Consider $a:=\{a_n\}_{n=0}^{\infty}\in\mathbb{R}^{\mathbb{N}}_{+}$ a decreasing sequence with $a_n\to 0$. Consider the following space of sequences $$ X_a:=\left\{x:=\{x_n\}_{n=0}^{\infty}\in\mathbb{C}^{\mathbb{N}}: \lim_{n\to\infty}\frac{x_n}{a_n}=0\right\},$$ endowed with the norm $\|x\|_a:=\sup_{n\in\mathbb{N}}\frac{|x_n|}{a_n}$. Show that $X_a$ is a Banach space.
My attempt:
I have verified that the given $\|\cdot\|_a$ is in fact a norm. It remains to show that $X_a$ is complete, which we can verify using sequences. Take a Cauchy sequence $\{x^{(n)}\}_n$ in $X_a$, where $x^{(n)}=\{x^{(n)}_k\}_k\in X_a$. I need to show that there is some $x\in X_a$ such that $x^{(n)}\to x$ as $n\to \infty$. From the definition of Cauchy sequences, we know that $\forall \varepsilon >0$, $\exists N\in \mathbb{N},\forall m,l\ge N$ we have $$ \sup_{k\in\mathbb{N}}\frac{|x_k^{(m)}-x_k^{(l)}|}{a_k}<\varepsilon.$$ In other words, $\forall k\in\mathbb{N}: |x_k^{(m)}-x_k^{(l)}|/a_k<\varepsilon$. This means that $\forall k\in \mathbb{N}$ the sequences $\{x_k^{(n)}/a_k\}_n$ are Cauchy in $\mathbb{C}$, hence $x_k^{(n)}/a_k\to x_k\in\mathbb{C}$ as $n\to\infty$. We claim now that $x^{(n)}\to (x_0a_0,x_1a_1,\dots)=: x$. Verification: $$ \|x^{(n)}-x\|_a=\sup_k\frac{|x^{(n)}_k-x_ka_k|}{a_k}=\sup_k \left|\frac{x_k ^{(n)}}{a_k}-x_k\right|\to 0\quad \text{ as } n\to \infty.$$
I have not used that $x^{(n)}\in X_a$ so I was wondering where this could be incorporated in the proof above.
Thank you.
If I'm reading this correctly, your argument boils down to $$\lim_{n\to\infty} \|x^{(n)}-x\|_a = \lim_{n\to\infty} \sup_{k\in\Bbb{N}} \left|\frac{x_k ^{(n)}}{a_k}-x_k\right| = \sup_{k\in\Bbb{N}} \underbrace{\lim_{n\to\infty} \left|\frac{x_k ^{(n)}}{a_k}-x_k\right|}_{=0} = 0$$ by definition of $x_k$. However, it isn't clear why you can swap the limit and the supremum. This is where you will need to use $x^{(n)} \in X_a$.
Also note that the map $$T : X_a \to c_0, \qquad T(x_n)_n := \left(\frac{x_n}{a_n}\right)_n$$ is an isometric isomorphism of your space $X_a$ to the Banach space $c_0$ of all sequences converging to $0$ equipped with the sup-norm. Therefore $X_a$ is also Banach.