Examples Discussion 11

(April 8, 2018) Examples discussion 11

Paul Garrett [email protected] http://www.math.umn.edu/egarrett/ [This document is http://www.math.umn.edu/˜garrett/m/real/examples 2016-17/real-disc-11.pdf]

[11.1] For T : V → V a continuous (=bounded) linear map of a Banach space V to itself, show that the operator norm is an upper bound for absolute values of all eigenvalues λ: |λ|C ≤ |T |op. Further, show that |T |op is an upper bound for all of the spectrum, that is, T − λ is invertible for |λ|C > |T |op. Discussion: First, for T v = λ · v for 0 6= v ∈ V , without loss of generality take |v| = 1, and then

|T |op = sup |T w| ≥ |T v| = |λ · v| = |λ|C · |v|V = |λ| |w|≤1

Second, for |λ|C > |T |op, we have |T/λ|op < 1, so S = 1 + T/λ + (T/λ)2 + ... = lim(1 + T/λ + (T/λ)2 + ... + (T/λ)N ) N is convergent in the operator norm on the Banach space of continuous/bounded linear operators on V , since the tails go to 0. Since 1 − T/λ is continuous, as expected

(1 − T/λ) · S = S · (1 − T/λ) = lim(1 − T/λ) · (1 + T/λ + (T/λ)2 + ... + (T/λ)N ) = lim 1 − (T/λ)N+1 = 1 N N

since (T/λ)N → 0: if there were any doubt,

N N |(T/λ) |op ≤ |T/λ| −→ 0

−1 since |T/λ|op < 1. Thus, the inverse S = (1 − T/λ) exists (as a continuous linear operator), and λ is not in any part of the spectrum. ///

[11.2] (Weyl’s criterion: approximate eigenvectors and continuous spectrum) Let T : V → V be a continuous linear operator on a Hilbert space V . For λ ∈ C, a sequence {vn} of vectors (normalized so that all their lengths are 1 or at least bounded away from 0) such that (T − λ)vn → 0 as n → +∞ is an approximate eigenvector for λ. Show that for λ not an eigenvalue for T , λ has an approximate eigenvector if and only if λ is in the spectrum of T . Discussion: In fact, this criterion is not reliable for detecting some types of residual spectrum. [1] Speciﬁcally, for λ not an eigenvector, we need to assume that (T − λ)V is not closed for existence of approximate eigenvectors to imply that λ is in the spectrum. This is implied by T being normal, for example self-adjoint. We give an example at the end of the discussion where this criterion fails. First, certainly, if λ is an eigenvector, with non-zero eigenvalue v, the constant sequence v, v, v, . . . ﬁts the requirement. [2] For general spectrum, let S = T − λ. For v1, v2,... with |vn| = 1 and Svn → 0, any alleged (continuous ) S−1 would give, interchanging S−1 and the limit by continuity,

−1 −1 0 = S (lim Svn) = lim S Svn = lim vn n n n

[1] Recall that residual spectrum of T is λ such that T − λ is injective, but does not have dense image.

[2] Recall that when there is an everywhere-deﬁned, linear inverse S−1 to S, necessarily S is a continuous bijection, and by the open mapping theorem S is open. That is, there is δ > 0 such that |Sv| ≥ δ · |v| for all v. This exactly asserts the boundedness of S−1, so S−1 is continuous.

1 Paul Garrett: Examples discussion 11 (April 8, 2018)

contradiction. Thus, existence of an approximate eigenvector for T − λ implies that T − λ is not invertible. Conversely, for S = T − λ not invertible, but λ not an eigenvector, then S is injective but not surjective. As noted above, we must assume that the image of S is not closed. In that case, S is injective, not surjective, and by non-closedness of the image there is vo (with |vo| = 1) not in the image of S, and v1, v2,... such that Sv1, Sv2,... → vo.

[11.1] Theorem: For λ not an eigenvalue for T , and for (T − λ)V not closed, λ is in the spectrum of T if and only if λ has an approximate eigenvector.

[11.2] Remark: This criterion is not uniformly reliable for detecting residual spectrum, which is why we must impose a further condition. [3] For example, we have seen that, for T : V → V a norma linear operator, for λ in the spectrum but not an eigenvalue, (T − λ)V is dense in V but is not all of V . Thus, the hypothesis of the theorem is met for normal T . We give an example of failure to detect residual spectrum after the proof.

Proof: Certainly if λ is an eigenvector, with non-zero eigenvalue v, the constant sequence v, v, v, . . . ﬁts the requirement. [4] For general spectrum, let S = T − λ. For v1, v2,... with |vn| = 1 and Svn → 0, any alleged (continuous ) S−1 would give, interchanging S−1 and the limit by continuity,

−1 −1 0 = S (lim Svn) = lim S Svn = lim vn n n n contradiction. Thus, existence of an approximate eigenvector for T − λ implies that T − λ is not invertible.

Conversely, for S = T − λ not invertible, but λ not an eigenvector, then S is injective but not surjective. We further assume that the image of S is not closed. [5] In that case, S is injective, not surjective, and by non-closedness of the image there is vo (with |vo| = 1) not in the image of S, and v1, v2,... such that Sv1, Sv2,... → vo. If {vn} were a Cauchy sequence, then it would have a limit, and by continuity of S

vo = lim Svn = S(lim vn) n n

and vo would be in the image of S, contradicting our assumption. Thus, {vn} is not Cauchy. In particular, we can replace {vn} by a subsequence such that there is δ > 0 such that |vm − vn| ≥ δ for all m 6= n. Then wn = vn − vn+1 forms an approximate 0-eigenvector, since their lengths are bounded away from 0, and

Swn = S(vn − vn+1) = Svn − Svn+1 −→ vo − vo = 0

as desired. ///

As noted, the case that λ is not an eigenvector, T − λ is not surjective, and/but the image of S = T − λ is closed, can only occur for non-normal T . For example, T : `2 → `2 by

T (c1, c2,...) = (c1, 0, c2, 0, c3, 0,...)

[3] Recall that residual spectrum of T is λ such that T − λ is injective, but does not have dense image.

[4] Recall that when there is an everywhere-deﬁned, linear inverse S−1 to S, necessarily S is a continuous bijection, and by the open mapping theorem S is open. That is, there is δ > 0 such that |Sv| ≥ δ · |v| for all v. This exactly asserts the boundedness of S−1, so S−1 is continuous.

[5] The image is not closed, for example, when T (hence S) has no residual spectrum, which is the case when T (hence S) is normal, or self-adjoint.

2 Paul Garrett: Examples discussion 11 (April 8, 2018)

is injective, not surjective, and has closed image. It is not invertible, but there is no approximate eigenvector for 0, so the criterion fails in this (non-normal) example. ///

[11.3] Show that the multiplication operator T : L2(R) → L2(R) by T f(x) = f(x) · sin x has empty discrete spectrum. Show that it is self-adjoint. Show that T has continuous spectrum the interval [−1, 1]. (We know that self-adjoint (or merely normal) operators have only point spectum and continuous spectrum, that is, no left-over residual spectrum.) Discussion: This operator is self-adjoint, since sin x is real-valued:

Z Z hT f, gi = f(x) · sin x · g(x) dx = f(x) · g(x) · sin x dx = hf, T gi R R

For a function f and ﬁxed λ ∈ C such that f(x) · sin x = λ · f(x) for almost all x, for x such that f(x) 6= 0, necessarily sin x = λ. Since sin x assumes any particular value at most countably many times, f = 0 almost everywhere. Thus, there are no eigenvalues. Since T is self-adjoint, it is normal, so there is no residual spectrum. Thus, Weyl’s criterion via approximate eigenvectors suﬃces to determine the remainder of the spectrum, which will be continuous. Given a value λ ∈ [−1, 1], let xo ∈ R be such that sin xo = λ. We claim that an approximate eigenvector for λ can be formed by functions concentrated ever-more-closely at xo, such as √  n (for |x − x | ≤ 1 )  o 2n vn(x) =  0 (otherwise)

Thus, T vn − λvn → 0, and {vn} is an approximate identity for λ, so every λ ∈ [−1, 1] is in the continuous spectrum. ///

[11.4] Let r1, r2, r3,... be an enumeration of the rational numbers inside the interval [0, 1]. Deﬁne 2 2 T : ` → ` by T (c1, c2,...) = (r1c1, r2c2,...). Show that T is a continuous/bounded linear operator, is self-adjoint, has eigenvalues exactly the r1, r2,..., and continuous spectrum the whole interval [0, 1] (with rationals removed, if one insists on disjointness of discrete and continuous spectrum).

Discussion: Since the set {|r1|, |r2|,...} is bounded by 1, the operator norm of T is at most 1, so it is bounded, hence continuous. Since the rn are all real, the operator is self-adjoint:

X hT (a1, a2,...), (b1, b2,...)i = h(r1a1, r2a2,...), (b1, b2,...)i = rnan · bn n

X = an · rnbn = h(a1, a2,...),T (b1, b2,...)i n

When λ · (c1, c2,...) = T (c1, c2,...) = (r1c1, r2c2,...), necessarily λ · cn = rn · cn for all n. When cn 6= 0, this implies λ = rn. Since the rn are distinct, there can be (at most) one index n for which cn 6= 0, and then λ = rn. Conversely, every rn is obviously an eigenvalue.

3 Paul Garrett: Examples discussion 11 (April 8, 2018)

Since we know that the whole spectrum is closed in C, it contains at least the closure of the rationals in [6] [0, 1], namely, [0, 1] itself. Since T is self-adjoint, its spectrum is contained in R. Since the spectrum is bounded by |T |op = 1, it is contained in [−1, 1].

To see that λ ∈ [−1, 0) is not in the spectrum, in that (T − λ)(c1, c2,...) = ((r1 − λ)c1, (r2 − λ)c2,...), we −1 −1 have |rn − λ| ≥ |λ| > 0, so the inverse (T − λ) can be written down immediately: (T − λ) (c1, c2,...) = −1 −1 −1 −1 [7] ((r1 − λ) c1, (r2 − λ) c2,...) and there is a uniform upper bound |(rn − λ) | ≤ |λ| . Finally, given 2 irrational λ ∈ [0, 1], let rn1 , rn2 ,... be rationals such that rni → λ. With standard basis {en} for ` , we claim

that {eni } is an approximate eigenvector for λ: given ε > 0, let N be suﬃciently large so that |rni − λ| < ε for i ≥ N. For ni ≥ N,

|(T − λ)e | = |(r − λ)e | = |r − λ| · |e | 2 = |r − λ| < ε ni ni ni ni C ni ` ni C

Thus, indeed, (T − λ)eni → 0, and the eni give an approximate identity for λ, so λ is in the spectrum. ///

2 2 [11.5] Let r1, r2, r3,... be a bounded sequence of complex numbers. Deﬁne T : ` → ` by T (c1, c2,...) = (r1c1, r2c2,...). Show that T is compact if and only if rn → 0.

2 Discussion: Let e1, e2,... be the standard (Hilbert-space) basis for ` . If the rn do not go to 0, then there

is a subsequence rn1 , rn2 ,... bounded away from 0. Since T is compact, the images T eni = rni eni must have 2 2 2 a convergent subsequence. But |rni eni − rnj enj | = |rni | + |rnj | for i 6= j, and this is bounded away from 0, so there is no convergent subsequence, contradicting the compactness of T . Thus, in fact, rn → 0. For the converse, perhaps the most economical approach is to observe that T is an operator-norm limit of ﬁnite-rank operators, hence compact:

Tn(c1, c2, . . . , cn, cn+1,...) = (c1, c2, . . . , cn, 0, 0,...)

The estimate on the operator norms is

|T − Tn|op = sup |(0,..., 0, rn+1vn+1,...)| = sup |rk| T o 0 |v|≤1 k≥n

Less eﬃciently, we can refer to deﬁnitions, and use the total boundedness criterion for compact closure. Given 2 ε > 0, let N be large enough so that |rn| < ε for n ≥ N. Write v = (v1, v2,...) ∈ ` as

v = (v1, . . . , vN , 0, 0,...) + (0,..., 0, vN+1, . . . , vN+2,...) | {z } | {z } v0 N | {z } v00

[6] The proof that self-adjoint operators T have spectrum inside R has more content than just the analogous assertion about eigenvectors. For T v = λv with v 6= 0, of course

λhv, vi = hλv, vi = hT v, vi = hv, T vi = hv, λvi = λhv, vi

shows that any eigenvalues are real. Since self-adjoint operators have no residual spectrum, to ﬁnd the rest of the spectrum it suﬃces to identify approximate eigenvectors. Note that for self-adjoint T always hT v, vi = hv, T vi = hT v, vi, so hT v, vi is real. Then for (T − λ)vn → 0, certainly h(T − λ)vn, vni → 0, so the imaginary parts go to 0. These are Imh(T − λ)vn, vni = ImhT vn, vni + Im(λ · hvn, vni) = 0 + Im(λ) · hvn, vni

Since |vn| are bounded away from 0, there can be an approximate identity only for λ ∈ R. ///

[7] For such a simple operator, a similar device shows that λ 6∈ R is not in the spectrum.

4 Paul Garrett: Examples discussion 11 (April 8, 2018)

Let B0 be the intersection of the unit ball B ⊂ `2 with the copy of CN ⊂ `2 with non-zero components only at the ﬁrst N places. Let B00 be the intersection of B with the subspace of `2 with 0 entries at the ﬁrst N places. Certainly B0 + B00 ⊃ B and B0 ⊥ B00.

By design, |T v00| ≤ ε for v00 ∈ B00. Since TB0 is a bounded subset of a ﬁnite-dimensional space CN , it has compact closure, so is totally bounded, so can be covered by ﬁnitely-many ε-balls U1,...,Uk. Then 0 00 00 00 00 TB ⊂ TB + TB ⊂ (U1 + TB ) ∪ ... ∪ (Uk + TB ), and every Ui + TB is contained in a 2ε-ball. Thus, TB is totally bounded, hence, has compact closure. ///

R x 2 n [11.6] Let V be the Volterra operator V f(x) = 0 f(t) dt on L [0, 1]. Show that |V |op → 0 as n → +∞. Show that the spectrum of V is just {0}.

n Discussion: In fact, |V |op → 0 is a weaker conclusion than was intended, in part because it would not help n 1/n showing that the spectrum of V is just {0}. Rather we would want to show something like limn |V |op = 0.

n Nevertheless, one way to estimate the behavior of V is to show that that |V |op < 1. For the latter, recall that a linear map T : L2(X) → L2(X) given by K(, ) ∈ L2(X × X) by Z T f(x) = K(x, y) f(y) dy X is a Hilbert-Schmidt operator, with operator norm bounded by the L2(X × X) norm of K(, ) (this is the Hilbert-Schmidt norm of the operator). Thus,  Z 1 Z 1  1 (for y < x) 2 Z 1 Z x Z 1 |V |2 ≤ dx dy = 1 dx dy = x dx = 1 < 1 op 2 0 0  0 (for y > x) 0 0 0

n n Thus, certainly, |V |op ≤ |V |op → 0. ///

R x We also recall the argument that V has no eigenvalues: when 0 f(t) dt = λ · f(x) for λ 6= 0, application of Cauchy-Schwarz-Bunyakowsky to the left-hand side shows that f is continuous. Then the equation shows that v ∈ C1 (and induction shows that f ∈ C∞). Diﬀerentiating the eigenfunction condition, f = λ · f 0. We know how to solve this equation: all solutions are multiples of x → ex/λ. However, the eigenfunction relation also shows that f(0) = 0 (and f is continuous, so this holds in a strong sense: we are not allowed to change its values on sets of measure 0), which does not hold for any of these functions. Thus, there are no eigenvalues. ///

n 1/n Now we return to demonstration of the stronger result, that |V |op → 0. To do so, we consider the integral/Schwartz kernel for the iterate V n:

Z x Z xn−1 Z x2 Z x1 n (V f)(x) = ... f(y) dy dx1 dx2 . . . dxn−2 dxn−1 0 0 0 0 Changing the order of integration to isolate the kernel, this is

Z x Z x Z xn−1 Z x2 n (V f)(x) = f(y) ... 1 dx1 dx2 . . . dxn−2 dxn−1 dy 0 y y y

(x − y)n−1 By induction, the inner integral is for 0 ≤ y ≤ x. That is, the kernel K (x, y) for V n is (n − 1)! n

 (x − y)n−1  (for y < x) Kn(x, y) = (n − 1)!  0 (for y > x)

5 Paul Garrett: Examples discussion 11 (April 8, 2018)

and its L2 norm squared is

Z 1 Z x Z 1 Z x 2(n−1) 2 (x − y) |Kn(x, y)| dy dx = 2 dy dx 0 0 0 0 ((n − 1)!)

Z 1 x2n−1) 1 1 2 = 2 dx = 2 < 0 ((n − 1)!) (2n − 1) ((n − 1)!) (2n − 1) (2n) n! Thus, its L2 norm is bounded by 1/n!. Since [8] (n!)1/n → +∞, the spectral radius is 0. ///

n To show that V − λ is invertible for all λ 6= 0, it is not sufficient to know that |V |op < 1 nor that |V |op → 0, n 1/n but knowing |V |op → 0 does suffice, as follows. To show T − λ = −λ ◦ (1 − T/λ) is invertible for λ 6= 0, it suffices to show that the series

1 + T/λ + (T/λ)2 + (T/λ)3 + ...

for the obvious candidate for (1 − T/λ)−1 converges in operator norm. That is, we want

2 3 1 + |T/λ|op + |(T/λ) |op + |(T/λ) |op + . . . < +∞

The left-hand side is −1 2 −2 3 −3 1 + |T |op · |λ| + |T |op · |λ| + |T |op · |λ| + ... Applying the root test,

1/n n −n n 1/n −1 −1 n 1/n −1 lim sup |T |op · |λ| ≤ lim sup(|T |op) · |λ| = |λ| · lim sup(|T |op) ≤ |λ| · 0 < 1 n n n so the series converges for all λ 6= 0. That is, (T − λ)−1 exists for all λ 6= 0. ///

n 1/n [11.3] Remark: For self-adjoint (or, more generally, normal) operators T , in fact limn |T |op = |T |op. n 1/n However, as we see for the Volterra operator, in general limn |T |op ≤ |T |op. For not-necessarily-normal n 1/n operators, the spectral radius is lim supn |V |op . The argument given for the Volterra operator shows that in general the spectral radius is an upper bound for the spectrum.

[11.4] Remark: In fact, the spectral radius is a sharp bound for the absolute values |λ| for λ in the spectrum. [... iou ...]

[8] That (n!)1/n → +∞ certainly follows from Stirling’s asymptotic, but also more elementary considerations. For example, for each ﬁxed 1 ≤ k ∈ Z, for n ≥ k

1/n 1 n−k+1 1/n 1− k−1 ≤ (k ) = k n −→ k n!

Since this holds for every ﬁxed k, the limit must be +∞.