The space $\ell^p(a_n)$

Question

Let $a_n$ be a positive sequence in $\mathbb{R}$ and

$$ \ell^p(a_n)=\{ x=(x_j) : \sum_{n=1}^{\infty}{a_n} \vert x_n \vert^p < \infty\} $$

with the norm $\Vert x \Vert = \Big( \sum_{n=1}^{\infty}{a_n} \vert x_n \vert^p \Big)^{\frac{1}{p}}$.

Prove that the space is a banach space.

I have already checked all the properties of a norm but I have no idea how to show that every cauchy sequence in the space converges.

Would appreciate any hints/help

Answer 1

Let $\{x_k\}_k \subset \ell^p(a)$ be a Cauchy sequence, i.e. $$\forall \epsilon > 0 \: \exists \: N \: \text{such that} \: h,k \geq N \Rightarrow ||x_h - x_k||^p = \sum_{n}a_n|x_h(n) - x_k(n)|^p \leq \epsilon^p$$ from this insequality we get $$a_n|x_h(n) - x_k(n)|^p \leq \sum_{n = 1}^M a_n|x_h(n) - x_k(n)|^p \leq \epsilon^p \tag{1}$$ for every $n$ and $M$, therefore $\{(a_n)^{1/p}x_k(n)\}_k$ is a Cauchy sequence of real numbers and for every $n$ $$\exists \lim_{k \to \infty} (a_n)^{1/p}x_k(n) = y(n)$$ Let us define $x \in \mathbb{R}^\mathbb{N}$ as $x(n) := \frac{y(n)}{(a_n)^{1/p}}$, taking the limit $h \to \infty$ in $(1)$ we get $$\sum_{n = 1}^M a_n|x(n) - x_k(n)|^p \leq \epsilon^p \tag{2}$$ for every $M$ and $k \geq N$. Now we will prove that $x \in \ell^p(a)$, let $k \geq N$ $$\Big(\sum_{n=1}^M a_n|x(n)|^p\Big)^{1/p} \leq \Big(\sum_{n=1}^M a_n|x(n) - x_k(n)|^p\Big)^{1/p} + \Big(\sum_{n=1}^M a_n|x(n)|^p\Big)^{1/p} \leq \epsilon + ||x_k||$$ for every $M$ and by passing to the limit $M \to \infty$ we conclude. Taking the limit $M \to \infty$ in $(2)$ gives $$||x - x_k|| \leq \epsilon$$ for every $k \geq N$ so $x_k \to x$ in $\ell^p(a)$

Answer 2

Your problem is just a corollary of the following fact:

Theorem: Let $(X,\mathcal{M},\mu)$ be a measure space and $p\in[1,\infty)$, then $L^{p}(\mu)$ is a Banach space.

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For your problem: Let $X=\mathbb{N}$, $\mathcal{M}=\mathcal{P}(\mathbb{N})$, and $\mu$ be defined such that $\mu(\{k\})=a_{k}$. Clearly, $(X,\mathcal{M},\mu)$ is a measure space. Let $f:X\rightarrow\mathbb{R}$ be a function. By Monotone Convergence Theorem, \begin{eqnarray*} \int|f|^{p}d\mu & = & \lim_{n\rightarrow\infty}\int|f|^{p}1_{\{1,2,\ldots,n\}}d\mu\\ & = & \lim_{n\rightarrow\infty}\int|f|^{p}\sum_{k=1}^{n}1_{\{k\}}d\mu\\ & = & \lim_{n\rightarrow\infty}\sum_{k=1}^{n}|f(k)|^{p}\mu(\{k\})\\ & = & \sum_{k=1}^{\infty}a_{k}|f_{k}|^{p}, \end{eqnarray*} where $f_{k}=f(k)$. Therefore, $l^{p}(a_{n})$ is precisely $L^{p}(\mu)$.