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I want to show that there is an isomorphism between the dual of $\ell^1$ and $\ell^\infty$. I don't really know where to start so any help is appreciated

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    Start with a map from $l^\infty$ to the dual of $l^1$ . How do you think you would define such a map? – 2017-01-31
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    I don't know, I'm having trouble understanding what exactly the dual of $l^1$ is. I have the definition but I don't understand it entirely – 2017-01-31
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    Let $F = \mathbb{R}$ or $\mathbb{C}$. If $(V, || \cdot ||)$ is a normed vector space over $F$, the dual $V^{\ast}$ of $V$ is defined to be the set of *bounded* linear functionals, i.e. all linear transformations $f: V \rightarrow F$ for which there exists a constant $N > 0$ such that $$|f(v)| \leq N ||v||$$ for all $v \in V$. Then $V^{\ast}$ is also a normed vector space over $F$. – 2017-02-01

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To a sequence $x = (x_1, x_2, ...) \in \ell^{\infty}(\mathbb{N})$, associate a linear functional

$$T_x: \ell^1(\mathbb{N}) \rightarrow \mathbb{C}$$

by $$T_x(y_1, y_2, ...) = \sum\limits_{i=1}^{\infty} x_iy_i$$

$T_x$ is a well defined function, because

$$\sum\limits_{i=1}^{\infty} |x_iy_i| \leq ||x||_{\infty} \sum\limits_{i=1}^{\infty} |y_i| = ||x||_{\infty} \space ||y||_1$$

and so that sum convergences absolutely. Now you need to prove that $T_x$ is bounded, $T_{cx+y} = cT_x + T_y$, $||x||_{\infty} = ||T_x||$, and that the map $x \mapsto T_x$ is a surjective linear transformation (injectivity follows from being norm preserving).

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    What are the relations that $T_x$ has to adhere to in order to be isomorphic? – 2017-01-31
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    What you wrote doesn't make sense. The *function* $\ell^1(\mathbb{N}) \rightarrow \ell^{\infty}(\mathbb{N})^{\ast}$ which sends $x$ to $T_x$ is the thing you want to show is an isomorphism. You have to show that this function is a linear transformation which is injective and surjective. – 2017-02-01
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    But you said $T_x$ is a well defined function, so is x mapped onto a function? – 2017-02-01
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    Yes $\space \space$ – 2017-02-01