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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?

joelittledrop
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  • How do you arrive at the contradiction? What if $E = \mathbb R^2$ with $l_1$ norm, $x=(1,0)=f$, $y=(0,1)$? – daw Jan 16 '23 at 14:58
  • I believe that your example is not a counterexample. You exhibited one particular functional for which the claim does not hold, but the statement is that there is one functional with the desired property. Following my idea of the supporting hyperplane, we would consider $f=(1,1)$, for which the claim holds. – joelittledrop Jan 16 '23 at 15:16
  • Hint. Consider the $2$-dimensional space spanned by $x$ and $y.$ Then show that $|(1-t)x\pm ty|=1.$ For $+$ sign use $|2x+3|=3$ and for $-$ sign use $|x-2y|=3.$ – Ryszard Szwarc Jan 16 '23 at 15:34
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    Is is an counterexample to your claim that there is a contradiction. – daw Jan 16 '23 at 18:30
  • @daw Oh, I think I understood what you mean. You give an example where ${ x\in E,\colon,\Vert x\Vert\leq1}\subseteq{ x\in E,\colon,f(x)\leq1}$ is true, but ${ x\in E,\colon,\Vert x\Vert=1}\subseteq{ x\in E,\colon,f(x)=1}$ is false. So that step in my proof is not valid. – joelittledrop Jan 16 '23 at 18:59

1 Answers1

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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.

Ryszard Szwarc
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