# Can $\sup\{\| x_1 - x_2 \| : x_1,x_2\in X, \|x_1+x_2\| = 2, \|x_1\| = \|x_2\| = 1\} = 0?$

Let $$(X,\|\cdot\|)$$ be a normed space Define $$r=\sup\{\| x_1 - x_2 \| : x_1,x_2\in X, \|x_1+x_2\| = 2, \|x_1\| = \|x_2\| = 1\}.$$

Question: Is $$r=0?$$

Clearly triangle inequality implies that $$r\leq 2.$$

I tried $$X = \mathbb{R}^2$$ with Euclidean norm, a Hilbert space. Let $$x_1 = (a_1,a_2)$$ and $$x_2 = (b_1,b_2)$$ with $$\|x_1\| = \|x_2\| = 1$$ and $$\|x_1+x_2\| = 2.$$ So we have $$a_1^2+ a_2^2 = 1 = b_1^2+b_2^2.$$ Since $$\|x_1+x_2\| = 2,$$ we have $$(a_1+b_1)^2 + (a_2+b_2)^2 = 4.$$ Combining all equations above gives us that $$a_1b_1 + a_2b_2 = 1.$$ Since the left expression is dot product between two norm $$1$$ vectors, we conclude that their angle is $$0,$$ which means that $$x_1=x_2.$$ This leads to $$\|x_1-x_2\| = 0.$$

Since $$x_1,x_2$$ are arbitrary, we conclude that $$r = 0.$$

However, I am not sure whether this is true in other spaces.

• The class of normed spaces satisfying this property is known as strictly convex. $L^p$ is strictly convex for $p\in (1,\infty)$ but not for $p=1,\infty$. – Fnacool Mar 6 '19 at 13:28

No, it is not true. Take for example $$X=\ell_{\infty}$$ or $$c_0$$ with the supremum norm, and take $$x_1$$ to be the sequence $$\{\frac{1}{n}\}_{n=1}^{\infty}$$ and $$x_2$$ the sequence $$\{\frac{1}{n^2}\}_{n=1}^{\infty}$$. Then $$\|x_1+x_2\|=2$$, and $$\|x_1\|=\|x_2\|=1$$, but $$\|x_1-x_2\|=\sup_{n}|\frac{1}{n}-\frac{1}{n^2}|\geq \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$
so $$r\geq\frac{1}{4}$$.