The question asks to show the equality/inequality of supremum/infimum of series.

Suppose that $\{a_{n,m}\}$ are nonnegative real numbers for all $n,m \in \mathbb{N}$.

  1. Suppose that for each $n$, $m \mapsto a_{n,m}$ is a nondecreasing function of $m$, i.e., $a_{n,m_1} \leq a_{n,m_2}$ when $m_1 \leq m_2$. Show that $$\sup_{m \in \mathbb{N}} \sum_{n = 1}^\infty a_{n,m} = \sum_{n =1 }^\infty \sup_{m \in \mathbb{N}} a_{n,m} $$ Regardless of whether or not the sides are finite or infinite.

  2. Suppose that for each $n$, $m \mapsto a_{n,m}$ is nonincreasing, i.e., $a_{n,m_1} \leq a_{n,m_2}$ for all $n \in \mathbb{N}$ and all $m_2 \leq m_1$. Show that $$ \sum_{n =1 }^\infty \inf_{m \in \mathbb{N}} a_{n,m} \leq \inf_{m \in \mathbb{N}} \sum_{n = 1}^{\infty} a_{n,m} $$ Give an example to show that the inequality is strict.

My attempt in part 1 is to show $\sup_{m \in \mathbb{N}} \sum_{n = 1}^\infty a_{n,m} \leq \sum_{n =1 }^\infty \sup_{m \in \mathbb{N}} a_{n,m}$ and also $\sum_{n =1 }^\infty \sup_{m \in \mathbb{N}} a_{n,m} \leq \sup_{m \in \mathbb{N}} \sum_{n = 1}^\infty a_{n,m}$. I am not sure how to apply the concept of supremum/infimum to equality/inequality of series. Any hints would be very helpful.


1 Answer 1


For any $n,m$, $a_{n,m} \leq \sup_r\,a_{n,r}$. Therefore $\sum_n{a_{n,m}} \leq \sum_n{\sup_r\,a_{n,r}}$ and the rest is textbook definition of sup to get the first inequality.

For the reverse inequality, show that for every $N \geq 1$, $\sum_{n=1}^N{\sup_r\,a_{n,r}} \leq \sup_r\,\sum_{n=1}^N{a_{n,r}}$ (show that in this case the sups are limits), then that the LHS is not greater than $\sup_r\,\sum_{n \geq 1}{a_{n,r}}$, and conclude.

For 2), you just need to show as above that the RHS is greater than any finite sum in the LHS.

For the counter-example, find for instance an example where all $a_{n,m}$ converge to zero, but all the $\sum_n{a_{n,m}}$ are infinite.


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