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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.

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    $\begingroup$ 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 $p=1,\infty$. $\endgroup$ – Fnacool Mar 6 '19 at 13:28
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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 $r\geq\frac{1}{4}$.

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