5
$\begingroup$

Let $X$ be a Banach space. $S_X=\{x\in X:\Vert x\Vert=1\}$. If $x, y, \frac{x+y}{2}\in S_X$, is it true that $\lambda x+(1-\lambda)y\in S_X$ for every $\lambda \in [0,1]$? I can show this if $X$ is a Hilbert space, but in general there is no inner product and no parallel law. Can you give a hint?

$\endgroup$
  • $\begingroup$ Did you mean the ball $\{ ‖ x ‖ \le 1 \}$? $\endgroup$ – Calvin Khor Apr 24 '17 at 12:33
  • 1
    $\begingroup$ @Calvin No. It is a sphere. $\endgroup$ – CSH Apr 24 '17 at 12:34
  • $\begingroup$ I'm also having a hard time understanding how the average of $x$ and $y$ can be on the sphere, if the Euclidean norm is used? Any convex combination lies on a line between the points, no? Sorry if it is obvious. $\endgroup$ – Bobson Dugnutt Apr 24 '17 at 12:36
  • 1
    $\begingroup$ This is a general Banach space. For Euclidean norm, I think it is hard to find such points. $\endgroup$ – CSH Apr 24 '17 at 12:38
  • $\begingroup$ For Hilbert spaces does it not follow that $\|x-y\|^2 = 2\|x\|^2 + 2 ‖y‖^2 - 4 ‖ (x+y)/2 ‖^2 = 0$? $\endgroup$ – Calvin Khor Apr 24 '17 at 12:41
3
$\begingroup$

May assume $$\lambda\in \left(\frac{1}{2},1\right).$$ If $$\Vert \lambda x+(1-\lambda)y\Vert<1,$$ then since $$\frac{1}{2}x+\frac{1}{2}y=\frac{1}{2\lambda}[\lambda x+(1-\lambda)y]+\left(1-\frac{1}{2\lambda}\right)y,$$ $$\Vert\frac{1}{2}x+\frac{1}{2}y\Vert<\frac{1}{2\lambda}\cdot 1+\left(1-\frac{1}{2\lambda}\right)\cdot 1=1,$$ a contradiction.

$\endgroup$
  • $\begingroup$ seems for $\lambda\in(0,1/2)$ you should use $x$ instead of $y$ in your convex combination (notice that $1/(2\lambda)>1$ -you don't want to use that number) $\endgroup$ – user8268 Apr 24 '17 at 14:25
  • $\begingroup$ I changed the range of $\lambda$ thanks. $\endgroup$ – CSH Apr 24 '17 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.