Let $(E,\mathcal E,\mu)$ be a probability space and $f\in L^2(\mu)$. Assume there is a $c\ge0$ with that we know that $$|\langle f,g\rangle_{L^2(\mu)}|\le c\left\|g\right\|_{L^2(\mu)}\tag1$$ for all $g\in L^2(\mu)$ with $g\ge0$ and $\int g\:{\rm d}\mu=1$.
Are we able to extend to all $g\in L^2(\mu)$ with $\left\|g\right\|_{L^2(\mu)}\le1$?
My idea was to consider the positive $g^+$ and negative part $g^-$ of $g$, i.e. $g^\pm=\max(0,\pm g)$, but I end up with $|\langle f,g\rangle_{L^2(\mu)}|\le c\left(\left\|g^+\right\|_{L^2(\mu)}+\left\|g^-\right\|_{L^2(\mu)}\right)$, which is not enough.
This cannot be done. In fact, I will prove the stronger statement that it cannot be done even if you strengthened your first assumption to hold for all $g \in L^2(\mu)$ with $g \geq 0$.
In this case, essentially you would be asking if $$\|f\|_{L^2(\mu)} = \sup_{g \in B} |\langle f,g \rangle| = \sup_{g \in B^+} |\langle f,g \rangle |$$ for an arbitrary $f$, where $B$ is the unit ball of $L^2(\mu)$ and $B^+$ is the set of non-negative elements of $B$.
Consider the case where $\mu$ is the Lebesgue measure on $[0,1]$ and let $f = 1_{[0,1/2)} - 1_{[1/2,1]}$. Then $\|f\|_{L^2(\mu)} = 1$ and so we would want to have that $\sup_{g \in B^+} |\langle f,g \rangle| = 1$.
Clearly that $\sup$ is the same as the supremum over the set of $g \in B^+$ such that $g = 0$ on $[1/2,1]$. Let $B_0^+$ be the set of $g \in B^+$ such that $g = 0$ on $[1/2,1]$. Then, by Holder's inequality, we have that $$\sup_{g \in B_0^+} |\langle f, g \rangle| = \sup_{g \in B_0^+} \int_0^{\frac12} g d\mu \leq \sup_{g \in B_0^+} \|g\|_{L^2(\mu)} \cdot \mu([0,1/2))^{\frac{1}{2}} = 2^{-\frac12} < 1.$$
Hence the two suprema are not equal.