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.

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 $f\ne 0$.
