Suppose that φ,γ∈L2([0,1],C) and define the operator φ⊗γ:L2([0,1],C)→L2([0,1],C) by setting φ⊗γ=⟨⋅,γ⟩φ. This is an integral operator with the kernel given by φ(x)¯γ(y) for each x,y∈[0,1]. Hence, the Hilbert-Schmidt norm of this operator is given by
‖φ⊗γ‖HS=(∫10∫10|φ(x)¯γ(y)|2dxdy)1/2=‖φ‖‖γ‖.
I want to find the singular values of this operator so that I can obtain the nuclear norm. Since the adjoint of φ⊗γ is given by γ⊗φ, we have that
(φ⊗γ)∗(φ⊗γ)(f)=⟨f,γ⟩‖φ‖2γ
and if we set f=γ/‖γ‖, we obtain
(φ⊗γ)∗(φ⊗γ)(γ‖γ‖)=‖φ‖2‖γ‖2γ/‖γ‖.
The composition of operators (φ⊗γ)∗(φ⊗γ) has one non-zero eigenfunction γ/‖γ‖ with the corresponding eigenvalue ‖φ‖2‖γ‖2. Hence, the operator φ⊗γ has one non-zero singular value given by ‖φ‖‖γ‖ and the nuclear norm of φ⊗γ is equal to ‖φ‖‖γ‖, which is also the Hilbert-Schmidt of this operator.
Are my calculations correct? Does the composition of operators (φ⊗γ)∗(φ⊗γ) have only one non-zero eigenfunction and the corresponding non-zero eigenvalue? Are the Hilbert-Schmidt and the nuclear norms really equal for this operator?
Any help is much appreciated!
Answer
Yes. The operators T=(φ⊗γ)∗(φ⊗γ) you are looking at, are rank one and positive. Concretely, you have that your operator is
Tf=‖φ‖2⟨f,γ⟩γ=‖φ‖2‖γ‖2⟨f,γ′⟩γ′,
where γ′=γ/‖γ‖.
Written this way, T is a scalar multiple of the projection operator onto the span of γ. So its spectrum consists of 0 and ‖φ‖2‖γ‖2. So, for the HS norm,
‖φ⊗γ‖2HS=Tr((φ⊗γ)∗(φ⊗γ))1/2=(0+‖φ‖2‖γ‖2)1/2=‖φ‖‖γ‖.
For the operator norm,
‖φ⊗γ‖=‖(φ⊗γ)∗(φ⊗γ)‖1/2=‖φ‖‖γ‖.
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