Hahn-Banach theorem exercise Let $X$ be a Banach space (over $\mathbb{R}$) and $u,v\in X$ such that $\|u\|=\|v\|=1$ and $\|2u+v\|=\|u-2v\|=3$. Show that there is $f\in X'$ of unit norm such that $f(u)=f(v)=1$.
My idea is building directly the functional on $Y=\operatorname{Span}\{u,v\}$, something like $f(\alpha u+ \beta v)=\alpha+\beta$, and then use HB theorem to extend it to the whole space, but I am missing the role of the second condition and (consequently) failing to prove $\|f\|=1$.
P.S: This is not homework, just an exercise taken from a course for independent study.

It suffices to prove that $\Vert \alpha u + \beta v\Vert=|\alpha + \beta|$. 
Maybe you can proceed through the following steps (I don't know):


*

*Prove that it is true for $\alpha$ and $\beta$ integer numbers.

*Prove that it is true for $\alpha$ and $\beta$ rational numbers, based on the last step.


Finally, since $(\alpha,\beta)\rightarrow\Vert\alpha u + \beta v\Vert$ is continuous on $R\times R$ and $Q\times Q$ is a dense subset, the clonclusion follows.
