Let $E$ be Banach space over $\mathbb{R}$. Let $u$ and $v$ be such that $||u||=||v||=1$ and $||2u+v||=||u-2v||=3$.
How do we show that there exists a linear functional $f$ defined on all of $E$ such that $||f||=1$, $f(u)=1$ and $f(v)=1$?
Let $E$ be Banach space over $\mathbb{R}$. Let $u$ and $v$ be such that $||u||=||v||=1$ and $||2u+v||=||u-2v||=3$.
How do we show that there exists a linear functional $f$ defined on all of $E$ such that $||f||=1$, $f(u)=1$ and $f(v)=1$?
Note first that $u$ and $v$ are linearly independent. Now consider the linear functional $g$ on the linear span $V$ of $u$ and $v$ such that $g(u) = 1$ and $g(v) = 1$. Thus $g(au + bv) = a+b$. What is the norm of this? Draw a picture of the unit ball of $V$, using the fact that it must be symmetric and convex...