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By Schenker J.

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N. Proof. Let Y = span{y1 , . . , yN } and define N ( on Y by N bj ) = j=1 bj α j . ) Clearly, is linear. Furthermore, since Y is finite dimensional is bounded. (Explicitly, since any two norms on a finite dimensional space are equivalent, we can find c such that |bj | yj ≤ c j bj y j . ) Thus is a linear functional on Y dominated by the norm · . , that is bounded. A closed subspace Y of a normed space X is itself a normed space. If X is a Banach space, so is Y . A linear functional ∈ X on X can be restricted to Y and is still bounded.

The proof is a simple extension of the corresponding result for linear functionals. An isometry of normed spaces X and Y is a map M : X → Y such that (1) M is surjective. (2) M (x) − M (y) = x − y . Clearly translations Tu : X → X, Tu (x) = x + u are isometries of a normed linear space. A linear map T : X → Y is an isometry if T is surjective and T (x) = x ∀x ∈ X. A map M : X → Y is affine if M (x) − M (0) is linear. So, M is affine if it is the composition of a linear map and a translation. 2 (Mazur and Ulam 1932).

2 (Mazur and Ulam 1932). Let X and Y be normed spaces over R. Any isometry M : X → Y is an affine map. Remark. The theorem conclusion does not hold for normed spaces over C. In that context any isometry is a real -affine map (M (x) − M (0) is real linear), but not necessarily a complex-affine map. For example on C([0, 1], C) the map f → f (complex conjugation) is an isometry and is not complex linear. Proof. It suffices to show M (0) = 0 =⇒ M is linear. To prove linearity it suffices to show 1 1 M (x + y) = (M (x) + M (y)) ∀x, y ∈ X.

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