Assume $(X_i,\mathcal{T}_i)$, $i \in I$ is a family of topological vector spaces and $X:=\prod\limits_{i\in I} X_i$ with the product topology $\mathcal{T}$. Let $\pi_i: X \rightarrow X_i$ be the canonical projections and $\iota_i: X_i \rightarrow X$ the canonical embeddings.
I want to prove that for all elements of the topological dual space $f \in (X,\mathcal{T})'$ there are finitely many indices $i_1,...,i_m \in I$ and $f_{i_1},...,f_{i_m} \in (X_i,\mathcal{T}_i)'$ with $f=\sum\limits_{j=1}^m f_{i_j} \circ \pi_{i_j}$.
My idea so far was to define $f_i:=f\circ \iota_i$ which is linear and continuous and then to show that the set $Y:=\{(x_i)_{i\in I}:x_i=0 ~\text{for almost all}~ i \in I\}$ is dense in the set $X$. Then I only need to show the the equality for the element in $Y$.
$\forall y \in Y: y = \sum\limits_{i\in I} (\iota_{i} \circ \pi_{i})(y)$
This sum is finite because there are only finitely many components which are not zero, therefore I can write the following:
$\forall y \in Y: f(y) = \sum\limits_{i\in I} f((\iota_{i}(\pi_{i}(y))) = \sum\limits_{i\in I} (f_i \circ \pi_{i})(y)))$
Can someone please help me how to get the indices $i_1,...,i_m$ and reduce the sum I have got so far to the expression which I need?
You have to use the continuity and hence you need to know how the neighbourhoods of $0$ in the product look like: They are of the form $\bigcap_{i\in F} \pi_i^{-1}(U_i)$ for a FINITE set $F\subseteq I$ and $0$-neighbourhoods $U_i$ in $X_i$. If a functional is bounded on such a set it only depends on the coordinates $(x_i)_{i\in F}$.