Confusion about Hahn-Banach

Question

I am confused about something related to Hahn-Banach. According to my book, one corollary of H-B is that for $X$ a real or complex normed space, there exists $f \in X'$ such that $\|f\| = 1$ and $f(x) = \|x\|$.

But, is $f$ then linear? Because for the norm we have $f(\alpha x) = \|\alpha x\| = |\alpha|\|x\|$, and $f(x+y) = \|x+y\| \leq \|x\| + \|y\|$, how can this give linearity? Since $X' = B(X,\mathbb{F}) \subseteq L(X,\mathbb{F})$, $f$ should be linear right?

Answer 1

It is not the case that $f(x)=\|x\|$ for every $x\in X$. The corollary goes like this: Given a particular $x_0\in X$, there exists $f\in X'$ such that $\|f\|=1 $ and $f(x_0)=\|x_0\|$.

Answer 2

The equation $f(x)=\|x\|$ holds for a particular $x$, not for all $x$. $f$ itself depends on the given vector $f$.