# How to argue that the map $\phi_{g}: L^{p}(X, \mu) \rightarrow \mathbb{F}$ is linear.

Let $$p,q>1 :1/p+1/q=1$$ and define the map $$\phi_{g}: L^{p}(X, \mu) \rightarrow \mathbb{F}$$ by $$\phi_{g}(f):=\int_{X} f g d \mu$$ where $$g\in L^q(X,\mu)$$ (with $$L^p$$ and $$L^q$$ function spaces). I already know that the function defined above is linear, but I would like to know if I can argue that this is true without a formal proof?

• Multiplication is bilinear and the integral is linear. Hence $\phi_g$ must be linear. Is this what you are looking for? Aug 5 at 17:06
• So effectively I have $\phi_{1}: L^{p}(X, \mu) \rightarrow L^{1}(X, \mu)$, defined by $\phi_{1}(f):=f g$, and $\phi_{2}: L^{1} \rightarrow \mathbb{F}$, defined by $\phi(f g):=\int_{X} f g d \mu$. Because $\phi_1$ and $\phi_2$ are both linear, the composition $\phi_{g}(f)=\left(\phi_{2} \circ \phi_{1}\right)(f)$ is linear?
– Logi
Aug 5 at 17:20
• $\phi_2(h) := \int_X h d\mu$ to be precise, but yes. Aug 5 at 17:25
• Great, thank you.
– Logi
Aug 5 at 17:37