Consider the set $$ \mathcal{U}\equiv \{U\in \mathbb{R}^K: T(U)< T'(U)\} $$ where $T$ and $T'$ are linear functionals of the vector $U$.
I want to show that $\mathcal{U}$ is convex using the definition of convex set, i.e., $\forall \beta \in (0,1)$ and $\forall \tilde{U},\hat{U}\in \mathcal{U}$, we have that $$ [\beta \tilde{U}+(1-\beta) \hat{U}] \in \mathcal{U} $$
I'm confused about how to proceed. In particular, I'm confused about which one of these two is the correct way to go:
1) Take any $\beta\in (0,1)$. By the linearity of $T,T'$, it follows that
$$\beta T(\tilde{U})+(1-\beta) T(\hat{U})<\beta T'(\tilde{U})+(1-\beta) T'(\hat{U})$$
Hence $\mathcal{U}$ is convex.
2) Take any $\beta\in (0,1)$. By the linearity of $T,T'$, it follows that
$$T(\beta \tilde{U}+(1-\beta)\hat{U})<T'(\beta \tilde{U}+(1-\beta)\hat{U})$$
Hence $\mathcal{U}$ is convex.
I think 2) is correct but I would like to know your opinion.
Yet 2) is correct but you will also have to use 1) along the way:
$$T(\beta \tilde{U}+(1-\beta)\hat{U})=\beta T(\tilde{U})+(1-\beta)T(\hat{U})<\beta T'(\tilde{U})+(1-\beta)T'(\hat{U})=T'(\beta \tilde{U}+(1-\beta)\hat{U})$$