Let $X \neq \lbrace 0 \rbrace$ be a reflexive space, and let $f \in X'$. Show that there exists $x \in X$ with $\|x\|=1$ and $f(x)=\|f\|$.
The definition of a reflexive space: $X$ is called reflexive if $J_X(X)=X''$ where $J_X: X \to X''$ defined by $J_Xx(f)=f(x)$ with $x \in X$ and $f \in X'$.
How to find this such $x$?
$X'$ is a Banach space, so by Hahn Banach there is a linear functional $l$ on $X'$ with $||l||=1$ and $l(f) = ||f||$. (This can be found in Folland's real analysis book, for example). Since $X$ is reflexive, this $l$ is of the form $\hat{x}$ for some $x \in X$. So, $||x|| = ||\hat{x}|| = 1$ and $||f|| = l(f) = \hat{x}(f) = f(x)$.
