Convolution operators

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Suppose f∈L1(∂D)f \in L^{1}(\partial \mathbf{D}). Define T:L2(∂D)→L2(∂D)T: L^{2}(\partial \mathbf{D}) \rightarrow L^{2}(\partial \mathbf{D}) by Tg=f∗gT g = f * g.

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(a)

Show that TT is a compact operator on L2(∂D)L^{2}(\partial \mathbf{D}).

(b)

Prove that TT is injective if and only if f^(n)≠0\widehat{f}(n) \neq 0 for every n∈Zn \in \mathbf{Z}.

(c)

Find a formula for T∗T^{*}.

(d)

Prove: TT is self-adjoint if and only if all Fourier coefficients of ff are real.

(e)

Show that TT is a normal operator.

Suppose y>0y > 0. Define on operator TT on L2(R)L^{2}(\mathbb {R}) by Tf=f∗PyT f = f * P_{y}.

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(a)

Show that TT is a self-adjoint operator on L2(R)L^{2}(\mathbb {R}).

(b)

Show that sp⁡(T)=[0,1]\operatorname {sp}(T) = [0,1].

Remark.
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Because the spectrum of each compact operator is a countable set (by 10.93), part (b) above implies that TT is not a compact operator. This conclusion differs from the situation on the unit circle-see Exercise 9 in Section 11B.

Let K:R→CK: \mathbb {R} \rightarrow \mathbb {C} be a 2π2 \pi-periodic Lebesgue measurable function and K∈L1([0,2π])K \in L_{1}(\left[0,2 \pi \right]). Prove that for 1<p<+∞1 < p < +\infty the operator

(Ax)(t)=∫02πK(t−s)x(s)ds,t∈[0,2π],x∈Lp([0,2π]) (A x)(t) = \int _{0}^{2 \pi } K(t-s) x(s) d s, t \in [0,2 \pi ], x \in L_{p}(\left[0,2 \pi \right])

is compact.

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Let u∈L1(R)u \in L_{1}(\mathbb {R}). Prove that the operator

(Ax)(t)=∫Ru(t−s)x(s)ds,t∈R,x∈L2(R) (A x)(t) = \int _{\mathbb {R}} u(t-s) x(s) d s, t \in \mathbb {R}, x \in L_{2}(\mathbb {R})

is linear and continuous in L2(R)L_{2}(\mathbb {R}). Under what conditions on the function uu is the operator AA compact?

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Let u∈L1([0,+∞))u \in L_{1}([0,+\infty )) and (Ax)(t)=∫0tu(t−s)x(s)ds,t∈[0,+∞),x∈L2([0,+∞))(A x)(t) = \int_{0}^{t} u(t-s) x(s) d s, t \in [0,+\infty ), x \in L_{2}([0,+\infty )). Prove that A∈L(L2([0,+∞)))A \in \mathcal{L}\left(L_{2}([0,+\infty ))\right). Under what conditions on the function uu is the operator AA compact?

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(1)

Find all λ∈R\lambda \in \mathbb {R}, for which the homogeneous integral equation x(t)=λ∫e−∣t−s∣x(s)ds,t∈Rx(t) = \lambda \int e^{-\left|t-s\right|} x(s) d s, t \in \mathbb {R}, has a nonzero solution x∈L∞(R)x \in L_{\infty }(\mathbb {R}).

(2)

Prove that for all λ<12\lambda < \frac{1}{2} and y∈L∞(R)y \in L_{\infty }(\mathbb {R}) the integral equation x(t)=λ∫e−∣t−s∣x(s)ds+y(t),t∈Rx(t) = \lambda \int e^{-\left|t-s\right|} x(s) d s+y(t), t \in \mathbb {R}, has a unique solution in the space L∞(R)L_{\infty }(\mathbb {R}), which is expressed by the formula

x(t)=λ1−2λ∫Re−1−2λ∣t−s∣y(s)ds+y(t),t∈R x(t) = \frac{\lambda }{\sqrt{1-2 \lambda }} \int _{\mathbb {R}} e^{-\sqrt{1-2 \lambda }\left|t-s\right|} y(s) d s+y(t), t \in \mathbb {R}

Let f∈L1(R)f \in L_{1}(\mathbb {R}). The operator A∈L(L2(R))A \in \mathcal{L}\left(L_{2}(\mathbb {R})\right) is defined by the formula Ag=f∗gA g = f * g (the convolution operator with function ff). Find σ(A)\sigma (A).

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Verify the statement in Example 4.12: Let h∈H=LC2(−π,π)h \in \mathcal{H} = L_{\mathbb {C}}^{2}(-\pi , \pi ) and define K:H→HK: \mathcal{H} \rightarrow \mathcal{H} by (Kf)(x)=(2π)−1/2∫−ππh(x−y)f(y) dy(Kf)(x) = (2\pi )^{-1/2} \int_{-\pi }^{\pi } h(x-y) f(y) \, dy. If λn=(2π)−1/2∫−ππh(x)exp⁡(−inx) dx=h^(n)\lambda_{n} = (2\pi )^{-1/2} \int_{-\pi }^{\pi } h(x) \exp (-inx) \, dx = \hat{h}(n), the nnth Fourier coefficient of hh, then Ken=λnenKe_{n} = \lambda_{n} e_{n}, where en(x)=(2π)−1/2exp⁡(−inx)e_{n}(x) = (2\pi )^{-1/2} \exp (-inx). (Note that the operator KK in this example is diagonalizable.)

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For functions f,g∈L2(R)f, g \in L^2(\mathbb {R}), define the convolution f∗gf*g by

f∗g(x)=∫−∞+∞f(y)g(x−y) dy, f*g(x) = \int _{-\infty }^{+\infty } f(y) g(x-y) \, dy,

where the integral is with respect to Lebesgue measure.

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(a)

Show that f∗g∈L∞(R)f*g \in L^{\infty }(\mathbb {R}). Do not neglect to check the measurability of f∗gf*g.

(b)

Suppose that ff has the property that for all g∈L2(R)g \in L^2(\mathbb {R}) the convolution f∗gf*g is also an element of L2(R)L^2(\mathbb {R}). Define Tf:L2(R)→L2(R)T_f : L^2(\mathbb {R}) \to L^2(\mathbb {R}) by Tf(g)=f∗gT_f(g) = f*g. It is evident that TfT_f is a linear operator; you need not check this. Show that TfT_f is a bounded linear operator.

Suppose that LL is a Banach space of functions on a locally compact abelian group GG with invariant measure μ\mu, with a norm ∥⋅∥L\left\| \cdot \right\|_{L} having the properties:

  1. ∥T(a)f∥L=∥f∥L\left\| T(a)f\right\|_{L}=\left\| f\right\|_{L}, where T(a)T(a) is the translation operator T(a)f(x)=f(x+a)T(a)f(x)=f(x+a);

  2. ∥T(a)f−f∥L→0\left\| T(a)f-f\right\|_{L}\to 0 as a→0a\to 0 in GG. Prove that the operator S(φ)S(\varphi ), φ∈L1(G,μ)\varphi \in L^{1}\left({G,\mu }\right), carries LL into itself and has a norm in this space not exceeding ∥φ∥L\left\| \varphi \right\|_{L}.

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Let f∈L1(G,μ)f\in L^{1}\left({G,\mu }\right). Prove that S(f)S(f) is a bounded operator in the Hilbert space L2(G,μ)L^{2}\left({G,\mu }\right). Compute the adjoint operator S(f)∗S(f)^*.

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Let GG be a locally compact abelian group with invariant measure μ\mu. Determine conditions on a function f∈L1(G,μ)f\in L^{1}\left({G,\mu }\right) under which the convolution operator S(f)S(f) is:

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(a)

selfadjoint;

(b)

unitary;

(c)

compact.

References

[Axl19] Axler S. Measure, Integration & Real Analysis. 3rd ed., Springer, 2019.
[BCK+24] Brayman V.B., Chaikovskyi A.V., Konstantinov O.Y., Kukush A.G., Mishura Y.S., Nesterenko O.N. Functional Analysis and Operator Theory. Springer, 2024.
[Con07] Conway J.B. A Course in Functional Analysis. 2nd ed., Springer, 2007.
[CJP+11] Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. Problems and Solutions in Mathematics. 2nd ed., World Scientific, 2011.
[KiGv88] Kirillov A.A., Gvishiani A.D. Теоремы и задачи функционального анализа. 2-е изд., Наука, 1988.