Bessel’s inequality

Let $𝑋$ be a Hilbert space, and $(| 𝑒 𝑖 ⟩) \infty 𝑖 = 1$ be an orthonormal sequence in $𝑋$ . Then for any $| 𝑥 ⟩ \in 𝑋$

\infty \sum 𝑛 = 1 | ⟨ 𝑒 𝑛 | 𝑥 ⟩ | 2 \leq ⟨ 𝑥 | 𝑥 ⟩

and thus the infinite series on the left converges. fun

Proof

Let
$| 𝑥 𝑘 ⟩ = 𝑘 \sum 𝑗 = 1 | 𝑒 𝑗 ⟩ ⟨ 𝑒 𝑗 | 𝑥 ⟩$
then
$⟨ 𝑥 | 𝑥 𝑘 ⟩ = 𝑘 \sum 𝑗 = 1 ⟨ 𝑥 | 𝑒 𝑗 ⟩ ⟨ 𝑒 𝑗 | 𝑥 ⟩$
meanwhile
$⟨ 𝑥 𝑘 | 𝑥 𝑘 ⟩ = (𝑘 \sum 𝑗 = 1 ⟨ 𝑥 | 𝑒 𝑗 ⟩ ⟨ 𝑒 𝑗 |) (𝑘 \sum 𝑖 = 1 | 𝑒 𝑖 ⟩ ⟨ 𝑒 𝑖 | 𝑥 ⟩) = 𝑘 \sum 𝑗 = 1 𝑘 \sum 𝑖 = 1 ⟨ 𝑥 | 𝑒 𝑗 ⟩ ⟨ 𝑒 𝑖 | 𝑥 ⟩ ⟨ 𝑒 𝑗 | 𝑒 𝑖 ⟩ = 𝑘 \sum 𝑗 = 1 𝑘 \sum 𝑖 = 1 ⟨ 𝑥 | 𝑒 𝑗 ⟩ ⟨ 𝑒 𝑖 | 𝑥 ⟩ 𝛿 𝑖 𝑗 = 𝑘 \sum 𝑗 = 1 ⟨ 𝑥 | 𝑒 𝑗 ⟩ ⟨ 𝑒 𝑗 | 𝑥 ⟩$
so $⟨ 𝑥 𝑘 | 𝑥 𝑘 ⟩ = ⟨ 𝑥 | 𝑥 𝑘 ⟩$ . Now
$‖ | 𝑥 ⟩ - | 𝑥 𝑘 ⟩ ‖ 2 = (⟨ 𝑥 | - ⟨ 𝑥 𝑘 |) (| 𝑥 ⟩ - | 𝑥 𝑘 ⟩) = ⟨ 𝑥 | 𝑥 ⟩ + ⟨ 𝑥 𝑘 | 𝑥 𝑘 ⟩ - 2 ℜ (⟨ 𝑥 | 𝑥 𝑘 ⟩) = ⟨ 𝑥 | 𝑥 ⟩ - ⟨ 𝑥 𝑘 | 𝑥 𝑘 ⟩$
so
$‖ | 𝑥 𝑘 ⟩ ‖ 2 = ‖ | 𝑥 ⟩ ‖ 2 - ‖ | 𝑥 ⟩ - | 𝑥 𝑘 ⟩ ‖ 2 \leq ‖ | 𝑥 ⟩ ‖ 2$
implying
$𝑘 \sum 𝑗 = 1 | ⟨ 𝑒 𝑗 | 𝑥 ⟩ | 2 = ⟨ 𝑥 𝑘 | 𝑥 𝑘 ⟩ \leq ⟨ 𝑥 | 𝑥 ⟩$
for all $𝑘$ .

statetidy | en | SemBr

Bessel’s inequality

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