Define $f: I \rightarrow \mathbb{R}$ as $f(x)= \sup {f_n(x) : n \geq n_0 }$ for $ x \in I$, It's convex? Suppose, that $f_n:I\rightarrow \mathbb{R}$ are convex functions for $n\geq n_0$ and 
$\forall_{x\in I} \exists_{y\in \mathbb{R}} \forall_{n\geq n_0} f_n(x)\leq y$ 
Define $f: I \rightarrow \mathbb{R}$ as $f(x)= \sup \{f_n(x) \mid n \geq n_0 \}$ for $ x \in I$
Please help me prove, that $f$ is convex . 
What I've tried - write fromulas from definition, try to glue then, but nothing work. Thanks in advance!
 A: Try using the characterisation of convexity which goes like this: $f$ is convex on $I$ exactly when
$$f(tx + (1-t)y) \le tf(x) + (1-t)f(y)$$ 
for any $x, y \in I$ and $t \in [0,1]$, assuming $I$ is convex. Also remember that for bounded sets $A, B \ne \{ \}$ we have
$$\sup (A+B) = \sup A + \sup B,$$
where $A + B = \{ a + b : a \in A, b \in B \}.$
A: If $\{f_\alpha \}_{\alpha \in A}$ is a collection of convex functions, then $\sup_{\alpha \in A} f_\alpha$ (defined pointwise) is convex.
One way to see this is to notice that $\operatorname {epi} \sup_{\alpha \in A} f_\alpha = \cap_{\alpha \in A} \operatorname {epi} f_\alpha$, and the intersection of an arbitrary collection of convex sets is convex.
A direct way is to notice that if $\lambda \in [0,1] $, then 
$$f_\alpha(\lambda x + (1-\lambda)y) \le \lambda f_\alpha(x) + (1-\lambda) f_\alpha(y), \ \ \ \forall \alpha$$
Hence we have
$$f_{\alpha'}(\lambda x + (1-\lambda)y) \le \lambda \sup_{\alpha \in A} f_\alpha(x) + (1-\lambda) \sup_{\alpha \in A}  f_\alpha(y), \ \ \ \forall \alpha'$$
Now take the $\sup$ over $\alpha'$ to get 
$$\sup_{\alpha' \in A} f_{\alpha'}(\lambda x + (1-\lambda)y) \le \lambda \sup_{\alpha \in A} f_\alpha(x) + (1-\lambda) \sup_{\alpha \in A}  f_\alpha(y)$$
Hence $\sup_{\alpha \in A} f_\alpha$ is convex.
In your case, $A = \{n | n \ge n_0 \}$.
