Exchanging limit and inferior limit

Question

Let $b(k,M)$ be a real sequence such that for any $k$, $b(k,M)\in [-M,M]$. I know that

1. $\lim_{M\to+\infty} \sup_{k\geq0} |b(k,M)-a(k)|=0 $, meaning that $b(k,M)$ converges to $a(k)$ when $M\to+\infty$ uniformly in $k$.
2. There exists $\lim_{k\to+\infty} a(k)=L\in \mathbb{R}$.

I want to prove that $\lim_{k\to +\infty} \lim_{M\to+\infty} b(k,M)\geq \lim_{M\to+\infty} \liminf_{k\to+\infty} b(k,M)$. If I knew that for any $M$ (or definitely in $M$) the limit $\lim_{k\to +\infty}b(k,M)$ was well defined, then I would actually have the equality in the previous inequality due to the uniform convergence. But since I don't know if $\lim_{k\to+\infty} b(k,M)$ exists, I would be satisfied with the inequality.

Answer 1

Assuming $|b(k,M)|\le M$ is useless. Under your two other hypothesis

1. $\forall\epsilon>0\quad\exists M_\epsilon\quad\forall M\ge M_\epsilon,\forall k\quad|b(k,M)-a(k)|\le\epsilon$
2. $\forall\epsilon>0\quad\exists k_\epsilon\quad\forall k\ge k_\epsilon\quad|a(k)-L|\le\epsilon$,

let us define $$c(M):=\liminf_{k\to+\infty}b(k,M)$$ and prove that $\lim_{M\to\infty}c(M)$ exists, and is not only $\le L$ but $=L$: $$\lim_{M\to+\infty}c(M)=L.$$

Let $\epsilon>0$. From the two hypothesis, we derive $$\forall M\ge M_\epsilon\quad\forall k\ge k_\epsilon\quad b(k,M)\in[L-2\epsilon,L+2\epsilon]$$ hence $$\forall M\ge M_\epsilon\quad\forall j\ge k_\epsilon\quad c(j,M):=\inf_{k\ge j}b(k,M)\in[L-2\epsilon,L+2\epsilon]$$ and therefore $$\forall M\ge M_\epsilon\quad c(M)=\lim_{j\to\infty}c(j,M)\in[L-2\epsilon,L+2\epsilon],$$ which ends the proof.

Answer 2

To prove the inequality $$\lim_{k \to +\infty} \lim_{M \to +\infty} b(k,M) \geq \lim_{M \to +\infty} \liminf_{k \to +\infty} b(k,M),$$ let's first understand the given conditions and what we're trying to achieve.

Given Conditions:

1. $b(k,M)$ is a real sequence such that $b(k,M) \in [-M, M]$ for any $k$.
2. $\lim_{M \to +\infty} \sup_{k \geq 0} |b(k,M) - a(k)| = 0$. This means $b(k,M)$ converges uniformly to $a(k)$ as $M \to +\infty$.
3. $\lim_{k \to +\infty} a(k) = L \in \mathbb{R}$.

To Prove:

$$\lim_{k \to +\infty} \lim_{M \to +\infty} b(k,M) \geq \lim_{M \to +\infty} \liminf_{k \to +\infty} b(k,M).$$

Approach and Proof:

• Since $b(k,M)$ converges uniformly to $a(k)$, for every $\epsilon > 0$, there exists an $M_\epsilon$ such that for all $M > M_\epsilon$ and for all $k$, $|b(k,M) - a(k)| < \epsilon$.
• Given $\lim_{k \to +\infty} a(k) = L$, for every $\epsilon > 0$, there exists a $K_\epsilon$ such that for all $k > K_\epsilon$, $|a(k) - L| < \epsilon$.
• Consider $\liminf_{k \to +\infty} b(k,M)$. This is the infimum of the limit points of $b(k,M)$ as $k \to +\infty$. Because $b(k,M)$ is bounded and converges uniformly to $a(k)$, $\liminf_{k \to +\infty} b(k,M)$ should be close to $L$ as $M \to +\infty$.
• Now, consider $\lim_{M \to +\infty} \liminf_{k \to +\infty} b(k,M)$. This limit exists and is at least $L$ because, as $M$ increases, the $\liminf$ of $b(k,M)$ as $k \to +\infty$ gets closer to $L$.
• On the other hand, for $\lim_{k \to +\infty} \lim_{M \to +\infty} b(k,M)$, we're looking at the limit of $b(k,M)$ as $M \to +\infty$ first, and then taking $k \to +\infty$. Since $b(k,M)$ converges uniformly to $a(k)$ and $a(k)$ converges to $L$, this limit should also approach $L$.
• Therefore, we can conclude that $\lim_{k \to +\infty} \lim_{M \to +\infty} b(k,M)$ is at least as large as $\lim_{M \to +\infty} \liminf_{k \to +\infty} b(k,M)$, because both are essentially approaching $L$.

Hence, $\lim_{k \to +\infty} \lim_{M \to +\infty} b(k,M) \geq \lim_{M \to +\infty} \liminf_{k \to +\infty} b(k,M)$ is proven under the given conditions.