Uniqueness of best approximation in strictly convex spaces

Remarks

The idea of the proof: let $d = \inf_{c \in C} \norm{x - c}$ and suppose $c_1 \ne c_2$ in $C$ both have $\norm{x - c_i} = d$. Then $d > 0$, since otherwise $c_1 = x = c_2$. The midpoint $m = (c_1 + c_2)/2$ lies in $C$, and $x - m$ is the midpoint of $x - c_1$ and $x - c_2$, two different vectors of norm $d$. Strict convexity gives $\norm{x - m} < d$, which contradicts the definition of $d$. The second sentence follows with the existence of best approximations from finite-dimensional subspaces.

Example: every Hilbert space and every $L^p$ space with $1 < p < \infty$ is strictly convex, so least squares and $L^p$ polynomial approximations are unique.

Strict convexity cannot be dropped. In $\R^2$ with the maximum norm, every point $(t, 0)$ with $\abs{t} \le 1$ is a best approximation of $(0, 1)$ from the horizontal axis. The space $C[a, b]$ with the supremum norm is not strictly convex either; there uniqueness of best polynomial approximation needs the Haar condition instead.

Convexity of $C$ is also needed: from the centre of a circle in the Euclidean plane, all points of the circle are nearest.

Going deeper

For a student

The theorem separates the geometry of the norm from the approximating set. For finite-dimensional subspaces, it says that strictly convex spaces make every such subspace a Chebyshev set. In non-strictly convex spaces such as $C(K)$ and $L^1$, uniqueness depends on the subspace, and Haar's theorem characterizes the finite-dimensional subspaces of $C(K)$ with unique best approximations.