I'm reading a paper where $\limsup$ and $\inf$ are interchanged:
$$\limsup_{D \backepsilon \text{ }x \rightarrow x_0} \bigg( \inf_n \textbf{E}^x \big[ \textbf{1}_{\{\forall t \in (0,h-1/n]: B(t) \in D)\}} \big] \bigg) \leq \inf_k \bigg(\limsup_{D \backepsilon \text{ }x \rightarrow x_0} \textbf{E}^x \big[ \textbf{1}_{\{\forall t \in (0,h-1/k]: B(t) \in D)\}} \big] \bigg)$$
I'm including context, but mainly my questions is about this swap. I'm having trouble justifying how the inequality arises. Thoughts?
Given a sequence $(x_n)_n$ of real numbers, remember that
$$\limsup_{n \to \infty} x_n = \inf_{n_0} \{ \sup_{n \ge n_0} x_n \} \tag 1 $$
and
$$\liminf_{n \to \infty} x_n = \sup_{n_0} \{ \inf_{n \ge n_0} x_n \} \tag 2$$
Claim: Let $(x_{m,n})_{m,n \in \mathbb{N}}$ be a double sequence of real numbers, then
$$\inf_n \big ( \limsup_{m \to \infty} x_{m,n} \big) \ge\limsup_{m \to \infty} \big ( \inf_n x_{m,n} \big) $$
Proof:
For all $m,n \in \mathbb{N}$, one has
$$x_{m,n} \ge \inf_n \big (x_{m,n}\big)$$
Therefore,
\begin{align}\inf_{m_0} \Big \{ \inf_{n} \big \{ \sup_{m\ge m_0} \big (x_{m,n}\big) \big \}\Big \} &\ge \inf_{m_0} \Big \{ \sup_{m\ge m_0}\big \{\inf_n \big (x_{m,n}\big)\big \} \Big \} \\ &\ge \limsup_{m \to \infty} \big ( \inf_n x_{m,n} \big) \end{align}
Where the last inequality follows from (1). But, for any double sequence $(y_{m,n})_{m,n \in \mathbb{N}}$ of real numbers
$$\inf_m \Big \{ \inf_n y_{m,n} \Big \}=\inf_{m,n} \big (y_{m,n}\big )= \inf_n \Big \{\inf_m y_{m,n} \Big \} $$
Thus, interchanging the infimums, it follows that
$$\inf_{n} \Big \{ \inf_{m_0} \big \{ \sup_{m\ge m_0} \big (x_{m,n}\big) \big \}\Big \}\ge\limsup_{m \to \infty} \big ( \inf_n x_{m,n} \big)$$
and again by (1) we get
$$\inf_n \big ( \limsup_{m \to \infty} x_{m,n} \big) \ge\limsup_{m \to \infty} \big ( \inf_n x_{m,n} \big) $$
as required.