I am trying to understand a step in this proof.
Let $\{f_j\}_{j \in J}$ be an arbitrary family of convex functions: $f_j: X \to \mathbb{R}$ where $X \subseteq \mathbb{R}^n $ is convex. Show that $f(x):= \sup\{f_j(x)| \; j\in J\}$ is convex.
Proof:
$$f(x) := \sup \{ f_j(x) | \; j \in J\}$$
$$f(\lambda x + (1-\lambda)y) := \sup \{f_j(\lambda x + (1-\lambda)y | \; j \in J\}$$
$$\sup \{f_j(\lambda x + (1-\lambda)y) | \; j \in J\} \leq \sup \{\lambda f_j(x) + (1-\lambda)f_j(y)| j \in J\} $$ $$\leq \sup\{\lambda f_j(x)|j\in J\}+\sup\{(1-\lambda)f_j(y)|j\in J\} $$
The first inequality seems to follow from the convexity of each $f_j$ - is this correct? But then where does the last inequality follow from, where the $\sup(\cdot)$'s are can be split apart? Should it not just be an equality? Any help is very much appreciated.