Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Lecture 36: Proof of Bessel's inequality and Hilbert space projections

Let H be a Hilbert space.
Let a1,a2,a3,… be an orthonormal sequence in H.
To show: ∑n∈ℤ>0∣⟨x,an⟩∣2≤‖x‖2.
To show: ∑n∈ℤ>0⟨x,an⟩2≤‖x‖2.
To show: limk→∞(∑n=1k⟨x,an⟩2)≤‖x‖2.
To show: If k∈ℤ>0 then ∑n=1k⟨x,an⟩2≤‖x‖2.
Assume k∈ℤ>0. Let xk=∑n=1k ⟨x,an⟩an so that ‖xk‖2= ∑n=1k ⟨x,an⟩2. To show: ‖xk‖2≤‖x‖2.
Then ⟨x-xk,xk⟩ = ⟨x,xk⟩- ⟨xk,xk⟩ = ∑n=1k ⟨x,an⟩ ⟨x,an⟩- ∑n=1k ⟨xk,xk⟩ = 0, and ‖x‖2 = ⟨x,x⟩ = ⟨xk+(x-xk),xk+(x-xk)⟩ = ⟨xk,xk⟩+ ⟨xk,(x-xk)⟩+ ⟨(x-xk),xk⟩+ ⟨x-xk,x-xk⟩ = ‖xk‖2+0+0+ ‖x-xk‖2. So ‖xk‖2≤‖x‖2.

Let W=span{a1,a2,…}.
To show: P:H→H given by P(x)=∑n∈ℤ>0 ⟨x,an⟩an is an orthogonal projection onto W‾.
To show:
(a) P:H→H is a function.
(b) If x∈H then P(x)∈W‾.
(c) If x∈H then x-P(x)∈(W‾)⊥.
(a) To show: If x∈H then P(x)=∑n∈ℤ>0⟨x,an⟩an exists in H.
Assume x∈H.
Let xk=∑n=1k ⟨x,an⟩an. To show: limk→∞xk exists in H.
Since H is complete, we need
To show: x1,x2,x3,… is a Cauchy sequence in H.
We know: ‖xk‖=∑n=1k ⟨x,an⟩2 so that ‖x1‖,‖x2‖,… is an increasing sequence in ℝ≥0 bounded by ‖x‖, (by Bessel's inequality).
So ‖x1‖,‖x2‖,… converges.
Let y=limk→∞‖xk‖.
To show: If ε∈ℝ>0 then there exists N∈ℤ>0 such that if m,n∈ℤ≥N then ‖xm-xn‖<ε.
Assume ε∈ℝ>0.
To show: There exists N∈ℤ>0 such that if m,n∈ℤ≥N then ‖xm-xn‖<ε.
Let N∈ℤ>0 such that if k∈ℤ≥0 then ∣y2-‖xk‖2∣<ε2.
To show: If m,n∈ℤ≥N then ‖xm-xn‖<ε.
Assume m,n∈ℤ≥N.
To show: ‖xm-xn‖<ε. ‖xm-xn‖2 = ‖ ∑j=1m ⟨x,aj⟩aj- ∑j=1n ⟨x,aj⟩aj ‖ 2 = ‖∑j=m+1n⟨x,aj⟩aj‖2 = ∑j=m+1n ⟨x,aj⟩2 = ∣ ‖xn‖2- ‖xm‖2 ∣ = ∣ ‖xn‖2- y2+y2- ‖xm‖2 ∣ ≤ ∣ ‖xn‖2- y2 ∣ + ∣ y2- ‖xm‖2 ∣ < ε2+ε2=ε. So x1,x2,… is a Cauchy sequence in H.
So limn→∞xn exists in H.
So ∑j∈ℤ>0⟨x,aj⟩aj exists in H.
(b) To show: If x∈H then P(x)∈W‾.
Assume x∈X.
To show: ∑n∈ℤ>0⟨x,an⟩an∈W‾.
To show: limk→∞xk∈W‾.
To show: xk∈W.
Since xk=∑j=1k⟨x,aj⟩aj∈span{a1,a2,…} then xk∈W.
So P(x)=limk→∞xk∈W‾.
(c) To show: If x∈H then x-P(x)∈W‾⊥.
Assume x∈H.
To show: x-P(x)∈W‾⊥.
To show: If b∈W‾ then ⟨x-P(x),b⟩=0.
Assume b∈W‾.
Let b1,b2,… be a sequence in W with limn→∞bn=b.
To show: ⟨x-P(x),b⟩=0. ⟨x-P(x),b⟩ = ⟨x-P(x),limn→∞bn⟩ = limn→∞ ⟨x-P(x),bn⟩ (since ⟨x-P(x),·⟩:H→ℂ is continuous) and there exist ℓ∈ℤ>0 and c1,…,cℓ∈ℂ such that ⟨x-P(x),bn⟩= ∑k=1ℓck ⟨x-P(x),ak⟩ . Then ⟨x-P(x),aj⟩ = ⟨x-limk→∞xk,aj⟩ = limk→∞ ⟨x-xk,aj⟩ (since ⟨·,aj⟩:H→ℂ is continuous) = limk→∞ ( ⟨x,aj⟩- ⟨x,aj⟩ ) = 0. So ⟨x-P(x),b⟩= limn→∞⟨x-P(x),bn⟩= limn→∞0=0. So x-P(x)∈W⊥.

Notes and References

These are a typed copy of Lecture 36 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on September 26, 2014.

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