Question:
For a bounded closed set $X$, let $B(x_0,r)$ be the smallest ball that contains $X$. Prove that $x_0\in Conv(X)$.
Attempt:
First, we prove that $B(x_0,r)$ is a convex set. Since it is given as the smallest ball which contains $X$, then it is the smallest convex set which contains $X$. However, we know $Conv(X)$ is the smallest convex set which contains $X$. Therefore, $B(x_0,r)=Conv(X)$. Hence, $x_0\in Conv(X)$.
I have a misgiving about my proof: Once we prove that $B(x_0,r)$ is a convex set, can we conclude it is the smallest convex set which contains $X$, since we know it is the smallest ball which contains $X$?.
Let $X$ be a bounded closed convex subset of the Euclidean space $E$ and let $B(x_0,r_0)$ be a closed ball containing $X$ such that $r_0$ is minimum. Then $x_0$ is unique and belongs to $X.$
Proof.
Part 1: Let $B(x_0,r)$ and $B(x'_0,r)$ two closed balls containing $X.$ If $x_0\neq x'_0$ let us show that there exists $h<r$ such that $ B(\frac{x_0+x'_0}{2},h)$ contains $X.$
Since $x_0\neq x'_0$ denote $2c=x_0-x'_0.$ We have $r\geq \|c\|$ since $B(x_0,r)\cap B(x'0,r)$ contains $X$ and therefore is not empty. Write $r^2=h^2+\|c\|^2.$ Then $B((x_0+x'_0)/2, h)$ contains $B(x_0,r)\cap B(x'_0,r).$ To see this, without loss of generality assume $(x_0+x'_0)/2=0$ and consider $x_1\in B(x_0,r)\cap B(x'_0,r).$ We have $\|x_1-x_0\| ^2\leq r^2$ and $\|x_1-x'_0\| ^2\leq r^2$, or $\|x_2\pm c\|^2\leq r^2$, or $\|x_2\|^2\pm 2\langle c,x_2\rangle +\|c\|^2\leq r^2$ implying $\|x_2\|^2\leq h^2..$ This proves already that $x_0$ is unique.
Part 2: To see that $x_0\in X$, suppose not. From Hahn Banach theorem there exists an hyperplane $H$ of $E$ such that $X\subset H_+$ and $x_0\in H_-$ where $H_{\pm}$ are the two open half spaces generated by $H.$ Let $x_0'$ be the orthogonal projection of $x_0$ on $H$. Let $x_1\in X$ and $x'_1$ its orthogonal projection on $H$. For simplicity write $\vec{u}=\frac{x'_0-x_0}{\|x'_0-x_0\|}.$ As a consequence $$\langle x_1-x_0,\vec{u}\rangle \geq \langle x_1-x'_0,\vec{u}\rangle>0$$ which implies that from Pythagoras $$r_0^2\geq \| x_1-x_0\|^2=(\langle x_1-x_0,\vec{u}\rangle )^2+\|x_1-x'_1\|^2\geq (\langle x_1-x'_0,\vec{u}\rangle)^2+\|x_1-x'_1\|^2=\| x_1-x'_0\|^2.$$
Comment:
A way to follow the unpalatable above calculation is to have the following geometric representation. Consider the vector $\vec{v}=\frac{x'_1-x'_0}{\|x'_11-x'_0\|}.$ Then $(\vec{u},\vec{v})$ is an orthogonal basis of the two dimensional linear space $\vec{P}$ which is the direction of the affine two dimensional plane $x'_0+\vec{P}$ containing the four points $x_1,x'_1,x_0,x'_0.$ A picture of these four points in the plane $P$ shows that the segment $x_1x_0$ is larger that the segment $x_1x'_0$: the coordinates of the $x_1,x'_1,x_0,x'_0$ in the basis $(\vec{u},\vec{v})$ are respectively
$$\langle x_1-x'_0,\vec{u}\rangle>0, \ \langle x_1-x'_0,\vec{v}\rangle$$ $$0,\ \langle x_1-x'_0,\vec{v}\rangle$$ $$\langle x_0-x'_0,\vec{u}\rangle<0, \ 0$$ $$0,\ 0$$
End of the proof: As a consequence $x_1\in B(x'_0,r_0)$ and more generally $X\subset B(x'_0,r_0).$ Since $r_0$ is minimal, from Part 1 $x_0=x'_0$ a contradiction with $x_0\in H_{-}.$ The result is proved.