Show that $\overline{\lim} (x_n+y_n)\leq \overline{\lim}x_n+\overline{\lim}y_b$, with $\overline{\lim}x_n=\inf_n \sup_{k\geq n}x_k$ by definition.
In Solution Royden, proves it how:
$\forall n\forall k\geq n, x_k+y_k\leq \sup_{k\geq n}x_k+\sup_{k\geq n}y_k$, then $\sup_{k\geq n}(x_k+y_k)\leq \sup_{k\geq n}x_k+\sup_{k\geq n}y_k$
Therefore $\inf_n\sup_{k\geq n}(x_k+y_k)\leq \inf_n\sup_{k\geq n}x_k+\inf_n\sup_{k\geq n}y_k$.
Why $\inf_n\sup_{k\geq n}(x_k+y_k)\leq \inf_n\sup_{k\geq n}x_k+\inf_n\sup_{k\geq n}y_k$? I dont see this.
thanks in advamce
Best Answer
Note that for two sequences $a_n$ and $b_n$, it is always true that:
$\sup (a_k+b_k) \leq \sup (a_k)+ \sup(b_k)$
Just by $\{ a_k+b_k :k\geq n \}\subseteq \{ a_k+b_l: k,l\geq n \}$
So you can conclude that $\sup_{k\geq n}(x_k+y_k)\leq \sup_{k\geq n}x_k+\sup_{k\geq n}y_k$.
Denote $c_n:=\underset{k\geq n}{\inf} a_k$ and $d_n:=\underset{k\geq n}{\inf} b_k$
Assuming that $\infty>\inf c_n>-\infty$ and $\infty>\inf d_n>-\infty$, for all $\epsilon>0$ there exists $i$ and $j$ such that:
$d_j<\inf d_n+\frac{\epsilon}{2}$ and $c_i\leq \inf c_n+\frac{\epsilon}{2}$. Then:
$\inf(c_n+d_n)\leq c_i+d_j \leq \inf c_n+ \inf d_n +\epsilon$
Since this is true for all $\epsilon>0$, this must mean that:
$\inf(c_n+d_n)\leq \inf c_n+ \inf d_n$, and thus:
$\inf_n\sup_{k\geq n}(x_k+y_k)\leq \inf_n\Big(\sup_{k\geq n}x_k+ \sup_{k\geq n}y_k \Big) \leq\inf_n\sup_{k\geq n}x_k+\inf_n\sup_{k\geq n}y_k$