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  "body": "\n## Statement\n\nLet $X$ be a [[strictly convex|GDTKW9KPH8]] normed space, let $C \\subseteq X$ be a nonempty [[convex set|02AS45T3H9]], and let $x \\in X$.\nThen $x$ has at most one [[best approximation|7N4988HN3P]] from $C$.\nIn particular, if $V \\subseteq X$ is a finite-dimensional linear subspace, every $x \\in X$ has exactly one best approximation from $V$.\n\n## Remarks\n\nThe idea of the proof: let $d = \\inf_{c \\in C} \\norm{x - c}$ and suppose $c_1 \\ne c_2$ in $C$ both have $\\norm{x - c_i} = d$.\nThen $d \u003e 0$, since otherwise $c_1 = x = c_2$.\nThe 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 $d$.\nStrict convexity gives $\\norm{x - m} \u003c d$, which contradicts the definition of $d$.\nThe second sentence follows with the existence of best approximations from finite-dimensional subspaces.\n\nExample: every Hilbert space and every $L^p$ space with $1 \u003c p \u003c \\infty$ is strictly convex, so least squares and $L^p$ polynomial approximations are unique.\n\nStrict convexity cannot be dropped.\nIn $\\R^2$ with the maximum norm, every point $(t, 0)$ with $\\abs{t} \\le 1$ is a best approximation of $(0, 1)$ from the horizontal axis.\nThe space $C[a, b]$ with the supremum norm is not strictly convex either; there uniqueness of best polynomial approximation needs the Haar condition instead.\n\nConvexity of $C$ is also needed: from the centre of a circle in the Euclidean plane, all points of the circle are nearest.\n\n## Explanations\n\n### L1 human\n\nIf the unit ball of a space has no flat pieces on its boundary, then from any point there cannot be two equally near points in a convex set.\nIf there were, the point halfway between them would be strictly nearer.\nThis is why least squares problems have unique solutions.\n\n### L2 human\n\nThe theorem separates the geometry of the norm from the approximating set.\nFor finite-dimensional subspaces, it says that strictly convex spaces make every such subspace a Chebyshev set.\nIn 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.\n",
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