Question

Prove that (i) \(\overline{\lim }\left(-x_{n}\right)=-\underline{\lim } x_{n}\) and (ii) \(\overline{\lim }\left(a x_{n}\right)=a \cdot \overline{\lim } x_{n}\) if \(0 \leq a<+\infty\).

Short answer

(i) True because negation reverses roles of sup and inf; (ii) True because scalar can be factored out if \(a \geq 0\).

Step-by-step solution

Understand the Supremum and Infimum Concepts

To solve these problems, it's important to understand the meaning of the terms \(\overline{\lim}\) and \(\underline{\lim}\). The \(\overline{\lim}\) (limit superior) of a sequence is the smallest value where values of the sequence above it appear infrequently. Conversely, the \(\underline{\lim}\) (limit inferior) is the largest value where values below it occur infrequently. These concepts are fundamentally explained through supremum and infimum over the tail ends of sequences.

Express the Limit Superior and Inferior

Given a sequence \(x_n\), the limit superior, \(\overline{\lim} x_n = \lim_{n \to \infty} \sup_{k \geq n} x_k\), is the limit of the supremum of the subsequences. Similarly, the limit inferior is \(\underline{\lim} x_n = \lim_{n \to \infty} \inf_{k \geq n} x_k\). The negative of \(x_n\) turns everything upside down, exchanging the roles of supremum and infimum.

Prove (i): Relation of Negatives and Limits

To prove \(\overline{\lim}(-x_n) = -\underline{\lim}x_n\), observe that when you take the negative of each element in a sequence, the infimum becomes the supremum of the negated sequence. Therefore, the limit of the supremum of the negated sequence \(-x_n\) becomes the negative of the limit inferior of the original sequence. Thus, \(\overline{\lim}(-x_n) = -\underline{\lim}x_n\).

Prove (ii): Scalar Multiplication and Limit Superior

For the second proof, note that by definition, \(\overline{\lim}(ax_n) = \lim_{n \to \infty} \sup_{k \geq n} (ax_k)\). If \(a\) is a non-negative scalar, supremum can be factored out as a constant multiplier, thus \(a \cdot \sup_{k \geq n} x_k = \sup_{k \geq n} (a x_k)\). Hence, \(\overline{\lim}(a x_n) = a \cdot \overline{\lim} x_n\) if \(0 \leq a < +\infty\).

Key concepts

Supremum and Infimum: Understanding the Limits

The concepts of supremum and infimum are essential in understanding the behavior of sequences in mathematics. The supremum, or the least upper bound of a set, is the smallest number greater than or equal to every number in the set. On the other hand, the infimum, or greatest lower bound, is the largest number that is less than or equal to every number in the set. These concepts become especially important when studying the limit superior (\(\overline{\lim}\)) and limit inferior (\(\underline{\lim}\)) of a sequence.

The limit superior of a sequence, often denoted as \(\overline{\lim} x_n\), captures the trend of the sequence where it frequently appears, or seldom surpasses, as the sequence progresses. It is essentially the limit of the supremum of the tails of the sequence as the index goes to infinity. Conversely, the limit inferior, denoted by \(\underline{\lim} x_n\), is determined by the infimum of the subsequences’ tails as the sequence stretches to infinity.

By examining these endpoints, we gain insight into the overall behavior of the sequence in the long run. They tell us how high values cluster (in the case of limit superior) and how low values bunch together (in the case of limit inferior), providing a comprehensive view of the sequence's bounds over an extended run.

Scalar Multiplication of Sequences: The Effect on Limits

Scalar multiplication, in the context of sequences, refers to the process of multiplying each element of a sequence by a constant number, known as the scalar. When dealing with limit superior or inferior, understanding the impact of scalar multiplication is crucial since it affects the sequence's overall trend and bounds.

If we consider a sequence \(x_n\) and a non-negative scalar \(a\), scalar multiplication results in a new sequence \(ax_n\). The interesting part happens when considering the limits at infinity for such sequences. Specifically, the limit superior for a scalar multiplied sequence, \(\overline{\lim}(ax_n)\), equates to multiplying the original sequence's limit superior by the scalar: \(a \cdot \overline{\lim} x_n\), assuming \(0 \leq a < +\infty\).

This relationship holds because when multiplying by a scalar, especially a non-negative one, the order and relative positions of sequence elements in terms of upper bounds remain unchanged, just magnified by the scalar. This property makes calculations involving limits more straightforward while preserving the sequence’s growth trend dictated by \(\overline{\lim} x_n\).

Sequence Negation in Limits: Flipping Bounds

When you negate a sequence, you are essentially creating a new sequence where each element is the negative of the original sequence. This operation affects how we calculate limit superior and inferior because the reversal alters the bounds we previously established.

Consider the sequence negation \(-x_n\), which transforms the setup because the largest values of \(x_n\) become the smallest values of \(-x_n\) and vice versa. This swapping means that the infimum of \(x_n\) turns into the supremum of \(-x_n\), leading us to \(\overline{\lim}(-x_n) = -\underline{\lim}x_n\). In essence, for the negated sequence, the limit superior, which literally speaks to the upper bound, now reflects the flipped order, essentially reversing to align with the negative trend.

This concept is valuable when comparing original sequences to their negations since the negation impacts how we interpret reaching boundaries and shifting trends over infinite stretches.