Proof. Existence Fix $x_{0}\in E$ and $R\define \norm{x_0}{}$. The set $E\cap \cl{B}(0,R)$ is closed and bounded hence compact, and nonempty since it contains $x_{0}$. The continuous function $x\mto \norm{x}{}$ attains its minimum at some $z\in E$. For $x\in E$ with $\norm{x}{}>R$ we have $\norm{x}{}>R\geq \norm{z}{}$ so $z$ minimizes $\norm{\cd}{}$ over all of $E$.

Uniqueness Suppose $z_{1}\neq z_{2}\in E$ both achieve the minimum value $m\define\MIN{x\in E}\norm{x}{}$. By convexity $\frac{z_{1}+z_{2}}{2}\in E$. By parallelogram identity

since $z_{1}\neq z_{2}$, contradicting minimality of $m$. Variational Inequality Let $x\in E$ and $t\in (0,1]$. By convexity $z+t(x-z)\in E$, so $\norm{z+t(x-z)}{}^{2}\geq \norm{z}{}^{2}$. Expanding we have

Letting $t\to 0^{+}$ gives $\inner{z}{x-z}\geq 0$ hence $\inner{z}{x}\geq \norm{z}{}^{2}$. Since $0\nin E,z\neq 0$ so $\norm{z}{}^{2}>0$.$\square$