Describe the topological dual of a norm space

Question

Let \begin{equation*} X=\left\{x=(x_{n})\in\mathbb{K}^{\mathbb{N}}\phantom{a}|\phantom{a}\sum_{n=1}^{\infty}{2^{n}|x_{n}|}<\infty\right\}. \end{equation*} be. Prove that the function \begin{align*} ||\cdot||_{X}&\colon X\to\mathbb{R}\\ &\phantom{a}x\longmapsto\sum_{n=1}^{\infty}{2^{n}|x_{n}|} \end{align*} is a norm in X. Also, describe the space $X^{*}$ and the norm $||\cdot||_{X^{*}}.$

Attempt: It is clear that $ || \cdot ||_{X} $ is a norm in X.

Let $$\phi\colon l^{\infty}\to X^{*}$$ be given by $$\phi_{y}(x)=\sum_{n=1}^{\infty}{x_{n}y_{n}},\phantom{a}\forall y\in l^{\infty}.$$

Note that for every $y\in l^{\infty},$ we have \begin{align*} |\phi_{y}(x)|=\left|\sum_{n=1}^{\infty}{x_{n}y_{n}}\right|&\leq\sum_{n=1}^{\infty}{|x_{n}y_{n}|}\\ &=\sum_{n=1}^{\infty}{|x_{n}||y_{n}|}\\ &\leq\sum_{n=1}^{\infty}{2^{n}|x_{n}||y_{n}|}\\ &\leq ||x||_{X}||y||_{\infty}. \end{align*}

The above argument, since $ x \in X, $ shows that $ \phi $ is well defined and linear. Therefore, we have $$||\phi_{y}||\leq ||y||_{\infty},\phantom{a}\forall y\in l^{\infty}.$$ Thus, $ \phi $ is continuous.

I have a doubt. Is the $ \phi$ function injective? I ask why I will use it later.

Let $x^{*}\in X^{*}$ be. Define the sequence $$y=(y_{n})=x^{*}(e_{n}),\phantom{a}\forall e_{n}\in X.$$

Notice that if $ m \in \mathbb {N},$ then \begin{equation*} \sum_{n=1}^{m}|x^{*}(e_{n})|=\sum_{n=1}^{m}{a_{n}x^{*}(e_{n})}, \end{equation*} with \begin{equation*} a_{n}=\left\{\begin{matrix} \dfrac{|x^{*}(e_{n})|}{x^{*}(e_{n})} & x^{*}(e_{n})\neq 0\\ 0 & \text{opposite case} \end{matrix}\right. \end{equation*} So, \begin{align*} \sum_{n=1}^{m}{|x^{*}(e_{n})|}&=\sum_{n=1}^{m}{a_{n}x^{*}(e_{n})}\\ &=x^{*}\left(\sum_{n=1}^{m}{a_{n}e_{n}}\right)\\ &\leq ||x^{*}||\left|\left|\sum_{n=1}^{m}{a_{n}e_{n}}\right|\right|_{X}\\ &\leq ||x^{*}|| \end{align*} Doing $ m \rightarrow \infty, $ we have \begin{equation*} \sum_{n=1}^{\infty}{|x^{*}(e_{n})|}\leq ||x^{*}||. \end{equation*}

The above argument guarantees that the series $\sum_{n=1}^{\infty}{x^{*}(e_{n})}$ is convergent. So, $y=(y_{n})=x^{*}(e_{n})\in l^{\infty}.$ Note that $||y||_{\infty}\leq ||x^{*}||.$ Let $x\in X$ be. Then \begin{equation*} T_{y}(x)=x^{*}(x)=x^{*}\left(\sum_{n=1}^{\infty}{x_{n}e_{n}}\right)=\sum_{n=1}^{\infty}{x_{n}x^{*}(e_{n})}=\sum_{n=1}^{\infty}{x_{n}y_{n}} \end{equation*} and $\phi$ is surjective.

Let $y\in l^{\infty}.$ Then $x^{*}=\phi(y).$ Since $ \phi $ is surjective, then $ y '\in l ^{\infty} $ exists, with $||y'||_{\infty}\leq ||x^{*}||,$ such that $\phi(y')=x^{*}.$ Since $ \phi $ is injective, we have $y=y'$ and $||y||_{\infty}\leq ||\phi_{y}||.$

Thus, $||\phi_{y}||=||y||_{\infty}$ and $\phi$ is an isometric isomorphism.

The argument is correct?

Any hints would be appreciated.

Answer 1

One of the (many) Riesz representation theorems says that

Let $(X,\mathcal A,\mu)$ be a measure space. Then for $1<p<\infty$ the dual of $L^p(\mu)$ is isometrically isomorphic to $L^q(\mu)$, where $q$ is the conjugate exponent of $p$. The same holds for $p=1$, provided that $\mu$ is a sigma-finite measure.

You fall in that case: Define a measure on $\mathbb{N}$ as $\mu(\{n\})=2^n$ and extend it to $P(\mathbb{N})$ by $\mu(E)=\sum_{n\in E}2^n$. Your space $X$ is just $L^1(\mu)$. It is immediate that $\mu$ is sigma-finite, so the dual is $L^\infty(\mu)$. Note that this does not differ from $\ell^\infty$ as a set and that the norms are equal, so the dual of $X$ is simply $\ell^\infty$ with the usual $\|\cdot\|_\infty$ norm.

PS: In case you do not know abstract measure theory, I suppose you are familiar with the result $(\ell^1)^*=\ell^\infty$ and its proof. You can always mimic this proof with $X$ replacing $\ell^1$; if you find trouble with the details, add them to your post!