Is a majorable subset in lp is relatively compact?

Question

Let A be a nonempty set in the sequence space lp with 1 ≤ p < +∞. I want to show that if A is majorable then it is relatively compact. A is majorable by definition if there exists y ∈ lp such that |x(k)| ≤ |y(k)|, for all x ∈ A, k ∈ N. I know that A is relatively compact if and only if it is totally bounded and if and only if every sequence in A has a subsequence that converges in lp. I thought that i could use this theorem to prove my thesis but i couldn't come up with a solution.

Answer 1

Let $\{x_n\}_n$ be any sequence in $A$. For every $k$ one has that the sequence of scalars $$ \{x_n(k)\}_{n\in {\mathbb N}} $$ is bounded by $|y(k)|$, so it admits a converging sub-sequence. By doing this for each $k$, and using Cantor's diagonal argument, we may find a sequence $n_1<n_2<\cdots \quad$ of natural numbers, such that $$ z(k) := \lim_{i\to \infty } x_{n_i}(k) $$ exists for every $k$. Since $|z(k)|\leq |y(k)|$, it is obvious that $z := (z(k))_k$ lies in $\ell^p$. Moreover $$ \lim_{i\to \infty }\|x_{n_i} - z\|^p = \lim_{i\to \infty }\sum_{k=1}^\infty |x_{n_i}(k) - z(k)|^p = 0, $$ by Lebesgue's Dominated Convergence Theorem. This proves that every sequence in $A$ has a converging sub-sequence (to a point in $\ell^p$), so $A$ is relatively compact.