In "Calculus of Variations" by F. Rindler I learned about the following result (direct method of calculus of variations):
If $X$ is a topological space, $f: X \to \mathbb{R}$ lower semicontinuous and coercive, then $f$ obtains a minimum in $X$. (1)
Here, $f$ is said to be coercive, if the set of all $u \in X$ satisfying $ f(u)\leq \Lambda $ is relatively sequentially precompact $\forall \Lambda \in \mathbb{R}$ (i.e. every sequence of this set admits a convergent subsequence). The proof is analogous to the one of Weierstraß' theorem.
Later in the book an analogous theorem is formulated:
If $X$ is a reflexive Banach space, $f: X \to \mathbb{R}$ weakly lower semicontinuous and weakly coercive, then $f$ obtains a minimum in $X$. (2)
(Weakly coercive is defined as above with weakly convergent subsequences.)
Now I don't see why we require $X$ to be a reflexive Banach space. I know this is useful for later applications, but do we really need this for the existence of a minimum? In the first theorem (1) we dealed with general topological spaces, so doesn't include (1) the second theorem (2) ?
You are right, when it is formulated in that way, it suffices if $X$ is a normed space. The reason one may want to assume that $X$ is a reflexive Banach space, is because the weak coercivity then follows from the property \begin{equation}\lim_{||x||\to\infty} f(x)=\infty,\end{equation} which is much easier to verify in practice. Namely, the set $$\{u\in X \, |\, f(u)\leq \Lambda\}$$ is then bounded, and hence, relatively sequentially precompact by the Banach-Alaoglu theorem.