 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2010/p/p3/d/i/solution

Let

v = \sum_{k = 1}^{n} ⟨ x, e_{k} ⟩ e_{k}

, using the convention that the inner product is linear in its first argument. Orthonormality gives

⟨ x - v, e_{j} ⟩ = ⟨ x, e_{j} ⟩ - \sum_{k} ⟨ x, e_{k} ⟩ ⟨ e_{k}, e_{j} ⟩ = 0

for every

j

. Hence

x - v

is orthogonal to

U_{n}

. For any

u \in U_{n}

, the Pythagorean identity gives

∥ x - u ∥^{2} = ∥ x - v ∥^{2} + ∥ v - u ∥^{2}

, so

v

is the unique best approximation. Therefore

P_{n} x = k = 1 \sum n ⟨ x, e_{k} ⟩ e_{k} .

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