# Proof of an equivalent definition of strictly convex?

$X$ 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?

• 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? Commented Dec 16, 2016 at 11:42
• 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. Commented Dec 16, 2016 at 13:59
• I fixed the definition. The previous one meant (literally) that $X$ is empty. Commented Dec 16, 2016 at 14:13
• Yes, the previous one may mislead the readers. Commented Dec 16, 2016 at 14:21

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.

• I think the first "less than" should be $\le$ in the sense of the triangle inequality. Commented Dec 16, 2016 at 13:56
• What do you mean by "1" in this context? Commented Mar 2, 2018 at 13:25
• The norm of x and y is $1$.
– daw
Commented Mar 2, 2018 at 14:10
• 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 Commented Mar 3, 2018 at 14:57
• @Sam see edit. .........
– daw
Commented Mar 5, 2018 at 7:42

$$\|(x+y)/2\|< 1 \implies \|x+y\|<2\tag1\label1$$

$$ax+(1-a)y = a(x+y)+(1-2a)y\tag2\label2$$ $$ax+(1-a)y = (1-a)(x+y)+(2a-1)x\tag3\label3$$ 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: $$\|cx\| = c\|x\| \iff c\geq0$$.