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$X$ is a normed space. If for all $x,y\in X$ such that $\|x\|=\|y\|=1, x\neq y$, we have that $\|\frac{x+y}{2}\|<1$, then we know that $X$ is strictly convex. How can I show that for all $\lambda \in (0,1)$, $\|\lambda x + (1-\lambda) y\|<1$ always holds?

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  • $\begingroup$ There is something wrong with what you wrote. As it stands, a set $X$ satisfying your definition is never even convex, unless it is either empty or a singleton. Did you miss a connective? $\endgroup$
    – tomasz
    Commented Dec 16, 2016 at 11:42
  • $\begingroup$ They are equivalent for the definition of strictly convex normed space. Though $\lambda \in (0,1)$ seems to be a stronger condtion, they are actually the same. $\endgroup$
    – Meta Fan
    Commented Dec 16, 2016 at 13:59
  • $\begingroup$ I fixed the definition. The previous one meant (literally) that $X$ is empty. $\endgroup$
    – tomasz
    Commented Dec 16, 2016 at 14:13
  • $\begingroup$ Yes, the previous one may mislead the readers. $\endgroup$
    – Meta Fan
    Commented Dec 16, 2016 at 14:21

2 Answers 2

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The point $\lambda x+ (1-\lambda) y$ is a convex combination of $\frac12(x+y)$ and of $x$ or $y$. Then the norm of $\lambda x+ (1-\lambda) y$ is, by convexity of the norm, $\le$ a convex combination of $\|\frac12(x+y)\|$ and $1$, which is strictly less that one.

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  • $\begingroup$ I think the first "less than" should be $\le$ in the sense of the triangle inequality. $\endgroup$
    – Meta Fan
    Commented Dec 16, 2016 at 13:56
  • $\begingroup$ What do you mean by "1" in this context? $\endgroup$
    – JustANoob
    Commented Mar 2, 2018 at 13:25
  • $\begingroup$ The norm of x and y is $1$. $\endgroup$
    – daw
    Commented Mar 2, 2018 at 14:10
  • $\begingroup$ The point $\lambda x+ (1-\lambda y)$ ..... i mean that 1. Or you maybe mean : The point $\lambda x+ (1-\lambda )y$?. If not, then i dont understand what you mean with 1 $\endgroup$
    – JustANoob
    Commented Mar 3, 2018 at 14:57
  • $\begingroup$ @Sam see edit. ......... $\endgroup$
    – daw
    Commented Mar 5, 2018 at 7:42
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$$\|(x+y)/2\|< 1 \implies \|x+y\|<2\tag1\label1$$

$$ax+(1-a)y = a(x+y)+(1-2a)y\tag2\label2$$ $$ax+(1-a)y = (1-a)(x+y)+(2a-1)x\tag3\label3$$ Break it into cases:

Case $1$: $a\leq1/2$

Use the triangle inequality on $\eqref2$ and substitute $\eqref1$.

Case $2$: $a\geq1/2$

Use triangle inequality on $\eqref3$ and substitute $\eqref1$.

Note: $\|cx\| = c\|x\| \iff c\geq0$.

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