Let $(X,\|\circ\|)$ be a banach space with
$$\forall x,y\in X, x\neq y, \|x\|=\|y\|=1 \Rightarrow \|\frac{x+y}{2}\|<1 $$
If M is convex and $z \in X$, then $\|x-z\|=dist(z,M)$ for at most one $x \in M$
$\text{dist}(z,M):=\inf\{d(x,y):y\in M\}$
I supposed that there are two points $x_1,x_2\in M$ which satisfy $\|x-z\|=dist(z,M)$. Then I know that also $x_1t+x_2(1-t)\in M$ But that doesn't help. I am struggling in how to use the property of X given above.
Any Ideas? Please don't post full solutions, only tipps please. Thanks in advance:)
We now that, if $\|u\|=\|v\|$ and $u\ne v$, then $\big\|\frac{1}{2}(u+v)\big\|<\frac{1}{2}\|u\|+\frac{1}{2}\|v\|$. To understand this, let $u_1=u/\|u\|$, $v_1=v/\|v\|$, then $\|u_1\|=\|v_1\|=1$, and as $u_1\ne v_1$, $$ \bigg\|\frac{1}{2}(u+v)\bigg\|=\bigg\|\frac{1}{2}\|u\|(u_1+v_1)\bigg\| =\|u\|\bigg\|\frac{1}{2}(u_1+v_1)\bigg\| \\ <\frac{1}{2}\|u\|(\|u_1\|+\|v_1\|)=\frac{1}{2}(\|u\|+\|v\|). $$
Now, if $$ \|x_1-z\|=\mathrm{dist}(z,M)\quad\text{and}\quad \|x_2-z\|=\mathrm{dist}(z,M) $$ then for $u=x_1-z$ and $v=x_2-z$, we obtain: $$ \big\|\tfrac{1}{2}(x_1+x_2)-z\big\|=\big\|\tfrac{1}{2}\big((x_1-z)+(x_2-z)\big)\big\|< \frac{1}{2}\big(\|x_1-z\|+\|x_2-z\|\big), $$ and as $\tfrac{1}{2}(x_1+x_2)\in M$, we reach to a contradicition.