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Let $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ be two sequences of $\color{red}{\overline{\mathbb{R}}:=\mathbb{R}\cup\{-\infty,\infty\}}$. Suppose that the sum $a_n+b_n$ exists for all $n\in\mathbb{N}$. Prove that if the sum $\inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$ exists, then

$$\inf_{n\in\mathbb{N}}\sup_{k\geq n}(a_k+b_k)\leq \inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$$

I want to prove the above inequality without using the notion of limits! Remember that the following operations ARE NOT defined: $(\infty)-(\infty)$, $(-\infty)+(\infty )$, $(\infty )+(-\infty )$ and $(-\infty)-(-\infty )$.

The order relation $\leq $ is defined in the following way: given any $x,y\in\overline{\mathbb{R}}$ we say that $x\leq y$ if, and only if, at least one of the following propositions is true:

  1. If $x,y\in\mathbb{R}$, then $x\leq y$ (here $\leq$ is the usual order of the real numbers)
  2. $y=\infty$
  3. $x=-\infty$

I tried to use the inequalities below, but I couldn't solve that problem.

  1. If the sum $\sup_{k\geq n}a_k+\sup_{k\geq n}b_k$ exists, then $\sup_{k\geq n}(a_k+b_k)\leq \sup_{k\geq n}a_k+\sup_{k\geq n}b_k$ for all $n\in\mathbb{N}$
  2. If the sum $\inf_{n\in\mathbb{N}}a_n+\inf_{n\in\mathbb{N}}b_n$ exists, then $\inf_{n\in\mathbb{N}}a_n+\inf_{n\in\mathbb{N}}b_n\leq \inf_{n\in\mathbb{N}}(a_n+b_n)$

Thank you for your attention!

rfloc
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  • https://math.stackexchange.com/questions/408135/prove-limsup-limits-n-to-infty-a-nb-n-le-limsup-limits-n-to-infty – stoic-santiago Feb 02 '21 at 03:00
  • @strawberry-sunshine The answers in that link are using the notion of limits! I want an proof without using the notions of limits! – rfloc Feb 02 '21 at 03:14
  • In particular, see this answer: https://math.stackexchange.com/a/3290688/425395 Just replace the last step of taking limits by taking the infimum over $n\in\mathbb N$ of both sides of the inequality. – stoic-santiago Feb 02 '21 at 03:34
  • @strawberry-sunshine If you read my question you'll see that I already proved that $\sup_{k\geq n}(a_k+b_k)\leq \sup_{k\geq n}a_k+\sup_{k\geq n}b_k$. However I don't know how to use this inequality to prove that $\inf_{n\in\mathbb{N}}\sup_{k\geq n}(a_k+b_k)\leq \inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$. – rfloc Feb 02 '21 at 03:47

1 Answers1

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Let us denote $$ \newcommand{\newsup}{\mathop{\smash{\mathrm{sup}}}} \begin{align} A &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}a_k \, ,\\ B &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}b_k \, , \\ C &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}(a_k+b_k) \, . \end{align} $$ If $A = +\infty$ or $B = +\infty$ then $A+B=+\infty$ (since the sum is assumed to be defined), so that $C \le A+B$ holds.

Now consider the case that $A, B \in \Bbb R \cup \{ - \infty \}$. The idea is to show that $$ \tag{*} \forall x, y \in \Bbb R: \left(x > A \text{ and } y > B \implies C \le x + y \right)\, . $$ It is not difficult to see that $(*)$ implies $C \le A+B$:

  • If $A = -\infty$ then $C < x + y$ for fixed $y \in \Bbb R$ and all $x \in \Bbb R$, so that necessarily $C = -\infty$.
  • Similarly if $B= -\infty$.
  • Otherwise both $A$ and $B$ are real numbers and $(*)$ implies $C \le (A+\epsilon)+( B+\epsilon)$ with arbitrarily small $\epsilon > 0$.

It remains to prove $(*)$: If $x > A$ and $y > B$ then, using the definition of the infimum, $$ \exists n_1 \in \Bbb N: \newsup_{k\geq n_1}a_k < x \, , \\ \exists n_2 \in \Bbb N: \newsup_{k\geq n_2}b_k < y \, . $$ It follows that $$ a_k + b_k \le x+y $$ for all $k \ge \max(n_1, n_2)$ and therefore $$ C \le \sup_{k\geq \max(n_1, n_2)}(a_k+b_k) \le x +y \, . $$

Martin R
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