Let $e,f$ be unit vectors in a real Banach space s.t $\|2e+f\|=\|e-2f\|=3$, show that $\|\lambda e+\mu f\|=|\lambda|+|\mu|$.

Question

Let $e,f$ be unit vectors in a real Banach space s.t $\|2e+f\|=\|e-2f\|=3$, show that $\|\lambda e+\mu f\|=|\lambda|+|\mu|$.

I have a hint which is to show there is are linear functionals of norm $1$ s.t $\phi(e)=\phi(f)=1$ and $\pi(e)=\pi(-f)=1$. I could not prove the hint.

I know that by hanh banach we can extend $\phi(\alpha e+\beta f)=\alpha +\beta$ to entire space. but I do not know how to show that it has norm $1$. I am almost certain I have to use $\|2e+f\|=\|e-2f\|=3$. I tried the following:

let $\alpha,\beta$ be s.t $\|\alpha e+ \beta g\|=1$ then given $\|\alpha e+ \alpha/2 f\|=3/2\alpha$ and so $\|\alpha e+ \beta g\|=\|\alpha e+\alpha f+(\beta-\alpha/2)f\|=1$. Then I tried reverse triangle inequality but that was not helpful. What do I do here?

Answer 1

By the Hahn Banach theorem there exists a norm-1 functional $\phi$ such that $\phi(2e+f)=\Vert 2e+f \Vert=3$. It follows that $3=2\phi(e)+\phi(f)$ that is $$2(1-\phi(e))+(1-\phi(f))=0$$ This implies that $\phi(e)=\phi(f)=1$ because $|\phi(e)|\le\Vert e\Vert=1$ and $|\phi(f)|\le\Vert f\Vert=1$.

Similarly, there exists a norm-1 functional $\pi$ such that $\pi(e-2f)=\Vert e-2f \Vert=3$. It follows that $3=\pi(e)-2\pi(f)$ that is $$(1-\pi(e))+2(1+\pi(f))=0$$ This implies that $\pi(e)=1=-\pi(f)$ because $|\pi(e)|\le1$ and $|\pi(f)|\le1$.

1. If $\lambda,\mu\ge0$ then $$\lambda+\mu=\phi(\lambda e+\mu f)\le \Vert\lambda e+\mu f\Vert\le \lambda \Vert e \Vert+\mu \Vert f\Vert=\lambda +\mu$$
2. If $\lambda\ge0\ge\mu$ then $$\lambda-\mu=\pi(\lambda e+\mu f)\le \Vert\lambda e+\mu f\Vert\le \lambda \Vert e \Vert-\mu \Vert f\Vert=\lambda -\mu$$
3. If $\mu\ge0\ge\lambda$ then $$-\lambda+\mu=\pi(-\lambda e-\mu f)\le \Vert\lambda e+\mu f\Vert\le -\lambda \Vert e \Vert+\mu \Vert f\Vert=-\lambda +\mu$$
4. If $0\ge\mu,\lambda$ then $$-\lambda-\mu=\phi(-\lambda e-\mu f)\le \Vert\lambda e+\mu f\Vert\le -\lambda \Vert e \Vert-\mu \Vert f\Vert=-\lambda -\mu$$

We conclude that in all cases we have $$\Vert \lambda e +\mu f\Vert=\vert \lambda\vert +\vert\mu\vert$$ as required. $\qquad\square$

Answer 2

This is a two-dimensional problem. We don't need the Hahn-Banach Theorem, i.e., the axiom of choice, to solve it. It is sufficient to work in the Banach subspace $V$ spanned by the two vectors $e, f\in{\cal B}$, wher ${\cal B}$ is the original "large" Banach space.

If $e$ and $f$ were linearly dependent we had $f=\lambda e$ with $\lambda\in\{-1,1\}$. But both choices lead to a contradiction with the norm conditions. It follows that $V$ has dimension $2$, and we may take the pair $(e,f)$ as a basis of $V$. We now have to determine the Banach unit ball $B\subset V$.

Let $$a:={1\over3}(2e+f),\qquad b:={1\over3}(e-2f)\ .$$ Then the $8$ points $$\pm e,\quad\pm f,\quad\pm a,\quad\pm b\tag{1}$$ have norm $1$, hence are lying on the unit sphere $\partial B$. Since $a$ is lying on the segment $[e,f]$, and similarly for the other given points in the four quadrants, the following figure makes it intuitively obvious that $B$ is the square containing the $8$ points on its boundary, i.e., $$B=\bigl\{\lambda e+\mu f\bigm| |\lambda|+|\mu|\leq1\bigr\}\ .\tag{2}$$ This means that $B$ is the (well known) unit ball of the $l^1$ norm in $V$ with basis $(e,f)$. In other words, $$\|\lambda e+\mu f\|=|\lambda|+|\mu|\ ,$$ as claimed.

In order to prove $(2)$ it is sufficient to look at the first quadrant. All points $$p_\tau:=(1-\tau)e+\tau f\qquad(0<\tau<1)$$of the segment $[e,f]$ have a norm $\leq1$. If there would be a $p_\tau$ with $\|p_\tau\|=:\rho<1$ the point $q:={1\over\rho} p_\tau$ would lie north-east of $[e,f]$ and have norm $1$. This implies that all points of the segments $[e,q]$ and $[q,f]$ have norm $\leq1$ and finally that $\|a\|<1$.

It follows that $\|p_\tau\|=1$ $\>(0<\tau<1)$, so that we now have full control over $\partial B$.