PROGRESS ON THE INVARIANT SUBSPACE PROBLEM - University Of Chicago

PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 7 supposing that there exists µ X such that µ Im(T ). Let Vn Im [T ]n X be a closed proper subspace. It follows that Vn 1 Vn is a closed proper subspace (in essence, that these sets are contracting). One proves this by induction, letting S (T ) be injective by assumption. Pick x Im(S) so that if we assume Im(S) Im(S 2 ) then Sx S 2 y for some y and thus x Sy gives a contradiction. The same holds for S n , S n 1 for any n N. We now apply the Riesz Lemma (2.1.2) to form a sequence, {xn }n N B, inf xn v v Vn 1 1 2 and so T xn T xm 2 for any n, m N which gives a sequence with no convergent subsequence. This contradicts the compactness of T . 3. Spectral Theory and Linear Methods This section will delve into some elementary functional calculus in order to better understand compact operators and derive some tenets of Spectral Theory. My goal is to avoid any strong complex analysis. From here, we will provide a proof of Gelfand’s Spectral Radius Formula and solve the Invariant Subspace Problem for Compact Operators. 3.1. Elementary Functional Calculus. Definition 3.1.1. Let (X, . ) be a Banach space on which T X X is a linear operator. The spectrum of the operator, (T ) C, is defined as the set of complex values, (T ), for which (T ) does not have a bounded inverse. Definition 3.1.2. The resolvent function, R(T, ), is defined as this bounded inverse, R(T, ) (T ) 1 , when it exists. The domain of the resolvent function, the resolvent set (T ), is the complement of (T ). Proposition 3.1.3. If T L(X) is compact and (T ) then is an eigenvalue. Theorem 3.1.4. The spectrum, (T ), is non-empty, compact, and contained in the complex disk, D {z C z T }. Proof. [B88] Define the map g( ) X X, g( ) k 0 Tk k 1

8 BEN GOLDMAN where for convention we say T 0 I. If T then k 0 T k k k 0 k 1 Tk and g( ) converges in norm. We can now extrapolate further, 1 1 1 (T )g( ) (T ) I 2 T 3 T 2 . . . I which gives us that g( ) R(T, ). We thus have that g( ) if and only if (T ) which gives us that (T ) D. The other claims follow from the maximum modulus principle. 3.2. Simple Invariant Subspace Case. Remark. Before seguing to the work of Gelfand, I have included a simple case of the invariant subspace problem that relies on these power series expansions. The motivation stems from the taylor expansion, ( 1)n tn n 1 (3.2.1) 1 t 1 k 1 n! n 1 k 1 2 Definition 3.2.2. A linear operator, T X X, has a hyperinvariant subspace, W X if S(W ) W for any operator, S, that commutes with T . Theorem 3.2.3 (Beauzamy). If T X X is a linear operator on X such that T , T I 1 2 and T 2 T 1 4 then T has a nontrivial, closed hyperinvariant subspace. Proof. [B88] Define the sum ( 4)n n 1 n k 1 T T 2 I K n! n 1 k 1 2 S I It follows that S (and therefore K) is well defined due to (3.2.1), and that since for t 1 we have 1 1 t 1 then we conclude K 1. Further reference to the Taylor Expansion gives: (3.2.4) S 2 I 4(T T 2 ) (K 2T ) (S I 2T ) 0

PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 9 Let F K 2T . We observe F is nonzero since F K 2T 2T K 0 We turn to the subspace U ker(F ). To show that U contains a nonzero vector, we first claim that S I 2T is nonzero since, if not, it would imply that S I 2(T I) which contradicts our assumption that T I 1 2. Since F commutes with T , U is a hyperinvariant subspace. Remark. This stronger notion of a hyperinvariant subspace will carry throughout the paper. As it turns out, these objects tend to arise more organically than the invariant subspaces, and we will often solve our simpler problem as a corollary of this broader notion. 3.3. Gelfand’s Spectral Radius Formula. Definition 3.3.1. Let (X, . ) be a Banach space. The spectral radius of an operator, T X X, r(T ), is given by r(T ) max (T ) Remark. We can further extrapolate about r(T ) using the function g( ) from before. The root test (from calculus) gives a further criteria for when g( ) is well 1 n defined, lim supn T n . It remains to be shown that this limit is well defined. Proposition 3.3.2. The identity (T )n (T n ) holds for n N where if A C then An {an , a A} Proof. We may write (T n n ) (T ) (T n 1 T n 2 . . . n 1 ) (T 1 ) (T 2 ) (T n 1 ) (T ) where 1 , 2 , . . . , n 1 C. Observe that ker(T ) ker(T n n ). We have that n (T n ) if and only if T n n has a bounded inverse, and therefore bijective. This occurs if and only if each of these factors is also bijective and bounded, and so Corollary 2.2.3 gives that each has a bounded inverse. Thus, if n (T n ) then (T ). Moreover, if n (T n ) then there exists some i such that i (T ). However, we deduce that in n and so n (T )n . Corollary 3.3.3. It follows that (T )n (T n ). Theorem 3.3.4 (Gelfand). The spectral radius of an operator, r(T ), is given by 1 n 1 n r(T ) lim supn T n limn T n

10 BEN GOLDMAN Proof. [B11] It remains to be shown that limn T n exists. Define a sequence, {an }n N , by an log T n Since T is bounded ann is non-increasing and bounded. Moreover, we use the inequalities below to deduce some important properties 1 n T n m T n T m an m an am T nm T n anm man By the Euclidean Algorithm, we deduce that for m n we have that m am qan ar an n , 0 r n, and maxr n arr . If inf n N ann it suffices to show where q m n limn an . For each " 0 select m 0 such that amm ". For n m, we have that ann amm m by using the above decomposition. Select N " so that for n max{N, m} we have that m an an an 1 n " lim , lim T n lim e n e n n n n n n Corollary 3.3.5. If (T ) {0} then T n 0 as n . Remark. This above corollary is notable as it mirrors the statement for finite dimensional spaces where the spectrum reduces to the set of eigenvalues. 3.4. Hilden’s Method. Theorem 3.4.1 (Lomonosov, Hilden). Let (X, . ) be a Banach space on which T L(X) is a compact operator. Then T necessarily has a hyperinvariant subspace. Our paper will present two similar proofs, though the latter will be for a slightly more general statement. However, they both begin with a nearly identical construction. We assume T does not have an eigenvector. Remark (Set Up). [M77] Assume T does not have an eigenvector. Select x X such that T (x ) 1 (and x B) and let B B(x , 1) for ease of notation. By definition, T (B) is relatively compact and is thus totally bounded. It also does not contain the origin. Let y X be a nonzero vector and define My X by My {Sy S is a bounded operator that commutes with T }

PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 11 so that My is hyperinvariant in T , as is My . Since each My contains a nonzero vector (for nonzero y), it suffices to prove that there is some y for which My is not the entire space. We proceed by contradiction, assuming My X for all y. For each S L(X) let U(S) {y Sy B} X. Since My is dense in B we have U(S) X {0} ST T S Observe that the U(S) sets form an open cover of T (B), and so there exists a finite subcollection of operators, {Si }ni 1 such that T (B) ni 1 U(Si ). Finally, we are ready to present Hilden’s method. Proof (Hilden). [M77] We continue under the assumption that T has no eigenvector. By Theorem 2.3.6, Proposition 3.1.3, and Corollary 3.3.5 we know that T n 0 (the same holds for any scalar multiple of T ). By definition, there exists some i such that T (x ) U(Si1 ) and thus Si1 (T (x )) B. Likewise, T (Si1 (T (x ))) T (B) so we find by the same logic Si2 such that Si2 T Si1 T (x ) B and thus extract an indefinite product for any j, j 1 Sij k T (x ) B k 0 Moreover, letting c max1 i n Si we get j 1 (cj T j ) k 0 Sij k (cj T j )(x ) B c but we know that 0 which implies 0 is a limit point of some sequence in B which is a contradiction. 4. Lomonosov’s Proof and Nonlinear Methods Hilden’s proof is easily the most accessible through traditional functional analysis, but Lomonosov’s original derivation was equally (if not more) ingenious, and therefore deserving of a discussion in this paper. We require another major result from analysis in order to continue. 4.1. Schauder’s Theorem. Definition 4.1.1. Let Y X be a nonempty subset. We define conv(Y ) to be the convex hull of Y , the smallest convex subset containing Y . Proposition 4.1.2. The convex hull of a compact set is compact, and the convex hull of an open set is open. Theorem 4.1.3 (Brouwer’s Fixed Point). Let F be a finite-dimensional space and Q F be a nonempty, compact convex set. If f Q Q is a continuous function then f has a fixed point.

12 BEN GOLDMAN We can now generalize this result to a general Banach space. The proof below is my own work, but is assigned as an exercise in Brezis [B11]. Theorem 4.1.4. Let E be a Banach Space where C is a nonempty, closed convex subset of E. Let F C C be a continuous function such that F (C) K where K is a compact subset of C. Then F has a fixed point in K. Proof. Pick " 0. Since K is compact (and therefore totally bounded), one extracts a finite 2" -net with indices i I so that " K B yi , i I, yi K 2 i I Set qi (x) max " F (x) yi , 0 . We define the continuous map, q i I qi , to be the sum of these distances. Picking some arbitrary, x C, set z F (x) so that z K and therefore it must be enclosed by at least one of the balls, z B(yi , 2" ). Thus, z yi 2" and so " z yi 2" . Notice that qi measures the ‘nearness’ of yi to z, and so one approximates z by the yi by taking a weighted average z qqi yi . We have that yi I i z yi " by our definition. Finally, we can show that this approximation is " fine: qi qi qi z yi (z yi ) " " q q q i I i I i I We can now define a function, F" , that will approximate F in general F" (x) where F" is continuous. i I qi (x)yi , F F" sup F F" " q(x) x C Let Q be the convex hull of {yi }i I . We know that Q is compact as it is a closed subset of K. Since the hull is generated by finitely many points, it has finite dimension. Our intent is to apply Brouwer’s Theorem to F" Q , giving us a fixed point, x" Q. Note that F" is the convex combination of vectors in Q, and thus F" (Q) Q. For ease of notation, pick an arbitrary sequence, {"i } descending to zero and set x"m xm , F"m Fm where m is equivalent to " 0. We have that Fm (xm ) is a sequence in K, and thus we can extract a convergent subsequence Fmk (xmk ) x. For any 0 we have that F (x) x F (x) Fmk (x) Fmk (x) Fmk (xmk ) xmk x

PROGRESS ON THE INVARIANT SUBSPACE PROBLEM for k sufficiently large. Thus, F (x) x by continuity of F . 13 4.2. Lomonosov’s Method. Remark. Before discussing this proof, it is worth understanding the extent to which Lomonosov’s result was a revelation for its time. We state the results that were known up to this point. Theorem 4.2.1 (Aronszajn and Smith, 1954). Any compact operator on a Banach space of dimension greater than one has an invariant subspace. [A54] The proof that they provide is fairly lengthy and relies heavily upon polynomial theory. Definition 4.2.2. An operator, S, is polynomially compact if S p(T ) where p is a finite degree polynomial and T is a compact operator. Theorem 4.2.3 (Bernstein and Robinson, 1964). Any polynomially compact operator on a Hilbert space has an invariant subspace. [L73] This proof is fairly notorious as one of the few uses of non-standard analysis in the mathematical canon. However, the longwinded nature of these proofs made it all the more shocking when Lomonosov unveiled a one page proof of greater generality than any of its predecessors. Theorem 4.2.4 (Lomonosov, 1973). Let (X, . ) be a Banach space on which T L(X) is a compact operator. Then T necessarily has a hyperinvariant subspace. Proof. [L73] We rely on the same setup in the lead up to Theorem 3.4.1., assuming that it does not have an eigenvector. Once again, we look at the finite covering given by T (B) ni 1 U(Si ). For each y T (B) let qi (y) max{0, 1 Si (y) x } and q(y) ni 1 qi (y) so that the weighted average (y) n i 1 qi (y) Si (y) q(y) is a map into the convex subset, B. Likewise, letting C be the convex hull of ( T )(B) gives that ( T )(C) maps into C (by the same logic we used in 4.1.4). Finally, Schauder’s Fixed Point Theorem gives a vector, x C, such that ( T )(x) x and i (y) thus ni 1 i Si (T (x)) x where i qq(y) . The subspace we look for is given by W ker i Si T I n i 1

14 BEN GOLDMAN and is hyperinvariant, which suffices to complete the proof. Lomonosov’s proof is slightly more substantial as this method of approximating linear functions is very useful in resolving other issues in functional analysis. 5. The Counterexample It is natural to ask, do all operators have an invariant subspace? Theorists initially questioned whether Lomonosov’s criteria (i.e. that it commutes with a compact operator) extended to all operators on Banach Spaces. As it turns out, there are indeed counterexamples to each of these claims. The first was discovered by Per Enflo in 1977 before his methods were simplified by Bernard Beauzamy and Charles Reed [B88]. 5.1. Preliminaries. Definition 5.1.1. Let (X, . ) be a Banach space on which T X X is an operator. A point, x X, is said to be cyclic in T if span(x, T x, T 2 x, . . . , T n x, . . . ) span(Ox ) X Proposition 5.1.2. An operator, T X X, does not have a non-trivial, closed invariant subspace if and only if every point is cyclic. Proof. If T has an invariant subspace, W X, then W must contain Ox for any x W . For a non-cyclic point, x X, then W is a valid invariant subspace. This reformulation gives a far more concrete way to work with the invariant subspace problem. Two simple examples are listed below. Example 3. Consider the right shift map, T L(l2 ), given by T (x1 , x2 , x3 , x4 , . . . ) (0, x1 , x2 , x3 , . . . ). It is fairly easy to see that e2 (0, 1, 0, 0, . . . ) l2 is a non-cyclic vector, and thus T has a closed, nontrivial invariant subspace. To verify this claim, one can show that the closed unit ball, B {x X x 1}, is such a subspace. Corollary 5.1.3. If (X, . ) is a non-separable Hilbert space (it does not have a countable, dense set), then a nonzero linear operator, T X X, has a nontrivial, closed invariant subspace. Sketch. One shows that no point in such a space can be a cyclic point. Assuming the contrary (i.e. that x X is cyclic), it suffices to show that span{Ox } is separable since span{Ox } X. In a Hilbert Space, this is a known property, and therefore gives a contradiction.

PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 15 Proposition 5.1.4. If x is a cyclic point then the nth order orbit, Oxn , is a linearly independent set. Proof. Suppose that T n i n i T i for some n N. For any i n, one can express T i in terms of these first n iterates. We prove this assertion by induction. For i n 1 we have T n 1 T i T i i T i 1 which suffices. For i n 2, we have i n i n T n 2 T i T i 1 T n 1 T n i T i 1 n 1 i T i 1 i T i 2 i n i n 1 In general, if it holds for i n j i.e. that T n j i n have T n j 1 T i n 1 iT i n 1 T n 1 iT i n 1 i n iT i 1 i i n 1 then for i n j 1 we n 1 i n iT i 1 which implies that Ox is finite dimensional which contradicts our assumption that x is cyclic. We next need to think about the space that we will be using for our counterexample. Definition 5.1.5. Our traditional norm, . 1 , for a polynomial, p, is given by p(z) 1 ai z i ai i 0 1 i 0 Shortly, we will derive a di erent countable basis {f0 , f1 , f2 , f3 , . . . } of vectors in the space of polynomials. The norm that we will be using will be given by p(z) i fi i i 0 i 0 We define P to be the completion of the space of polynomials equipped with the norm, . . Remark. Our goal is going to be show that each element of the completion is a cyclic vector. This is not merely the finite degree polynomials, but also those mixed in by taking the completion.

16 BEN GOLDMAN 5.2. Constructing the Norm. Remark. Before we construct the norm, it is worth establishing a bit more background on the space of polynomials. Definition 5.2.1. Let (P , . ) be (as before) the Banach space of polynomials with complex coefficients. The subset, Pn , is defined as the collection of polynomials with degree not exceeding n. Let p(z) i 0 ai z i be a polynomial in P . We name the index of the first nonzero coefficient to be the valuation of the polynomial i.e. val(p) min{j aj 0} Lemma 5.2.2. Let n 0 and ", , M 0 and g Pn be a polynomial such that g 1 M . There exists some constant K 0 such that for any 0 m n if Pm (g) 1 then there exists q Pn such that Pn (gq) z m 1 " and q 1 K. Proof. [B88] Define the compact subset (for each 0 m n) m p Pn p 1 M and Pm (p) 1 If g m , then polynomial division gives a quotient and remainder, q, r, such that z m gq r and val(r) n. Thus gq z m r and Pn (gq) z m . We now look at the open cover of m given by neighborhoods of g, N (g, ), that are sufficiently small ( 1) so that if p N (g, r) then Pn (pq) z m 1 ". One extracts a finite subcover, {Ug1 , Ug2 , . . . , Ugj }, composed of polynomials where the remainder theorem gives respective quotients, {qi }i I {q1 , q2 , q3 , . . . , qj }. One bounds these quotients by Km maxi I qi 1 . Repeating this process for each m n allows us to take K maxm n Km which completes the proof. Now, we are ready to approach the construction. Each of the known counterexamples relies on some bookkeeping for a few (typically two) rapidly increasing sequences. I believe it is best to take Beauzamy’s approach, holding o on writing exactly how fast each sequence must grow until we arrive at the parts of the proof that require these relationships. In the meantime, we will begin with some essentials. Let {an }n N , {bn }n N be sequences in N such that a0 b 0 1 a1 b 1 a2 b 2 an b n . . . and let {vn }n 1 be given by vn (n 1)(an bn ) and v 1 v0 0. We define the basis vectors, {f0 , f1 , f2 , f3 , . . . } P from 5.1.5 as follows: p

BEN GOLDMAN Contents 1. Introduction 2 1.1. Finite Dimensions 2 2. Fundamentals of Functional Analysis 3 2.1. Banach Spaces 3 2.2. The Open Mapping Theorem 4 2.3. Compact Operators 5 3. Spectral Theory and Linear Methods 7 3.1. Elementary Functional Calculus 7 3.2. Simple Invariant Subspace Case 8 3.3. Gelfand's Spectral Radius Formula 9 3.4. Hilden's Method 10 4. Lomonosov's Proof and .