Let $f_n:[0,\infty[\,\to\mathbb{R}$ be a convex decreasing functions of $x$, for all $n\in\mathbb{N}$.
Suppose that there exists $\lim_{n\to\infty}\lim_{x\to\infty}f_n(x)=\lambda$.
Suppose also that there exists $\lim_{n\to\infty}f_n(x)=f(x)$.
Can I interchange the limits and say that $\lim_{x\to\infty}f(x)=\lambda$ ?
I know that every convergent sequence of convex functions on an open interval is uniformly convergent on compact sub-intervals. Here can I do something to deal with the fact that $x\to\infty$? Maybe is it possible to conclude adding some hypothesis?
You cannot interchange the limits here. A counterexample is $f_n(x) = \exp(-x/n)$, which gives you $\lim_{x\to\infty} f_n(x) = 0$, but $\lim_{n\to\infty} f_n(x) = 1$.
This counterexample also shows that even with very nice functions (strictly convex with derivatives of all orders jointly bounded), uniform convergence cannot be concluded if the domain is not (totally) bounded. For that reason, I am inclined to say that the fact that the functions are convex is probably of no use at all.