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?

  • $\begingroup$ Did you mean the ball $\{ ‖ x ‖ \le 1 \}$? $\endgroup$ – Calvin Khor Apr 24 '17 at 12:33
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    $\begingroup$ @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
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    $\begingroup$ 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

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.

  • $\begingroup$ 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

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