 Solution

ID: past-exam-of-the-algebra-and-analysis-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/2/a/solution

Let

p = \sum_{j = 1}^{m} ⟨ u, v_{j} ⟩ v_{j}

⟨ u - p, v_{k} ⟩ = 0

for every

k

, so

u - p

p

∥ u ∥^{2} = ∥ u - p ∥^{2} + ∥ p ∥^{2} = ∥ u - p ∥^{2} + \sum_{j = 1}^{m} ⟨ u, v_{j} ⟩^{2} .

Since a squared norm is nonnegative,

⟨ u, u ⟩ \geq j = 1 \sum m ⟨ u, v_{j} ⟩^{2} .

This is Bessel's inequality, with equality exactly when

u

belongs to the span of the orthonormal set.

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