# $a (\sup q_n)=(\sup q_n)b$ provided that $aq_n=q_nb$?

Let $A$ be a unital Baer*-ring.Let $a,b$ be in $A$.

# Sup preserving maps between distributive lattices

I have been looking at categories of sup semilattices and sup preserving maps.If $A$ and $B$ are two such, the set I denote $[A,B]$ is sup preserving homomorphisms between is also a sup semilattice with pointwise sup.

# Multiplication of lim sups

Let $s_n$ and $t_n$ be bounded sequences of nonegative numbers.Prove $\text{lim sup} (s_nt_n) \leq (\text{lim sup} s_n)(\text{lim sup}$ t_n).

# When does Max=Sup ?

I know the following statement is true.$\forall S,$ (S is compact and S$\neq\emptyset \to \max(S)=\sup(S))$

In what else situations can be Max=Sup?

# How to compare sums of infimum (or supremum) of two sets?

$\sup\{x_n+y_n\}⩽\sup\{x_n\}+\sup\{y_n\}$
$\sup\{x_n+y_n\}⩾\sup\{x_n\}+\sup\{y_n\}$
$\sup\{x_n−y_n\}⩽\sup\{x_n\}−\sup\{y_n\}$
$\sup\{x_n−y_n\}⩾\sup\{x_n\}−\sup\{y_n\}$
$\sup\{x_n+y_n\}⩽\sup\{x_n\}+\inf\{y_n\}$
$\sup\{x_n+y_n\}⩾\sup\{x_n\}+\inf\{y_n\}$

Which of this expression are correct?Why or why not?

# Uniform or Holder estimates for positive Schrodinger operators

Let $\phi_\lambda(x) = \phi(\lambda x)$.Observe that $\sup |\phi_\lambda u| \leq \sup |u|$, but
$$\lim_{\lambda \to 0} \sup |\phi_\lambda u| = \sup |u|. # How do we show \sup\{ks_n:n>N\}=k\sup\{s_n:n>N\}? Let M=\sup A=\sup \{s_n : n>N\} .and B=\{ks_n : n>N\} . # Proving that sup(S+T)=sup(S)+sup(T) Let S and T be nonempty sets of real numbers and define$$S+T = \{s+t| s \in S, t \in T\}$$show that sup(S+T)=sup(S)+sup(T) if S and T are bounded above.I am thinking about a proof by contradiction such as: Let's suppose that sup(S+T) \neq sup(S) +sup(T), then either sup(S+T) > sup(S) +sup(T) OR sup(S+T) < sup(S) +sup(T) If$$sup(S+T) > sup(S) +sup(T) sup(S+T) -2\epsilon > sup(S) -\epsilon +sup(T) -\epsilon $$with \epsilon >0  \exists x_3 \in(S+T) ,  \exists x_2 \in S ,  \exists x_1 \in T such that$$ x_3 >sup(S+T) -2\epsilon > x_1 + x_2> sup(S) -\epsilon +sup(T) -\epsilon 
(using the property of supremum) statement A

Also, if $x_2 \in S$ and $x_1 \in T$, then $x_2,x_1 \in (S+T)$

and, as $x_3 > x_2+x_1$ , using the theorem of supremum, $x_2+x_1 > sup(S+T)- 2\epsilon$ statement B

However, statement B is in contradiction with statement A.

# How to show that the“$\le$” in $\lim\sup(s_t+t_n)\le\lim\sup s_n+\lim\sup t_n$ cannot be replaced by $=$?

How to show that the"$\le$" in $\lim\sup(s_t+t_n)\le\lim\sup s_n+\lim\sup t_n$ cannot be replaced by $=$?I think this means we need to show an example that $\lim\sup(s_t+t_n)\lt\lim\sup s_n+\lim\sup t_n$.