$ and $\Lambda^{\otimes N}$ both defined on $\underbrace{C_c(X)\otimes...\otimes C_c(X)} \to \mathbb{C}$ (where we view $C_c(X)\otimes...\otimes C_c(X)$ as a subspace of $C_c(X\times...\times X)$ in the obvious way). If we then plug $f\otimes ...\otimes f$ into the inequality we get
$$|\Phi(f)|^N \leq 2 \Lambda(|f|)^N \implies |\Phi(f)|\leq 2^{1/N}\Lambda(|f|)$$
Letting $N\to\infty$ we get the desired inequality.
