# For $f$ in dual space, there exists $x$ with norm 1 and $f(x)=\|f\|$ if space is reflexive (and nontrivial)

Let $X\ne\{0\}$ be a reflexive space and let $f\in X^*$, where $X^*$ is the dual of $X$. I want to know: in general, does there exist an $x\in X$ with $\|x\|=1$, and $f(x)=\|f\|$, where $\|f\|$ is defined as $\sup\{|f(x)|:x\in X,\|x\|=1\}$?

I know this is true for $\mathbb{R}^n$ with the norm from the standard inner product, but I'm wondering if it is true in general.

If the space is reflexive than the immersion $$\iota:X\to X^{**}, x\mapsto x(L):=L(x)$$ is a linear bijection. Now let $f\in X^*$ and define the following map $$L: \mathbb{R}f:=\{g\in X^*: g=\alpha f, \alpha\in \mathbb{R}\}\to \mathbb{R}, g=\alpha f\mapsto \alpha\|f\|.$$ It is well defined and continuous, moreover its norm is $1$ (check directly), so we can apply Hahn-Banach extension theorem to find an extension $$\tilde{L}:X^{**},\ \tilde{L}|_{\mathbb{R}f} = L,\ \|\tilde{L}\|= \|L\|=1.$$ By reflexivity there exists $x\in X$ such that $L(g) = x(g)$ for all $g\in X^*$ and $\|L\|=\|x\|=1$. But we done since $$f(x) = x(f) = \tilde{L}(f) = L(f) = 1\|f\|=\|f\|.$$
Yes if $X$ is reflexive, as already noted. No in general:
Let $X=C([0,2])$ and define $$\lambda f=\int_0^1f - \int_1^2 f.$$ Then $\lambda\in X^*$ and $||\lambda||=2$, but $|\lambda f|<2||f||$ if $f\ne 0$.