lim(sup) bevis
Posted: 16/09-2011 21:52
Hei.
Jeg er litt usikker på om jeg har løst følgende oppgave riktig. Setter derfor veldig stor pris på innspill/kommentarer.
Oppgave: For any two real sequences [tex]\{a_n\}, \{b_n\}[/tex], prove that
[tex]\lim_{n \to \infty} sup(a_n + b_n) \leq \lim_{n \to \infty} sup(a_n) + \lim_{n \to \infty} sup(b_n)[/tex]
provided the sum on the right is not on the form [tex]\infty - \infty[/tex].
Ok. Min løsning (på engelsk):
Consider a sequence [tex]\{n_k\}[/tex] of positive integers, such that [tex]n_1 < n_2 < n_3 < . . .[/tex]. For [tex](a_n + b_n)[/tex] there must exist a subsequence, [tex](a_n_i + b_n_i)[/tex] such that [tex]\lim_{n \to \infty}(a_n_i + b_n_i) = \lim_{n \to \infty} sup(a_n + b_n)[/tex]
However, it is not certain that [tex]\lim_{n \to \infty} (a_n_i)[/tex] as defined above is an upper limit for [tex]\{a_n\}[/tex] or that [tex]\lim_{n \to \infty} (b_n_i)[/tex] is an upper limit for [tex]\{b_n\}[/tex]. Thus we may define [tex]\lim_{n \to \infty} (a_n_i) \leq \lim_{n \to \infty}(a_n_j) = \lim_{n \to \infty} sup(a_n)[/tex] and we may define [tex]\lim_{n \to \infty} (b_n_i) \leq \lim_{n \to \infty} (b_n_j) = \lim_{n \to \infty} sup(b_n)[/tex].
This gives us:
[tex]\lim_{n \to \infty}(a_n_i + b_n_i) = \lim_{n \to \infty} sup(a_n + b_n) \leq \lim_{n \to \infty}(a_n_j) + \lim_{n \to \infty}(b_n_j) = \lim_{n \to \infty} sup(a_n) + \lim_{n \to \infty} sup(b_n)[/tex].
And the desired inequality:
[tex]\lim_{n \to \infty} sup(a_n + b_n) \leq \lim_{n \to \infty} sup(a_n) + \lim_{n \to \infty} sup(b_n)[/tex]
follows.
Setter som sagt stor pris på kommentarer/rettelser.
Jeg er litt usikker på om jeg har løst følgende oppgave riktig. Setter derfor veldig stor pris på innspill/kommentarer.
Oppgave: For any two real sequences [tex]\{a_n\}, \{b_n\}[/tex], prove that
[tex]\lim_{n \to \infty} sup(a_n + b_n) \leq \lim_{n \to \infty} sup(a_n) + \lim_{n \to \infty} sup(b_n)[/tex]
provided the sum on the right is not on the form [tex]\infty - \infty[/tex].
Ok. Min løsning (på engelsk):
Consider a sequence [tex]\{n_k\}[/tex] of positive integers, such that [tex]n_1 < n_2 < n_3 < . . .[/tex]. For [tex](a_n + b_n)[/tex] there must exist a subsequence, [tex](a_n_i + b_n_i)[/tex] such that [tex]\lim_{n \to \infty}(a_n_i + b_n_i) = \lim_{n \to \infty} sup(a_n + b_n)[/tex]
However, it is not certain that [tex]\lim_{n \to \infty} (a_n_i)[/tex] as defined above is an upper limit for [tex]\{a_n\}[/tex] or that [tex]\lim_{n \to \infty} (b_n_i)[/tex] is an upper limit for [tex]\{b_n\}[/tex]. Thus we may define [tex]\lim_{n \to \infty} (a_n_i) \leq \lim_{n \to \infty}(a_n_j) = \lim_{n \to \infty} sup(a_n)[/tex] and we may define [tex]\lim_{n \to \infty} (b_n_i) \leq \lim_{n \to \infty} (b_n_j) = \lim_{n \to \infty} sup(b_n)[/tex].
This gives us:
[tex]\lim_{n \to \infty}(a_n_i + b_n_i) = \lim_{n \to \infty} sup(a_n + b_n) \leq \lim_{n \to \infty}(a_n_j) + \lim_{n \to \infty}(b_n_j) = \lim_{n \to \infty} sup(a_n) + \lim_{n \to \infty} sup(b_n)[/tex].
And the desired inequality:
[tex]\lim_{n \to \infty} sup(a_n + b_n) \leq \lim_{n \to \infty} sup(a_n) + \lim_{n \to \infty} sup(b_n)[/tex]
follows.
Setter som sagt stor pris på kommentarer/rettelser.