Exercise 1.4

Question

Let $\{a_n\}$ and $\{b_n\}$ be sequences in $[-\infty,\infty]$, and prove
    the following assertions:
    \begin{enumerate}[label=(\alph*)]
    \item $\limsup_{n 	o \infty}(-a_n) = - \liminf_{n 	o \infty} a_n$.
    \item $\limsup_{n 	o \infty}(a_n + b_n) \le \limsup_{n 	o \infty} a_n +
      \lim$ $\sup_{n	o\infty} b_n$
      provided none of the sums is of the form $\infty - \infty$.
    \item If $a_n \le b_n$ for all $n$, then
      \begin{equation*}
        \liminf_{n 	o \infty} a_n \le \liminf_{n 	o \infty} b_n.
      \end{equation*}
    \end{enumerate}
    Show by an example that strict inequality can hold in $(b)$.

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $(a)$]
      We have
      \begin{align*}
        \limsup_{n \to \infty}(-a_n) &= \inf_{k \ge 1} \left\{\sup_{n \ge k}
        -a_n\right\}
        = \inf_{k \ge 1} \left\{- \inf_{n \ge k} a_n \right\}\
        &= -\sup_{k \ge 1} \left\{ \inf_{n \ge k} a_n \right\}
        = - \liminf_{n \to \infty} a_n.\qedhere
      \end{align*}
    \end{proof}
    \begin{proof}[Proof of $(b)$]
      We have
      $
        \sup_{n \ge k} (a_n + b_n) \le \sup_{n \ge k} a_n + \sup_{n \ge k} b_n
      $
      and letting $k \to \infty$ gives the claim.
      \par For strict inequality, if we have $a_{2n} = b_{2n+1} = 1$ and
      $a_{2n+1} = b_{2n} = 0$ for each $n$, then
      \begin{gather*}
        \limsup_{n \to \infty} (a_n + b_n) = 1\
        \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n = 2.\qedhere
      \end{gather*}
    \end{proof}
    \begin{proof}[Proof of $(c)$]
      We have the inequality $\inf_{n \ge k} a_n \le \inf_{n \ge k} b_n$ for all
      $k$. Letting $k \to \infty$ gives the claim.
    \end{proof}

Exercise 1.4

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Exercise 1.4

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