Prove the existence of a certain continuous linear functional I have the following homework problem:
Let $E$ be real normed space and consider $x,y\in E$ such that
$$\Vert x\Vert=\Vert y\Vert=1$$
$$\Vert 2x+y\Vert=\Vert x-2y\Vert=3$$
Prove that there is $f\in E^{\star}$ such that $\Vert f\Vert=1$ and $f(x)=f(y)=1$.
My attempt:
Setting $x=y$ in those relations give us a contradiction. Thus, $x\neq y$. Note that:
$$\left\Vert \frac{2}{3}x+\frac{1}{3}y\right\Vert=1$$
That is, we have two distinct points $x$ and $y$ at the boundary of the unit ball, such that a convex combination of the two still lies at the boundary of the unit ball. My proof came from the geometric observation that a supporting hyperplane for the unit ball at $x$ will contain the whole line segment $[x,y]$.
By the Hahn-Banach theorem, there is $f\in E^{\star}$ such that $\Vert f\Vert=1$ and $f(x)=1$. Now for any $z\in E$ with $\Vert z\Vert\leq1$, we have:
$$f(z-x)+1=f(z)\leq\Vert f\Vert\Vert z\Vert\leq1$$
Hence, $f(z)\leq f(x)=1$. Thus, $\{z\in E\,\colon\,\Vert z\Vert\leq1\}\subseteq\{z\in E\,\colon\,f(z)\leq1\}$. In particular, $f(y)\leq1$. Remains to show that $f(y)=1$. Suppose, by contraction, that $f(y)<1$. Then:
$$f\left(\frac{2}{3}x+\frac{1}{3}y\right)=\frac{2}{3}f(x)+\frac{1}{3}f(y)<1$$
But since $\{z\in E\,\colon\,\Vert z\Vert\leq1\}\subseteq\{z\in E\,\colon\,f(z)\leq1\}$, then considering the boundary of those sets, we get $\{z\in E\,\colon\,\Vert z\Vert=1\}\subseteq\{z\in E\,\colon\,f(z)=1\}$, and we derive a contradiction. Therefore, $f(x)=f(y)=1$.
Can someone verify my proof?
 A: Let $$v_t=(1-t)x+ty,\quad 0\le t\le 1$$ Then $\|v_t\|\le 1.$ We are going to show that $\|v_t\|=1.$
For $t\le {1\over 3}$ we have
$${2\over 3}x+{1\over 3}y={2\over 3(1-t)}v_t+{1-3t\over 3(1-t)}y$$
Thus
$$1={1\over 3}\|2x+y\|\le {2\over 3(1-t)}\|v_t\|+{1-3t\over 3(1-t)}$$
Hence $\|v_t\|\ge 1.$
For $t>{1\over 3}$ we have
$${2\over 3}x+{1\over 3}y={1\over 3t}v_t+{3t-1\over 3t}x$$
Thus
$$1 ={1\over 3}\|2x+y\|\le {1\over 3t}\|v_t\|+{3t-1\over 3t}$$
Hence $\|v_t\|\ge 1.$
Let $u_t=(1-t)x+t(-y).$ Then $\|u_t\|\le 1.$ Basing on $\|2(-y)+x\|=1$ we can show, similarly as above, that $\|u_t\|=1.$
Consider the linear functional $f$ on $E_0={\rm span}\,\{x,y\}$ defined by $f(x)=f(y)=1.$ We are going to show that $\|f\|=1.$
As $f(x)=1$ we get $\|f\|\ge 1.$
Let $0\neq w\in X_0.$ Then
$w=\alpha x+\beta y,$ where $|\alpha|+|\beta|>0.$ We will show that $\|w\|\ge |\alpha+\beta|=|f(w)|,$ i.e. $\|f\|\le 1.$ The conclusion is obvious when $\alpha=0$ or $\beta =0.$ Consider the remaining case $\alpha\neq 0$ and $\beta\neq 0.$ By homogeneity we may assume that $\alpha>0$ and $\alpha+|\beta|=1.$ If $t:=\beta>0$ then $w=v_t.$ On the other hand if $t=-\beta>0,$ then $w=u_t.$ In both cases we get
$$\|w\|=1=\alpha+|\beta|\ge |\alpha+\beta|$$
By the Hahn-Banach theorem the functional $f$ can be extended to a linear functional $\tilde{f}:E\to \mathbb{R},$ so that $\|\tilde{f}\|=1.$
Remark The idea behind the proof is the following. If $\|u\|=\|v\|=1$ and $w$ belongs to the line through $u$ and $v,$ outside the segment connecting $u$ and $v,$ then $\|w\|\ge 1.$ For example for $t\le {1\over 3}$ the element $v_t$ belongs to the line through $y$ and ${2\over 3}x+{1\over 3}y,$ but does not belong to the segment between these elements.
