For context I am in the first week of a linear analysis course.
Consider the following:
Consider $\ell_p$ for $p\in [1, \infty)$. What is $\ell_p^*$? Suppose $q$ is the \textcolor{\autocolortwo}{conjugate exponent} of $p$, i.e. $$ \frac{1}{q} + \frac{1}{p} = 1. $$ (if $p = 1$, define $q = \infty$) It is easy to see that $\ell_q \subseteq \ell_p^*$ by the following: Suppose $(x_1, x_2, \cdots) \in \ell_p$, and $(y_1, y_2, \cdots)\in \ell_q$. Define $y(\mathbf{x}) = \sum_{i = 1}^\infty x_i y_i$. We will show that $y$ defined this way is a bounded linear map. Linearity is easy to see, and boundedness comes from the fact that $$ \|y\|_{\ell_p^*} = \sup_{\mathbf{x}}\frac{\|y(\mathbf{x})\|}{\|\mathbf{x}\|_{\ell_p}} = \sup_{\mathbf{x}} \frac{\sum x_i y_i}{\|\mathbf{x}\|_{\ell_p}} \leq \sup \frac{\|\mathbf{x}\|_{\ell_p}\|\mathbf{y}\|_{\ell_q}}{\|\mathbf{x}\|_{\ell_p}} = \|\mathbf{y}\|_{\ell_p}, $$ by the H"older's inequality. So every $(y_i) \in \ell_q$ determines a bounded linear map. In fact, we can show $\ell_p^*$ is isomorphic to $\ell_q$.
I am OK with everything up to "In fact". For the last part, I assume that the map discussed before will give us the isomorphism. However, I do not immediately see how this is an isomorphism. Is this something trivial or is this something I should revisit later on after I have finished the course?
