Let \(X\) be a real Banach space such that for all \(x,y\in X\), \[ \|x+y\|^2+\|x-y\|^2\le 2\|x\|^2+2\|y\|^2. \] Let \(f:X\to\mathbb R\) be a linear functional with \(\|f\|=1\). Prove that there exists a unique \(x\in X\) such that \[ \|x\|=1, \qquad f(x)=1. \]
Solution
1. The norm satisfies the parallelogram identity.
Apply the assumed inequality to the pair \(x+y,x-y\). Since \[ (x+y)+(x-y)=2x, \qquad (x+y)-(x-y)=2y, \] we obtain \[ 4\|x\|^2+4\|y\|^2 \le 2\|x+y\|^2+2\|x-y\|^2. \] Dividing by \(2\) gives \[ 2\|x\|^2+2\|y\|^2 \le \|x+y\|^2+\|x-y\|^2. \] Combining this with the original inequality yields equality: \[ \|x+y\|^2+\|x-y\|^2 =2\|x\|^2+2\|y\|^2. \]
2. The norm comes from a Hilbert-space inner product.
By the Jordan–von Neumann theorem, the parallelogram identity implies that \[ \langle x,y\rangle =\frac14\bigl(\|x+y\|^2-\|x-y\|^2\bigr) \] defines a real inner product whose induced norm is the given norm. Since \(X\) is complete for that norm, \(X\) is a Hilbert space.
3. Riesz representation gives existence and uniqueness of the norming point.
By the Riesz representation theorem, there exists a unique \(u\in X\) such that \[ f(x)=\langle x,u\rangle \qquad(x\in X), \] and \[ \|u\|=\|f\|=1. \] Therefore \[ f(u)=\langle u,u\rangle=\|u\|^2=1, \] so \(u\) is a required point.
If \(x\in X\) also satisfies \(\|x\|=1\) and \(f(x)=1\), then \[ 1=\langle x,u\rangle\le\|x\|\,\|u\|=1. \] Equality holds in Cauchy–Schwarz. In a real Hilbert space this forces \(x\) and \(u\) to be linearly dependent, and because both have norm \(1\) and \(\langle x,u\rangle=1\), one must have \(x=u\). Thus the norming point is unique.