Inequality for Limit inferior

Question

I was wondering if anyone can give a mathematical explanation for the claim below.

If $F(x-\epsilon)\le F_n(x)+p_n$ where $\lim_{n\to\infty}p_n=0.$ (1)

Then $F(x-\epsilon)\le \liminf\limits_{n\to\infty}F_n(x)$. (2)

My intuition says it should be based on the definition of $\lim\inf$ of a sequence which is defined as $\lim\limits_{n\to\infty}\inf\limits_{k\ge n}x_k$.

As my first attempt, I took the limit inferior on both sides of eq. (1). The left hand side doesn't change as it is a constant, and since $\lim_{n\to\infty}p_n=0$, therefore the limit inferior is also $0$. Hence we can justify the claim. But I am not sure if this is the right approach. Could anyone provide any help/suggestion?

Answer 1

In general, we have only: $$\liminf\bigl( F_n(x)+p_n(x)\bigr)\ge \liminf F_n(x)+\liminf p_n(x),$$ which would not allow to conclude. However, if one of the sequences has a limit, the inequality becomes an equality.

Answer 2

use the fact that $$ a_n\leqslant b_n\implies \liminf a_n\leqslant \liminf b_n \quad\text{and}\quad\liminf c= c $$ We prove that if $\lim b_n$ exists, then $$ \liminf (a_n+b_n)= \liminf a_n+\lim b_n \tag 1 $$ Let $a_{n_k}$ be a sequence that $$ \liminf_{n\to\infty} a_n=\lim_{k\to\infty} a_{n_k}\tag 2 $$ Since $\lim b_n$ exists $$ \liminf_{n\to\infty} b_n=\lim_{n\to\infty} b_n=\lim_{k\to\infty} b_{n_k}\tag 3 $$ So $a_{n_k}+b_{n_k}\to\liminf (a_n+b_n)$. And by $(2)$ and $(3)$ $$ \liminf_{n\to\infty} (a_n+b_n)=\lim_{k\to\infty} (a_{n_k}+b_{n_k})=\lim_{k\to\infty} a_{n_k}+\lim_{k\to\infty} b_{n_k}=\liminf_{n\to\infty} a_n+\lim_{n\to\infty} b_{n} $$ So we have $(1)$ hold. For each $x$ we have \begin{align} F(x-\epsilon)&=\liminf_{n\to\infty} F(x-\epsilon) \\ &\leqslant \liminf\limits_{n\to\infty}(F_n(x)+p_n) \\ &= \liminf\limits_{n\to\infty}F_n(x)+ \lim\limits_{n\to\infty}p_n \\ &=\liminf\limits_{n\to\infty}F_n(x) \end{align}