Showing that the dual of a (general) product of normed spaces is isomorphically isometric to the product of the duals This is Problem III.5.4 in Conway's $\textit{Functional Analysis}$:
Let $\{X_i\}_{i\in I}$ be a collection of normed spaces. If $1 \leq p < \infty$ and $q$ is the conjugate exponent to $p$, then $\left( \oplus_p X_i \right)^*$ is isometrically isomorphic to $\oplus_q X_i^*$.
Here, $\oplus_p X_i = \left\{ \textbf{x} \in \prod\limits_{i \in I}X_i : \lVert \textbf{x} \rVert := \left[ \sum\limits_{i\in I} \lVert x_i\rVert^p \right]^{1/p} < \infty  \right\}$.
I'm fairly certain that I have to do this directly by producing a candidate operator $$T : \left( \oplus_p X_i \right)^* \to \oplus_q X_i^*$$ then showing it satisfies what I need, but I'm having difficulty doing so. Any help would be appreciated!
 A: It's easier to start by defining a map going the other way. Define $S: \oplus_q X_i^* \to (\oplus_p X_i)^*$ by setting for $f = (f_i)_{i \in I} \in \oplus_q X_i^*$,
$$Sf(x) = \sum_{i \in I} f_i(x_i)$$
where we notice that the sum converges absolutely since $\sum_{i \in I} |f_i(x_i)| \leq \|f\|_q \|x\|_p$ by Holder's inequality. Notice that this also proves that $\|S\| \leq 1$.
The rest of this answer will show that $S$ is an isometry. I'm going to only deal with the case $1< p < \infty$. The case $p = 1$ follows by adaptations to the following arguments.

Lemma: $S$ is an isometric embedding. 

Proof: Pick a non-zero $f$ in $\oplus_q X_i^*$. Let $A = \{i \in I: f_i \neq 0\}$ and for each $i \in A$, pick $y_i \in X_i$ such that $\|y_i\| = 1$ and $f_i(y_i) > (1- \varepsilon) \|f_i\|$ and let $x_i = \|f_i\|^{q-1}y_i$. For $i \not \in A$, let $x_i = 0$. Then $$\|x\|_p^p = \sum_{i \in A} \|f_i\|^{p(q-1)} = \sum_{i \in I} \|f_i\|^q = \|f\|_q^q < \infty$$
so that $x \in \oplus_p X_i$ and $\|x\|_p = \|f\|_q^{q/p} = \|f\|_q^{q-1}$. Then
\begin{align*}
\|Sf\| \geq \frac{|Sf(x)|}{\|x\|_p} = \|x\|_p^{-1} \sum_{i \in I} f_i(x_i) > (1-\varepsilon) \|x\|_p^{-1} \|f\|_q^q = (1- \varepsilon) \|f\|_q.
\end{align*}
Since $\varepsilon > 0$ was arbitrary and $\|S\| \leq 1$ we conclude that $\|Sf\| = \|f\|_q$ which proves the Lemma.

Lemma: $S$ is surjective

Proof: Fix $\phi \in (\oplus_p X_i)^*$. We want to show that there is an $f \in \oplus_q X_i^*$ such that $Sf = \phi$. By the definition of $S$, the only candidate for $f$ is defined by $f_i = e_i^* \phi$ where $e_i: X_i \to \oplus_i X_i$ is the obvious isometric embedding. To conclude the proof we need to check that $f$ defined this way is an element of $\oplus_q X_i^*$ which means we need to check that 
$$\sum_{i \in I} \|f_i\|^q < \infty.$$
To do this, it suffices to find a constant $C$ such that for $J \subset I$ with $|J|< \infty$, we have that $\sum_{i \in J} \|f_i\|^q \leq C$. 
As before, we can find $x_i \in X_i$ with $\|x_i\| = \|f_i\|^{q-1}$ and $f_i(x_i) > \frac12 \|f_i\|^q$. Hence for a finite subset $J \subset I$, we have
\begin{align*}\sum_{i \in J} \|f_i\|^q < 2 \sum_{i \in J} f_i(x_i) =& 2 \phi( \sum_{i \in J} e_i x_i ) \\\leq& 2 \|\phi\| \bigg\| \sum_{i \in J} e_i x_i \bigg \| \\=&  2 \|\phi\|  \bigg( \sum_{i \in J} \|f_i\|^{p(q-1)} \bigg)^{1/p} \\=& 2\|\phi\|  \bigg( \sum_{i \in J} \|f_i\|^{q} \bigg)^{1/p}
\end{align*}
which implies that 
$$\bigg(\sum_{i \in J} \|f_i\|^q \bigg)^{1/q} = \bigg(\sum_{i \in J} \|f_i\|^q \bigg)^{1 - 1/p} \leq 2\|\phi\|.$$
Hence $f \in \oplus_q X_i^*$ and $Sf = \phi$ so that $S$ is surjective, as desired. 
