$\bullet~$ Problem: Let $R$ be the Ring of all functions in $\mathscr{C}^{0}[0, 1]$. Prove that the map $\varphi : R \to \mathbb{R}$ defined by
$$ \varphi(f) = \int_0^1 f(t) dt \quad \text{for any } f \in \mathscr{C}^{0}([0, 1])$$
is a Homomorphism for additive groups but not a Ring Homomorphism.
$\bullet~$ My Solution: We are given the ring $R$ of all continuous real-valued functions on the closed interval $[0, 1]$.
According to question let's consider the map \begin{align*} \varphi : &~ R \rightarrow \mathbb{R}\\ & f(t) \mapsto \int_{0}^{1} f (t) dt \end{align*} $\circ~$ Now we'll check if it's an Additive Group Homomorphism. \begin{align*} \varphi(f + g)(t) & = \varphi( f (t) + g(t) ) = \int_{0}^{1} ( f (t) + g (t) )dt\\ & = \int_{0}^{1} f(t)dt + \int_{0}^{1} g (t)dt\\ & = \varphi(f) + \varphi(g) \end{align*} Hence it is an additive group homomorphism.
$\circ~$ Edit: Consider $f(t) = t \in \mathscr{C}^0[0, 1]$ and $g(t) = t^2 \in \mathscr{C}^{0}[0, 1]$ then $(f\cdot g)(t) = t^3 \in \mathscr{C}^0[0, 1] $ .
Now we'll check for the multiplicative one. \begin{align*} \varphi( fg )(t) & = \varphi( f(t) g(t))\\ & = \int_{0}^{1} f (t) g (t) dt\\ & = \int_0^1 t^3 dt\\ & = \frac{1}{4} \neq \frac{1}{6}\\ & = \int_0^1 t dt \cdot \int_0^1 t^2 dt\\ & = \bigg( \int_{0}^{1} f (t) dt \bigg) \bigg( \int_{0}^{1} g (t) dt \bigg) \\ & = \varphi( f) \cdot \varphi( g ) \end{align*} And hence it's clear that, our map is not a ring homomorphism.
Please check the solution for glitches and give new ideas :)
The fact that $\phi(|f||g|) \leq \phi(|f|)\phi(|g|)$ is perfectly consistent with the claim that $\phi(|f||g|) = \phi(|f|)\phi(|g|)$. You need to come up with a specific example where equality is violated.