So our Analysis teacher has given us this homework:
Let $B=\{b_k\}_{k=1}^\infty $ be a sequence of real numbers. Prove that the following are equivalent:
a) The sequence is bounded (i.e., $B\in \ell^\infty(\mathbb N))$.
b) For every sequence $A =\{a_k\}_{k=1}^\infty \in \ell^1(\mathbb N)$ (i.e., $\sum_{k=1}^{\infty}|a_k|$ convergent) we have $A\cdot B=\{a_kb_k\}_{k=1}^\infty \in \ell^1(\mathbb N)$
And one can easily do the "$ a)\implies b) $" but the backwards implication is the one I'm trying to figure out. Our teacher insists on proving it by contradiction i.e.:
Assuming $B\notin \ell^\infty(\mathbb N)$ ($B$ not bounded) then there exists $A \in \ell^1(\mathbb N)$ so that $A\cdot B \notin \ell^1(\mathbb N)$.
I imagine such an A sequence is constructed from the B values using that it's unbounded, but no matter how many I try I can't find a general way of constructing A.
In addition the second part of the exercise is proving an 'equivalent' result but for continuous functions and I have the same issue...any idea?
Let $g $ be a Lebesgue measurable function defined on $\mathbb R$. Prove that the following are equivalent:
a) The function is bounded (i.e., $g\in L^\infty(\mathbb R)$.
b) For every function $f \in L^1(\mathbb R)$ we have $f(x) g(x)\in L^1(\mathbb R)$
Anyways thank you very much!!
Answer for the first part. Suppose $(b_k)$ is not bounded. There is a subsequence $(k_j)$ such that $|b_{k_j}| >j$ for all $j$. Let $a_k=0$ if $k \notin \{k_1,k_2,..\}$ and $a_{k_j}=\frac 1 {j b_{k_j}}$. Then $\sum |a_n| <\infty$ but $\sum |a_nb_n| =\infty$.