Spectral radius formula (Gelfand)

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Find the norm, the spectral radius, the set of eigenvalues and the spectrum of an operator A:l2→l2A: l_{2} \rightarrow l_{2}.

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(1)

Ax=(0,x1,x22,…,xnn,…)A x = \left(0, x_{1}, \frac{x_{2}}{2}, \ldots , \frac{x_{n}}{n}, \ldots \right);

(2)

Ax=(x2,x3,…,x100,0,0,…)A x = \left(x_{2}, x_{3}, \ldots , x_{100}, 0,0, \ldots \right);

(3)

Ax=(2x2,32x3,…,n+1nxn+1,…)A x = \left(2 x_{2}, \frac{3}{2} x_{3}, \ldots , \frac{n+1}{n} x_{n+1}, \ldots \right);

(4)

Ax=(4x2,94x3,…,(n+1)2n2xn+1,…)A x = \left(4 x_{2}, \frac{9}{4} x_{3}, \ldots , \frac{(n+1)^{2}}{n^{2}} x_{n+1}, \ldots \right).

Let HH be a Hilbert space, A∈L(H)A \in \mathcal{L}(H) be a normal operator (for example, self-adjoint or unitary). Prove that

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(1)

σr(A)=∅\sigma_{r}(A) = \varnothing;

(2)

r(A)=∥A∥r(A) = \left\| A\right\|.

An operator A∈L(X)A \in \mathcal{L}(X), where XX is a Banach space, is called quasinilpotent if lim⁡n→∞∥An∥1n=0\lim_{n \rightarrow \infty }\left\| A^{n}\right\|^{\frac{1}{n}} = 0. Prove that AA is quasinilpotent if and only if σ(A)={0}\sigma (A) = \left\{ 0\right\}.

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Consider the operator A∈L(L2([0,1]))A \in \mathcal{L}\left(L_{2}(\left[0,1\right])\right), which is defined by the formula (Ax)(t)=∫0tx(s)ds,t∈[0,1](A x)(t) = \int_{0}^{t} x(s) d s, t \in [0,1].

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(1)

find AnA^{n} for all n≥1,r(A)n \geq 1, r(A) and σ(A)\sigma (A);

(2)

find Rλ(A)R_{\lambda }(A);

(3)

find A+A∗A+A^{*} and prove that it is a projector onto a one-dimensional subspace, which consists of constant functions.

Find the spectrum of the Volterra integral operator in L2([0,1])L_{2}(\left[0,1\right]) with a continuous kernel KK, i.e., the operator (Ax)(t)=∫0tK(t,s)x(s)ds,t∈[0,1],x∈L2([0,1])(A x)(t) = \int_{0}^{t} K(t, s) x(s) d s, t \in [0,1], x \in L_{2}(\left[0,1\right]), where K∈C({(t,s)∣0≤s≤t≤1})K \in C(\left\{ (t, s) \mid 0 \leq s \leq t \leq 1\right\} ).

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Let XX be a Banach space, A,B∈L(X)A, B \in \mathcal{L}(X) and AB=BAA B = B A. Prove that

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(1)

r(AB)≤r(A)r(B);r(A B) \leq r(A) r(B) ;

(2)

r(A+B)≤r(A)+r(B)r(A+B) \leq r(A)+r(B).

Are these statements correct if AA and BB do not commute?

Let X=C([0,12])X = C\left(\left[0, \frac{1}{2}\right]\right) and (Ax)(t)=tx(t2),t∈[0,12],x∈X(A x)(t) = t x\left(t^{2}\right), t \in \left[0, \frac{1}{2}\right], x \in X. Prove that r(A)=0r(A) = 0.

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The operator AA in C([0,1])C(\left[0,1\right]) is defined by the formula

(Ax)(t)=1t∫0tx(s)ds,0<t≤1,(Ax)(0)=x(0) (A x)(t) = \frac{1}{t} \int _{0}^{t} x(s) d s, 0 < t \leq 1,(A x)(0) = x(0)

Prove that r(A)=1r(A) = 1.

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Prove that r(AB)=r(BA)r(A B) = r(B A) for all A,B∈L(X)A, B \in \mathcal{L}(X).

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Prove Theorem 12.1.

Theorem 12.1: Let XX be a Banach space. If A∈B(X)A \in \mathcal{B}(X), then the limit lim⁡n→∞∥An∥1/n\lim_{n \rightarrow \infty }\left\| A^{n}\right\|^{1 / n} exists and

r(A)=lim⁡n→∞∥An∥1/n, r(A) = \lim _{n \rightarrow \infty }\left\| A^{n}\right\| ^{1 / n},

where r(A)=max⁡λ∈σ(A)∣λ∣r(A) = \max_{\lambda \in \sigma (A)}\left|\lambda \right| is the spectral radius of the operator AA (Definition 12.2).

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Find the spectrum, spectral radius, and resolvent of the operator (Ax)(t)=∫0tx(s)ds(A x)(t) = \int_{0}^{t} x(s) d s in the spaces:

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(a)

Lp[0,1],p∈[1,∞]L_{p}[0,1], p \in [1, \infty ];

(b)

C[0,1]C[0,1].

Find the spectrum, spectral radius, and resolvent of the operator A:lp→lp,p∈[1,∞)A: l_{p} \rightarrow l_{p}, p \in [1, \infty ):

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(a)

Ax=(0,x1,0,x3,…,0,x2n−1,0,…)A x = \left(0, x_{1}, 0, x_{3}, \ldots , 0, x_{2 n-1}, 0, \ldots \right);

(b)

Ax=(0,x1,0,x2,…,0,xn,0,…)A x = \left(0, x_{1}, 0, x_{2}, \ldots , 0, x_{n}, 0, \ldots \right);

(c)

Ax=(0,x1,x2,0,x4,x5,0…,0,x3n+1,x3n+2,0,…)A x = \left(0, x_{1}, x_{2}, 0, x_{4}, x_{5}, 0 \ldots , 0, x_{3 n+1}, x_{3 n+2}, 0, \ldots \right);

(d)

Ax=(x2,x1,x4,x3,…,x2n+2,x2n+1,…)A x = \left(x_{2}, x_{1}, x_{4}, x_{3}, \ldots , x_{2 n+2}, x_{2 n+1}, \ldots \right);

(e)

Ax=(x2,x3,x1,x5,x6,x4,…,x3n+2,x3n+3,x3n+1,…)A x = \left(x_{2}, x_{3}, x_{1}, x_{5}, x_{6}, x_{4}, \ldots , x_{3 n+2}, x_{3 n+3}, x_{3 n+1}, \ldots \right);

(f)

Ax=(x1,x2,…,xn,0,0,…)A x = \left(x_{1}, x_{2}, \ldots , x_{n}, 0,0, \ldots \right).

Let HH be a Hilbert space, AA a normal operator. Prove that r(A)=∥A∥r(A) = \left\| A\right\| (see problem 8.78).

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Let XX be a Banach space, A,B∈B(X)A, B \in \mathcal{B}(X) and AB=BAA B = B A. Prove that

r(A+B)⩽r(A)+r(B),r(AB)⩽r(A)r(B) r(A+B) \leqslant r(A)+r(B), \quad r(A B) \leqslant r(A) r(B)
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Prove that the integral operator

(Ax)(t)=∫atK(t,s)x(s)ds (A x)(t) = \int _{a}^{t} K(t, s) x(s) d s

is a Volterra operator in the space

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(a)

C[a,b]C[a, b], if the function K(t,s)K(t, s) is continuous for a⩽s⩽ta \leqslant s \leqslant t;

(b)

L2[a,b]L_{2}[a, b], if the function K(t,s)K(t, s) is measurable and bounded for a⩽s⩽ta \leqslant s \leqslant t.

Let AA be a Volterra operator and λ∈C\lambda \in \mathbb {C}. Prove that the equation (I−λA)x=y(I-\lambda A) x = y in a Banach space XX is solvable for every right-hand side and every λ\lambda. Prove that the solution is representable as the series x=∑k=0∞λkAkyx = \sum_{k = 0}^{\infty } \lambda^{k} A^{k} y.

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Let XX be a complex infinite-dimensional Banach space, A∈B(X)A \in \mathcal{B}(X) and σ(A)={0}\sigma (A) = \left\{ 0\right\}. Is it true that AA is compact?

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Let XX be a Banach space, A∈B(X)A \in \mathcal{B}(X) and ∥A∥∈σ(A)\left\| A\right\| \in \sigma (A). Prove that ∥I+A∥=1+∥A∥\left\| I+A\right\| = 1+\left\| A\right\|.

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Let HH be a Hilbert space and let the operator A∈B(H)A \in \mathcal{B}(H) be normal. Prove that ∥A2∥=∥A∥2\left\| A^{2}\right\| = \left\| A\right\|^{2} and, more generally, ∥A2n∥=∥A∥2n\left\| A^{2^{n}}\right\| = \left\| A\right\|^{2^{n}} for every natural n.n.1

Footnotes

  1. In fact, for every normal operator A∈B(H)A \in \mathcal{B}(H) the relation ∥An∥=∥A∥n\left\| A^{n}\right\| = \left\| A\right\|^{n} holds for all n∈Nn \in \mathbb {N}. ↩

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If A\mathcal{A} is a Banach algebra with identity and r:A→[0,∞)r: \mathcal{A} \to [0, \infty ) is the spectral radius, show that rr is upper semicontinuous. If a∈Aa \in \mathcal{A} such that r(a)=0r(a) = 0, show that rr is continuous at aa.

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Verify the statements made in the following example. A measurable function k:[0,1]×[0,1]→Ck: [0,1] \times [0,1] \to \mathbb {C} is called a Volterra kernel if kk is bounded and k(x,y)=0k(x, y) = 0 when x<yx < y. If 1≤p≤∞1 \le p \le \infty and kk is a Volterra kernel, define Vk:Lp(0,1)→Lp(0,1)V_k: L^p(0,1) \to L^p(0,1) by

Vkf(x)=∫01k(x,y)f(y) dy=∫0xk(x,y)f(y) dy. V_k f(x) = \int _0^1 k(x, y) f(y) \, dy = \int _0^x k(x, y) f(y) \, dy.

Then Vk∈B(Lp)V_k \in \mathcal{B}(L^p) and ∥Vk∥≤∥k∥∞\left\| V_k\right\| \le \left\| k\right\|_\infty. If k,hk, h are Volterra kernels and (hk)(x,y)=∫01h(x,t)k(t,y) dt(hk)(x, y) = \int_0^1 h(x, t) k(t, y) \, dt, then hkhk is a Volterra kernel, ∥hk∥∞≤∥h∥∞∥k∥∞\left\| hk\right\|_\infty \le \left\| h\right\|_\infty \left\| k\right\|_\infty, and Vhk=VhVkV_{hk} = V_h V_k. If kk is the characteristic function of {(x,y)∈[0,1]×[0,1]:y<x}\left\{ (x,y) \in [0,1] \times [0,1] : y < x\right\}, then VkV_k is the Volterra operator. If kk is any Volterra kernel, then σ(Vk)={0}\sigma (V_k) = \{ 0\}.

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(McCabe) If X\mathcal{X} is a complex Banach space and T∈B(X)T \in \mathcal{B}(\mathcal{X}), show that the following statements are equivalent.

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(a)

r(T)<1r(T) < 1.

(b)

∥Tm∥<1\left\| T^m\right\| < 1 for some positive integer mm.

(c)

∑n∥Tn(x)∥<∞\sum_n \left\| T^n(x)\right\| < \infty for every xx in X\mathcal{X}.

Define A:L2(R)→L2(R)A : L^2(\mathbb {R}) \to L^2(\mathbb {R}) by (Af)(x)=exp⁡(−x2)f(x−1)(Af)(x) = \exp (-x^2) f(x-1) for all ff in L2(R)L^2(\mathbb {R}).

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(a)

Show that A∈B(L2(R))A \in \mathcal{B}(L^2(\mathbb {R})).

(b)

Find ∥An∥\left\| A^n\right\| and show that r(A)=0r(A) = 0 so that σ(A)={0}\sigma (A) = \left\{ 0\right\}.

(c)

Show that AA is injective.

(d)

Find A∗A^* and show that ran⁡A\operatorname {ran} A is dense.

(e)

Define B=A−1B = A^{-1} with dom⁡B=ran⁡A\operatorname {dom} B = \operatorname {ran} A and show that B∈C(L2(R))B \in \mathcal{C}(L^2(\mathbb {R})) with σ(B)=□\sigma (B) = \square.

For each operator AA,

r(A)=lim⁡n∥An∥1/n, r(A) = \lim _n \left\| A^n\right\| ^{1/n},

in the sense that the indicated limit always exists and has the indicated value.

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Express the norm and the spectral radius of a weighted shift in terms of its weights.

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For which sequences {pk}\left\{ p_k\right\} of positive numbers does there exist an operator AA such that pk=∥Ak∥p_k = \left\| A^k\right\|, k=0,1,2,…k = 0, 1, 2, \ldots?

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Remark.
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The power norms of an operator AA, that is the numbers ∥Ak∥\left\| A^k\right\|, constitute an interesting sequence.

If p={pn}p = \left\{ p_n\right\} is a sequence of positive numbers such that {pn+1/pn}\left\{ p_{n+1}/p_n\right\} is bounded, what, in terms of {pn}\left\{ p_n\right\}, is the spectral radius of the shift on l2(p)l^2(p)?

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Remark.
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Let p={p0,p1,p2,…}p = \left\{ p_0, p_1, p_2, \ldots \right\} be a sequence of strictly positive numbers, and let l2(p)l^2(p) be the set of complex sequences ⟨ξ0,ξ1,ξ2,…⟩\left\langle \xi_0, \xi_1, \xi_2, \ldots \right\rangle with ∑n=0∞pn∣ξn∣2<∞\sum_{n=0}^{\infty } p_n \left|\xi_n\right|^2 < \infty. With respect to the coordinatewise linear operations and the inner product defined by

⟨⟨ξ0,ξ1,ξ2,…⟩,⟨η0,η1,η2,…⟩⟩=∑n=0∞pnξnηn∗, \left\langle \left\langle \xi _0, \xi _1, \xi _2, \ldots \right\rangle , \left\langle \eta _0, \eta _1, \eta _2, \ldots \right\rangle \right\rangle = \sum _{n=0}^{\infty } p_n \xi _n \eta _n^{*},

the set l2(p)l^2(p) is a Hilbert space; it may be called a weighted sequence space.

Construct a quasinilpotent operator whose point spectrum is empty.

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Remark.
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An operator is nilpotent if some positive integral power of it is zero (and the least such power is the index of nilpotence); an operator AA is quasinilpotent if lim⁡n∥An∥1/n=0\lim_n \left\| A^n\right\|^{1/n} = 0. Nilpotence implies quasinilpotence. The spectral mapping theorem implies that if AA is nilpotent, then spec⁡A={0}\operatorname {spec} A = \left\{ 0\right\}. The expression for the spectral radius in terms of norms implies that if AA is quasinilpotent, then spec⁡A={0}\operatorname {spec} A = \left\{ 0\right\}, and, moreover, the converse is true. A nilpotent operator always has a non-trivial kernel, and hence a non-empty point spectrum; for quasinilpotent operators that is not so. (Observe that on finite-dimensional spaces such a construction is clearly impossible.)

Is every quasinilpotent operator analytic?

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Remark.
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If AA is a quasinilpotent operator, and if ff is a function that is analytic in a neighborhood of 00, then ff has a power series expansion, f(z)=∑n=0∞αnznf(z) = \sum_{n=0}^{\infty } \alpha_n z^n, convergent when, say, ∣z∣<ε\left|z\right| < \varepsilon; in that case write f(A)=∑n=0∞αnAnf(A) = \sum_{n=0}^{\infty } \alpha_n A^n. This operator series converges (in the norm), since ∥An∥1/n<ε\left\| A^n\right\|^{1/n} < \varepsilon for nn sufficiently large, so that ∥An∥<εn\left\| A^n\right\| < \varepsilon^n for nn sufficiently large. The correspondence f↦f(A)f \mapsto f(A) between functions analytic at 00 and operators f(A)f(A) (for a fixed quasinilpotent operator AA) is called a functional calculus; it is an extension of the trivial functional calculus for polynomials.

An operator AA is called algebraic if there exists a non-zero polynomial pp such that p(A)=0p(A) = 0; by a natural extension of language it seems proper to say that AA is analytic if there exists a non-zero function ff analytic at 00 such that f(A)=0f(A) = 0. Every nilpotent operator is algebraic.

Is the spectrum of a direct sum of operators always the closure of the union of their spectra?

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Remark.
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The spectrum of the direct sum of two operators is the union of their spectra, and the same is true of the point spectrum, the approximate point spectrum, and the compression spectrum. The extension of this result from two direct summands to any finite number is a trivial induction.

Is it true that to each operator AA and to each positive number δ\delta there corresponds a positive number ε\varepsilon such that if ∥A−B∥<ε\left\| A-B\right\| < \varepsilon, then ∣r(A)−r(B)∣<δ\left|r(A) - r(B)\right| < \delta? Equivalently: if An→AA_n \to A, does it follow that r(An)→r(A)r(A_n) \to r(A)?

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Remark.
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Note that the example of Problem 102 gives no information; in that case the spectral radius is equal to 11 for each term of the sequence and also for the limit.

If AA and BB are operators such that spec⁡(A+nB)=spec⁡A\operatorname {spec}(A + nB) = \operatorname {spec} A for n=0,1,2,…n = 0, 1, 2, \ldots, does it follow that BB is quasinilpotent?

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Remark.
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If small is interpreted in the sense of norm, a perturbation that produces only a small change in the spectrum does not have to be small: if, for instance, A=(α00β)A = \begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix} and B=(0100)B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, then spec⁡(A+B)=spec⁡A={α,β}\operatorname {spec}(A+B) = \operatorname {spec} A = \left\{ \alpha , \beta \right\}, but BB is not small; in fact spec⁡(A+nB)=spec⁡A\operatorname {spec}(A+nB) = \operatorname {spec} A for all nn. The answer is clearly yes in some degenerate cases: if, for instance, AA is a scalar, and if spec⁡(A+nB)=spec⁡A\operatorname {spec}(A+nB) = \operatorname {spec} A for just one non-zero value of nn, then spec⁡B=0\operatorname {spec} B = 0.

If r(A)≤1r(A) \leq 1, is AA similar to a contraction?

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A Volterra operator with a bounded kernel is quasinilpotent.

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Remark.
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A Volterra kernel is a kernel KK in L2(μ×μ)L^{2}\left(\mu \times \mu \right) (where μ\mu is Lebesgue measure on the unit interval) such that K(x,y)=0K(x,y) = 0 when x<yx < y; equivalently, a Volterra kernel is a Hilbert--Schmidt kernel that vanishes above the diagonal (x=yx = y) of the unit square. The integral operator AA (a Volterra operator) induced by a Volterra kernel KK is given by

(Af)(x)=∫0xK(x,y)f(y) dy. (Af)(x) = \int _0^x K(x,y) f(y) \, dy.

"Bounded" refers to the kernel, not the operator: the assumption is that KK is bounded almost everywhere in the unit square. An operator AA is quasinilpotent if lim⁡n∥An∥1/n=0\lim_n \left\| A^n\right\|^{1/n} = 0.

Is every Volterra operator quasinilpotent?

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Remark.
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A Volterra kernel is a kernel KK in L2(μ×μ)L^{2}\left(\mu \times \mu \right) (where μ\mu is Lebesgue measure on the unit interval) such that K(x,y)=0K(x,y) = 0 when x<yx < y. The integral operator induced by a Volterra kernel KK is a Volterra operator, given by (Af)(x)=∫0xK(x,y)f(y) dy(Af)(x) = \int_0^x K(x,y) f(y) \, dy. An operator AA is quasinilpotent if lim⁡n∥An∥1/n=0\lim_n \left\| A^n\right\|^{1/n} = 0.

If AA is hyponormal, then ∥An∥=∥A∥n\left\| A^n\right\| = \left\| A\right\|^n for every positive integer nn.

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Remark.
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An operator AA such that A∗A≥AA∗A^{*}A \geq AA^{*} is called hyponormal.

Compute the spectral radius of the Volterra operator AA on L2(0,1)L^{2}\left({0,1}\right) defined by the formula

Af(x)=∫0xf(t) dt. Af(x)=\int _0^xf(t)\, dt.
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Let AA be a commutative Banach algebra with unit. The radical of AA is defined to be the intersection of all maximal ideals of AA. Prove that the following three statements about an element x∈Ax \in A are equivalent:

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(a)

xx is in the radical of AA.

(b)

lim⁡n→∞∥xn∥1/n=0\lim_{n \rightarrow \infty }\left\| x^{n}\right\|^{1 / n} = 0

(c)

h(x)=0h(x) = 0 for every complex homomorphism of AA.

Find an element xx in a Banach algebra AA (for instance, a bounded linear operator on a Hilbert space) such that xn≠0x^{n} \neq 0 for all n>0n > 0, but lim⁡n→0∥xn∥1/n=0\lim_{n \rightarrow 0}\left\| x^{n}\right\|^{1 / n} = 0.

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Suppose AA is a commutative Banach algebra with unit, and let Δ\Delta be the set of all complex homomorphisms of AA, as in Sec. 18.16. Associate with each x∈Ax \in A a function x^\hat{x} on Δ\Delta by the formula

x^(h)=h(x)(h∈Δ) \hat{x}(h) = h(x) \quad (h \in \Delta )

x^\hat{x} is called the Gelfand transform of xx. Prove that the mapping x→x^x \rightarrow \hat{x} is a homomorphism of AA onto an algebra A^\hat{A} of complex functions on Δ\Delta, with pointwise multiplication. Under what condition on AA is this homomorphism an isomorphism? (See Exercise 12).

Prove that the spectral radius ρ(x)\rho (x) is equal to

∥x^∥∞=sup⁡{∣x^(h)∣:h∈Δ} \left\| \hat{x}\right\| _{\infty } = \sup \left\{ \left|\hat{x}(h)\right|: h \in \Delta \right\}

Prove that the range of the function x^\hat{x} is exactly the spectrum σ(x)\sigma (x).

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Define a shift operator SS and a multiplication operator MM on ℓ2\ell^{2} by

(Sx)(n)={0 if n=0x(n−1) if n≥1(Mx)(n)=(n+1)−1x(n) if n≥0 \begin{aligned} (S x)(n) & = \begin{cases} 0 & \text{ if } n = 0 \\ x(n-1) & \text{ if } n \geq 1\end{cases} \\ (M x)(n) & = (n+1)^{-1} x(n) \quad \text{ if } n \geq 0 \end{aligned}

Put T=MST = M S. Show that TT is a compact operator which has no eigenvalue and whose spectrum consists of exactly one point. Compute ∥Tn∥\left\| T^{n}\right\|, for n=1,2,3,…n = 1,2,3, \ldots, and compute lim⁡n→∞∥Tn∥1/n\lim_{n \rightarrow \infty }\left\| T^{n}\right\|^{1 / n}.

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Use the identity (xy)n=x(yx)n−1y(x y)^{n} = x(y x)^{n-1} y to prove that xyx y and yxy x always have the same spectral radius.

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Problem 3.10.19 from Rudin W. "Functional Analysis" [Rud91]More problems in Chapter 3.10 "Banach Algebras"

Let SRS_{R} be the right shift, acting on ℓ2\ell^{2}, as in Exercise 2. Let {cn}\left\{ c_{n}\right\} be a sequence of complex numbers such that cn≠0c_{n} \neq 0 but cn→0c_{n} \rightarrow 0 as n→∞n \rightarrow \infty. Define M∈B(ℓ2)M \in \mathscr {B}\left(\ell^{2}\right) by

(Mf)(n)=cnf(n)(n≥0) (M f)(n) = c_{n} f(n) \quad (n \geq 0)

and define T∈B(ℓ2)T \in \mathscr {B}\left(\ell^{2}\right) by T=MSRT = M S_{R}.

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(a)

Compute ∥Tm∥\left\| T^{m}\right\|, for m=1,2,3,…m = 1,2,3, \ldots.

(b)

Show that σ(T)={0}\sigma (T) = \left\{ 0\right\}.

(c)

Show that TT has no eigenvalue. (Its point spectrum is therefore empty, although its spectrum consists of a single point!)

(d)

Show that (λI−T)−1(\lambda I-T)^{-1} does not have a pole at 0.

(e)

Show that TT is a compact operator.

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(a)

Prove that AA is commutative if there is a constant M<∞M < \infty such that ∥xy∥≤M∥yx∥\left\| x y\right\| \leq M\left\| y x\right\| for all xx and yy in A. Hint: ∥w−1yw∥≤M∥y∥\left\| w^{-1} y w\right\| \leq M\left\| y\right\| if ww is invertible in AA. Replace ww by exp⁡(λx)\exp (\lambda x), where x∈Ax \in A and λ∈C\lambda \in \mathbb {C}. Continue as in Theorem 12.16.

(b)

Prove that AA is commutative if ∥x2∥=∥x∥2\left\| x^{2}\right\| = \left\| x\right\|^{2} for every x∈Ax \in A. Hint: Show that ∥x∥=ρ(x)\left\| x\right\| = \rho (x). Use Exercise 1 to deduce that ∥w−1yw∥=∥y∥\left\| w^{-1} y w\right\| = \left\| y\right\|. Continue as in (a).

If xy=yxx y = y x for some xx and yy in a Banach algebra, prove that

ρ(xy)≤ρ(x)ρ(y) and ρ(x+y)≤ρ(x)+ρ(y). \rho (x y) \leq \rho (x) \rho (y) \quad \text{ and } \quad \rho (x+y) \leq \rho (x)+\rho (y).
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Problem 3.11.13 from Rudin W. "Functional Analysis" [Rud91]More problems in Chapter 3.11 "Commutative Banach Algebras"

Let tt be a large positive number, and define a norm on φ2\varphi^{2} by

∥w∥=∣w1∣+t∣w2∣ if w=(w1,w2) \left\| w\right\| = \left|w_{1}\right|+t\left|w_{2}\right| \quad \text{ if } w = \left(w_{1}, w_{2}\right)

Let AA be the algebra of all complex 2-by-2 matrices, with the corresponding operator norm:

∥y∥=max⁡{∥y(w)∥:∥w∥=1}(y∈A) \left\| y\right\| = \max \left\{ \left\| y(w)\right\| : \left\| w\right\| = 1\right\} \quad (y \in A)

For y∈Ay \in A, let y∗y^{*} be the conjugate transpose of yy. Consider a fixed x∈Ax \in A, namely,

x=[0t210].  x =\left[\begin{smallmatrix} 0 & t^{2} \\ 1 & 0 \end{smallmatrix}\right] \text{. }

Prove the following statements.

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(a)

∥x(w)∥=t∥w∥\left\| x(w)\right\| = t\left\| w\right\|; hence ∥x∥=t\left\| x\right\| = t.

(b)

σ(x)={t,−t}=σ(x∗)\sigma (x) = \left\{ t,-t\right\} = \sigma \left(x^{*}\right).

(c)

σ(xx∗)={1,t4}=σ(x∗x)\sigma \left(x x^{*}\right) = \left\{ 1, t^{4}\right\} = \sigma \left(x^{*} x\right).

(d)

σ(x+x∗)={1+t2,−1−t2}\sigma \left(x+x^{*}\right) = \left\{ 1+t^{2},-1-t^{2}\right\}.

(e)

Therefore commutativity is required in Theorem 11.23 and in Exercise 12.

(f)

If F(y)F(y) is the sum of the four entries in yy, for y∈Ay \in A, then FF is a positive functional on AA.

(g)

The equality ∥F∥=F(e)\left\| F\right\| = F(e) [see (e) of Theorem 11.31] does not hold, because F(e)=2F(e) = 2 and F(x)=1+t2F(x) = 1+t^{2}, so that ∥F∥>t\left\| F\right\| > t.

(h)

If KK is the set of all positive functionals ff on AA that satisfy f(e)≤1f(e) \leq 1 (as in Theorem 11.33), then KK has many extreme points, although 0 is the only multiplicative linear functional on AA. Commutativity is therefore required in the implication (c)→(a)(c) \rightarrow (a) of Theorem 11.33.

Suppose AA is a Banach algebra, mm is an integer, m≥2,K<∞m \geq 2, K < \infty, and

∥x∥m≤K∥xm∥ \left\| x\right\| ^{m} \leq K\left\| x^{m}\right\|

for every x∈Ax \in A. Show that there exist constants Kn<∞K_{n} < \infty, for n=1,2,3,…n = 1,2,3, \ldots, such that

∥x∥n≤Kn∥xn∥(x∈A) \left\| x\right\| ^{n} \leq K_{n}\left\| x^{n}\right\| \quad (x \in A)

(This extends Theorem 11.12).

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Put ϕ(t)=exp⁡(−t2)\phi (t) = \exp \left(-t^{2}\right). Define S∈B(L2)S \in \mathscr {B}\left(L^{2}\right), where L2=L2(R)L^{2} = L^{2}(R), by

(Sf)(t)=ϕ(t)f(t−1)(f∈L2) (S f)(t) = \phi (t) f(t-1) \quad \left(f \in L^{2}\right)

so that (S2f)(t)=ϕ(t)ϕ(t−1)f(t−2)\left(S^{2} f\right)(t) = \phi (t) \phi (t-1) f(t-2), etc. (Note that SS is presented in its polar decomposition S=PUS = P U).

Find S∗S^{*}. Compute that

∥Sn∥=exp⁡{−(n−1)n(n+1)12}(n=1,2,3,…) \left\| S^{n}\right\| = \exp \left\{ -\frac{(n-1) n(n+1)}{12}\right\} \quad (n = 1,2,3, \ldots )

Conclude that SS is one-to-one, that R(S)\mathscr {R}(S) is dense in L2L^{2}, and that σ(S)={0}\sigma (S) = \left\{ 0\right\}. Define TT, with domain D(T)=R(S)\mathscr {D}(T) = \mathscr {R}(S), by

TSf=f(f∈L2) T S f = f \quad \left(f \in L^{2}\right)

Prove that σ(T)\sigma (T) is empty.

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Let AA be a normed algebra over a complete normed field KK. Prove that if x∈Ax \in A, then the limit exists

ρ(x)=lim⁡n→∞∥xn∥1/n \rho (x)=\lim _{n \rightarrow \infty }\left\| x^{n}\right\| ^{1 / n}
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Prove that lim⁡k→∞Ak=O⟺r(A)<1\lim_{k \rightarrow \infty } \mathcal{A}^{k}=\mathcal{O} \Longleftrightarrow r(\mathcal{A})<1.

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