How to show that if either $\{a_k\}$ or $\{b_k\}$ converges, equality holds. (lim sup, lim inf)

Question

I am studying Real analysis by the textbook, Measure and Integral: an introduction to real analysis by Wheeden and Zygmund. This is my first time to study mathematical analysis. However, instructor gave me assignments which is solving even problems in Chapter 1's exercises.

I solved $2$, and $4$ (a), (b), (c) but failed to solve $4$ (d). Can someone help me?

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4.

(b) Show that $\limsup_{k\to\infty}(a_k+b_k)\le\limsup_{k\to\infty}a_k + \limsup_{k\to\infty}b_k$.

(c) If $\{a_k\}$ and $\{b_k\}$ are nonnegative, bounded sequences, show that $\limsup_{k\to\infty}(a_kb_k)\le(\limsup_{k\to\infty}a_k)(\limsup_{k\to\infty}b_k)$.

(d) (1) Give examples for which the inequalities in parts (b) and (c) are not equalities. (2) Show that if either $\{a_k\}$ or $\{b_k\}$ converges, equality holds in (b) and (c).

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I solved (b) and (c) and (d)-(1).

Please let me know how to prove (d)-(2). Thank you very much.

Answer 1

Suppose that $a_k \to A$. Let $\epsilon > 0$ be fixed, and choose $N$ so that $k \ge N$ implies $|a_k - A| < \epsilon$. Let $n \ge N$ and let $k \ge n$. Then $A - \epsilon < a_k$ so that $$ b_k = (b_k + A - \epsilon) - A + \epsilon \le (b_k + a_k) + \epsilon \le \sup_{j \ge n} (a_j + b_j) - A + \epsilon.$$ The right-hand-side does not depend on $k$, so take the supremum over all $k \ge n$ to find $$ A + \sup_{k \ge n} b_k \le \sup_{j \ge n} (a_j + b_j) + \epsilon. $$ Both sides are nondecreasing in $n$ so we may let $n \to \infty$ to find $$ A + \limsup_{k \to \infty} b_k \le \limsup_{k \to \infty} (a_k + b_k) + \epsilon.$$ Here $\epsilon > 0$ is arbitrary so take $\epsilon \to 0^+$ to get $$ A + \limsup_{k \to \infty} b_k \le \limsup_{k \to \infty} (a_k + b_k).$$

Answer 2

Use the definition of $\limsup$:

$g_k=\sup _{n\geq k}\left \{ a_n \right \}$

$h_k=\sup _{n\geq k}\left \{ b_n \right \}$

If $a_n$ and $b_n$ converge to $L$ and $M$, resp. then $g_k$ and $h_k$ converge to $L$ and $M$, resp. as well. (why?), so now using the sum formula for limits we get

$\lim_{k\to \infty }(g_k+h_k)=\lim_{k\to \infty }g_k+\lim_{k\to \infty }h_k$,

which is what we want.