We are looking at directed graphs with no loops or parallel edges, but given two vertices $x$ and $y$, we allow the presence of both the edge $(x, y)$ and $(y, x)$. Thus, if $G$ is a directed graph on $n$ vertices, $|E(G)| \le n(n-1)$.

A tournament is a directed complete graph where for any pair of vertices $x$ and $y$, exactly one of the edges $(x, y)$ and $(y, x)$ is present.

The question is, given a fixed tournament $T$, how many edges (as a function of $n$) must a graph on $n$ vertices have in order to force the existence of $T$ as a subgraph?

An obvious obstruction is the following. Assume $T$ has $k$ vertices, and let $G$ be any undirected graph which does not have $K_k$ as a subgraph. Let $\bar{G}$ be obtained from $G$ by adding two directed edges, one in each direction, for every edge of $G$. Then $\bar{G}$ does not have $T$ as a subgraph. Thus, the extremal function for $T$ is at least twice the extremal function for $K_k$ in undirected graphs. Could this be the correct bound?

An easy induction argument shows that if $T$ has 3 vertices, then $|E(G)| > n^2/2$ implies $G$ contains $T$ as a subgraph. Thus, the extremal function for $T$ is exactly twice the extremal function for the presence of a triangle.

The final question is, the above question seems quite basic, and I would be surprised if no one had considered it before. Does anyone know of a reference looking at these questions?