See the Bessel inequality for what we'll prove.
Since we're approximating x, let's write down the exact approximation and error term:
yn = ∑i=1n⟨x,ei⟩ei
rn = x - yn
Now a bit of simple magic: rn is x without its projections on e1,...,en, so it should be orthogonal to all of them
Indeed, for j=1,...,n, we have that
⟨rn,ej⟩ = ⟨x,ej⟩ - ∑i=1n⟨⟨x,ei⟩ei,ej⟩ =
(All elements of the sum disappear except for i=j, since ⟨ei,ej⟩=0...)
⟨x,ej⟩ - ⟨x,ej = 0.
Thus we have a shorter orthogonal sequence of elements e1,...,en,rn; in particular, rn⊥yn
But (applying what is essentially the Pythagorean theorem...) we know that
||x||2 = ||rn+yn||2 =
⟨rn+yn,rn+yn⟩ =
||rn||2+||yn||2
The first term ||rn||2 is nonnegative, so
||yn||2 ≤ ||x||2.
Computing ||yn||2=∑i=1n|⟨x,ei⟩|2 yields the inequality.
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