2
$\begingroup$

$\{x_n\}$ be a bounded sequence such that for every bounded function $\{y_n\}$,

$$\limsup\{x_{n}+y_{n}\} =\limsup\{x_{n}\}+\limsup\{y_{n}\}$$ then prove that {$x_{n}$} is convergent. This is a question asked in in real analysis exam. I started answering like this : If $\limsup\{x_n\} = L$ then for every $\epsilon$ we can find natural number N such that $x_{n} <L+ \epsilon$ for n $\geq N$ A similar result holds for given bounded sequence $\{x_n\}$ but I cannot say that $L- \epsilon <x_{n}$, $n \geq N $.

Can we prove this result from the same definition of limit ? Or should we use some other results which connects limit superior and limit of a sequence? What is the importance of boundedness of the sequence (is it given to say that limsup is not $\infty$ ?)

$\endgroup$
  • $\begingroup$ LaTeX is quite happy to do brackets, e.g. \{ and \} give $\{ \text{ and } \}$ while if n is bigger than, or equal to, N then why not out n \ge N into LaTeX to give $n \ge N$. Moreover, how about \limsup to give $\limsup$? $\endgroup$ – Fly by Night Apr 28 '14 at 16:57
3
$\begingroup$

Choosing $y_n=-2x_n$ we conclude from $$ \limsup_{n\to\infty} (x_n+y_n)=\limsup_{n\to\infty} x_n +\limsup_{n\to\infty} y_n $$ that $$ \limsup_{n\to\infty} (-x_n)=\limsup_{n\to\infty} x_n +\limsup_{n\to\infty}(-2x_n) $$ that is $$ -\liminf_{n\to\infty} x_n =\limsup_{n\to\infty} x_n -2\liminf_{n\to\infty} x_n $$ or equivalently $$ \liminf_{n\to\infty} x_n=\limsup_{n\to\infty} x_n. $$ and we are done. $\qquad\square$

Remark: We used strongly the fact that the sequence $\{x_n\}$ is bounded to insure that both $\liminf_{n\to\infty} x_n$ and $\limsup_{n\to\infty} x_n$ and we can do arithmetic operations on them. Consider what happens when $x_n=(-1)^n n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.