Proof. Let \(d=\inf _{x\in A}||x||\). The proof is the parallelogram law and nothing else, which is why the result holds in an inner product space and fails in a general normed space.

Existence

Choose \(x_n\in A\) with \(||x_n||\rightarrow d\). Applying the parallelogram law to \(x_n\) and \(x_m\), \[||x_n-x_m||^2=2||x_n||^2+2||x_m||^2-4\left \|\frac {x_n+x_m}{2}\right \|^2 .\] Convexity puts the midpoint \(\frac {x_n+x_m}{2}\) in \(A\), so its norm is at least \(d\), giving \[||x_n-x_m||^2\leq 2||x_n||^2+2||x_m||^2-4d^2 .\] As \(n,m\rightarrow \infty \) the right-hand side tends to \(2d^2+2d^2-4d^2=0\), so \((x_n)\) is Cauchy. Completeness of \(A\) gives a limit \(x_0\in A\), and continuity of the norm gives \(||x_0||=d\).

Uniqueness

If \(||x_0||=||x_1||=d\) with both in \(A\), the same identity gives \[||x_0-x_1||^2=2d^2+2d^2-4\left \|\frac {x_0+x_1}{2}\right \|^2\leq 4d^2-4d^2=0 ,\] so \(x_0=x_1\).

Convexity is essential: without it the midpoint need not lie in \(A\) and the bound fails. So is the parallelogram law — in \(l_1\), say, a closed convex set can contain more than one point of least norm. □