PROGRESS on the INVARIANT SUBSPACE PROBLEM Contents 1

PROGRESS ON THE INVARIANT SUBSPACE PROBLEM

BEN GOLDMAN

Contents 1. Introduction 2 1.1.FiniteDimensions 2 2. FundamentalsofFunctionalAnalysis 3 2.1. Banach Spaces 3 2.2. The Open Mapping Theorem 4 2.3. Compact Operators 5 3. SpectralTheoryandLinearMethods 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’sProofandNonlinearMethods 11 4.1. Schauder’s Theorem 11 4.2. Lomonosov’s Method 13 5. The Counterexample 14 5.1. Preliminaries 14 5.2.ConstructingtheNorm 16 5.3.TheRemainingLemmas 17 5.4. The Proof 21 6. Acknowledgements 24 References 24

Date: August, 2020. 1 2 BEN GOLDMAN

1. Introduction An important tool in working on any abstract space is to search for invariants, notions which do not change after a shift. The most elementary of these is the fixed point, an element which remains stationary after a transformation of space. The study of linear operators then introduces an invariant equipped with nonzero dimensionality, the eigenvector. In this paper, we endeavor to study how these simple principles can be extended to a more abstract notion, the invariant subspace. We begin with some very elementary, motivating examples. 1.1. Finite Dimensions. Let V,W be complex vector spaces. Definition 1.1.1. Recall that an operator, T V W ,islinearifforallv,µ V and ↵, R we have that T ↵v µ ↵T ∶ v →T µ ∈ ∈ Definition 1.1.2. Recall that, v,isaneigenvectorofalinearoperator,T ,witheigen- ( + ) = ( ) + ( ) value, C,ifT v v. Example 1. Consider the linear transformation, T 2 2,givenbyT x, y ∈ ( ) = R R 2x, 3y .Thishaseigenvectors,1, 0 and 0, 1 with real eigenvalues, 2 and 3, respectively. ∶ → ( ) = ( ) ( ) ( ) Example 2. Now, consider the linear transformation, T R2 R2,whichrotatesthe plane by some angle, ✓.ItisfairlyeasytoobservethatT has no real eigenvalues. However, one observes that the unit disk, D v R2 ∶ v → 1 ,isinvariantinT (i.e. T D D). This motivates our definition, below: = { ∈ = } Definition 1.1.3. A subspace, W V ,isinvariantinalinearoperator,T V V ,if ( ) = T W W . ⊆ ∶ → Remark. Granted, not all invariant subspaces are created equal. For instance, the ( ) ⊆ origin in V is invariant under any linear transformation, but this does not tell us anything significant about the operator nor the space. Similarly, for any surjective linear operator, T V V ,wehavethatT V V ,whichisanotherunimportant example. We dub these the trivial invariant subspaces for an operator. ∶ → ( ) = Definition 1.1.4. The orbit of a vector, v V ,withrespecttoalinearoperator, T V V ,isgivenby ∈ O v,Tv,T 2v,T 3v,T 4v,...,T nv,... ∶ → v One notes that O is also an invariant subspace, but a fairly disappointing one. v = { } Remark. In order to avoid the aforementioned simple cases, we state the Invariant Subspace Problem: Under which conditions does T have a nontrivial, closed invariant subspace? PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 3

It is fairly easy to prove this for the case of a finite dimensional complex vector space. Theorem 1.1.5. Any nonzero operator on a finite dimensional, complex vector space, V , admits an eigenvector. Proof. [A16] Let n dim V and suppose T V V is a nonzero linear oper- n ator. Select nonzero v V so that by finite dimensionality, we know that Ov v,Tv,T 2v,...,T nv =must( be) a linearly dependent∶ set.→ Thus, ∈ = ↵ T nv ↵ T n 1v ...↵ Tv ↵ v ⌘ T T ... T v 0 { n n 1 } 1 0 1 2 n − where we use − to denote I (the identity map). At least one of these factors + i C + i + = ( − )( − ) ( − ) = must have a nontrivial kernel which suces to prove the theorem. One can adapt the proof above to∈ real valued vector spaces, but the irreducible polynomials end up being quadratic instead. ⇤ Corollary 1.1.6. Any nonzero operator on a finite dimensional, complex vector space, V , admits a closed, nontrivial invariant subspace.

2. Fundamentals of Functional Analysis 2.1. Banach Spaces. Deﬁnition 2.1.1. Recall that a Banach Space is a complete Vector Space with norm, . .

Lemma 2.1.2 (Riesz). Let x E, x 1 be the unit sphere in an inﬁnite dimensional Banach Space, E, . .IfY E is a closed proper subspace of E then there exists x x " suchB = that{ ∈ = } ∗ ∗ ( ) 1 ⊂ = ( ) ∈ B y x y Y," 0 ∗ 1 " Proof. Pick an arbitrary x − so> that d ∀ inf∈ > x y . Select y so that + y Y ∗ x y d d"0 d d" and set ∈ ∈ B = − ∗ 1 1 − = + ∈ ( − ) > 0+ + + ⇤ ∗ Theorem= 2.1.3+ ( (Riesz)+ ) . If E, . is an inﬁnite dimensional banach space then the unit sphere, , is not compact. ( ) B 4 BEN GOLDMAN

Proof. Pick x1 and set Y1 span x1 . Apply the previous lemma to yield x2 1 such that infy Y1 x2 y 2 and let Y2 span x1,x2 .Continuethisprocess ∈ B = { } inductively to∈ yield a sequence, xn and Yn ,suchthatxn has no convergent subsequence. − > = { } ⇤ { } { } This stands in stark contrast to finite dimensional vector spaces which have compact unit spheres. We conclude this subsection with a fact about linear operators on Banach Spaces. Definition 2.1.4. We define a norm on the space of linear operators between Banach spaces, T X, . X Y, . Y ,by

∶ ( ) → ( ) T sup T x Y x

One notes that there is not necessarily = a∈B vector, ( )x0 ,forwhich T x0 T due to the loss of compactness. ∈ B ( ) = Theorem 2.1.5. If T X, . X Y, . Y is a linear map between Banach Spaces then it is continuous if and only if T ∶ ( ) → ( ) " Proof. If T then for all " 0if −1 < T x T y Y x y X T x y T x y X " x y X Y ( ) − ( ) = − ( − ) ≤ − < We will use the manipulation above i.e. −T x Y T x X quite often. If T is 1 unbounded then for each n N select xn X such that xn X n and T xn Y n. ( ) ≤ This produces a sequence, xn 0suchthat T xn Y 1foralln. And thus, lim supx 0 T x Y 1, and∈ so T is not continuous∈ at the origin. = ( ) > ⇤ → ( ) ≥ → ( ) ≥

2.2. The Open Mapping Theorem. Theorem 2.2.1 (Baire Category Theorem). Complete metric spaces are Baire spaces.

More precisely, if X,⇢ is a complete metric space and n n N is a countable col- lection of open dense subsets then the intersection, , is also dense. [R10] n N n∈ ( ) {O } ∈ We require the above theorem from non-functionalI analysis= O in order to justify the following results.

Theorem 2.2.2 (Open Mapping Theorem). If T X, . X Y, . Y is a surjective, continuous linear map then T is an open map. ∶ ( ) → ( ) PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 5

Proof. [R10] Let BX BX 0, 1 be the unit ball in X and BY be the unit ball in Y .ObservethatX r N rBX and that Y r N T rBX .BytheBaireCategory = ( ) Theorem, there exists R∈ N such that T RBX ∈ T BX 0,R has nonempty inte- rior. = ( ) = ( ) ∈ ( ) = ( ( ))

Pick BY y0,r0 T BX 0,R so that for all y BY we have that y0 r0y BY y0,r0 . Thus, ( ) ⊆ ( r0(y ))T RBX T RBX∈ T 2RBX + ∈ ( ) and we thus have that B T R B where R 2R r . Y∈ ( X) − ( ) ⊆ ( 0 ) ∗ ∗ For any y Y and " 0,⊆ we( find a) point x =R BX such that T x y ". R We can refine this approach to get a stronger result.∗ Select x1 B 0, for which 2∗ ∈ 1 > R ∈ ( ) ( 1) − < Tx1 y and xn B 0, n such that T x1 x2 xn y n .Letting 2 2 ∗ ∈ 2 x n N xn where x n N xn 2R gives that T x y.Thus,wehave 1 shown ∗ − that < B T 2R∈∗B and so B 0, ∗ ( T+ B +.⋅⋅⋅+∗ ) − < ∈ Y X ∈ Y 2R X = ∑ ∗ ≤ ∑ ≤ ∗ ( ) = Finally, one shows⊆ ( that T )is open. Since T is ⊆ linear,( it) suces to prove that the image of the unit ball is open (other open sets are achieved by translations/scalings). Take y T BX and select x BX such that Tx y.Pick" 0suchthat BX x, " BX and pick r suciently small so that BY y,r T BX x, " so that B y,r ∈T B( X .) ∈ = > ⇤ ( ) ⊆ ( ) ⊆ ( ( )) We finish this subsection with a handful of useful statements that are immediate ( ) ⊆ ( ) consequences of the Open Mapping Theorem.

Corollary 2.2.3. If T X, . X Y, . Y is a bijective, continuous operator between two banach spaces then T 1 is continuous. [R10] ∶ ( −) → ( ) Corollary 2.2.4. If . 1 and . 2 are two norms on a Banach space, X, such that . 1 C . 2 then they are equivalent. [R10] ≤ 2.3. Compact Operators. Notation 2.3.1. Let X be a norm space. We call X the space of continuous linear operators mapping a space, X, into itself. If T X then T X X. L( ) For obvious reasons, we will be concentrating our e↵orts on these types of operators ∈ L( ) ∶ → for the remainder of the paper. The loss of compactness in Banach spaces makes it dicult to study these linear operators to the same degree as one could in a Euclidean setting. However, we deﬁne a special class of operators which has very nice properties (and that we will see, later on, satisﬁes the invariant subspace problem). 6 BEN GOLDMAN

Deﬁnition 2.3.2. An operator, T X ,iscompactifitmapstheunitballintoa set with compact closure. In other words, T BX is a relatively compact set. ∈ L( ) Remark. If v is a bounded sequence in X then there exists a subsequence, v , n ( ) nk such that T vnk v for some v X. { } ( ) → ∈

Deﬁnition 2.3.3. Let X, . be a norm space. A sequence, vn X,issaidto converge weakly, vn v,if' vn ' v for every real-valued continuous linear map, ' X R.Wealternatelywritethat( ) ' X ,thedualof{X.} ⊂ ⇀ ( ) → ( ) Proposition 2.3.4. If v v then v v. ∗ ∶ → n n ∈ Theorem 2.3.5. Let X be a Banach space on which x x is a weakly convergent → ⇀ n sequence. It follows that T xn T x is strongly convergent if T is compact. ⇀ Proof. If T is compact then for any subsequence x ,thereexistsafurthersubse- ( ) → ( ) nk quence, x , such that T x T x . nkj nkj ⇤ Remark. In general, solutions( ) to→ the( invariant) subspace problem will require a broad construction of a suitable space, tailored to each type of operator. We will provide two di↵erent approaches to prove that compact operators have invariant subspaces which rely on this approach. We conclude this with a robust statement pertaining to compact operators due to Fredholm. Theorem 2.3.6 (Fredholm). Let X, . be a complex Banach space on which T X is a compact operator. If C is not an eigenvalue then T has a bounded inverse. ( ) ∈ L( ) ∈ ( − ) Our proof hinges on the lemma below. Lemma 2.3.7. If is not an eigenvalue then there exists a positive constant, C 0, such that T x C x for all x X. > Proof. [T11] We will prove this via contradiction by showing that will otherwise ( − )( ) ≥ ∈ be an eigenvalue. By assumption, there exists some sequence on the unit sphere, 1 xn n N ,suchthat T xn n for all n and thus, T xn 0. Moreover, we have that T x 0byassumption.SinceT is compact, we ∈ n { } ⊂ B ( − )( ) ≤ ( − )( ) → can pass to a subsequence, xnk ,suchthatT xnk x where x is nonzero and ( ) → > T T x T T x 0 T x ker T ( ) → ( − ) ○ ( ) = ○ ( − )( ) = ⇒ ( ) ∈ ( − ) ⇤ Proof. [T11] By Corollary 2.2.3, it suces to prove that T is surjective. In other words, we need to show that Im T X. We approach this via contradiction, ( − ) ( − ) = PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 7

n supposing that there exists µ X such that µ Im T .LetVn Im T 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 S2 then Sx S2y for some y and thus x Sy gives a contradiction. The same holds= ( − for)Sn,Sn 1 for any n N. ∉ ( ) ( ) = ( ) += = We∈ now apply the Riesz Lemma (2.1.2) to form a sequence, 1 xn n N , inf xn v v Vn 1 2 ∈ { } ⊆ B ∈ + − > and so Txn Txm 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,isdefinedasthesetofcomplex values, T ,forwhich( T) does not have a bounded inverse.∶ → ( ) ⊆ Definition∈ 3.1.2( ). The resolvent( −function,) R T, ,isdefinedasthisboundedinverse, R T, T 1,whenitexists.Thedomainoftheresolventfunction,the resolvent set ⇢ T ,isthecomplementof− (T . ) ( ) = ( − ) Proposition 3.1.3.( ) If T X is compact( ) and T then is an eigenvalue. Theorem 3.1.4. The spectrum,∈ L( ) T , is non-empty,∈ compact,( ) and contained in the complex disk, z C z T . ( ) Proof. [B88] DefineD = { the∈ map∶ ≤ } T k g X X, g ∞ k 1 k 0 + ( ) ∶ → ( ) = = 8 BEN GOLDMAN

where for convention we say T 0 I.If T then T k T k = > ∞ k 1 ∞ k k 0 k 0 + and g converges in norm. We= can now≤ = extrapolate< +∞ further, 1 1 1 ( ) T g T I T T 2 ... I 2 3 which gives us that( −g ) ( ) =R( T,− ).Wethushavethat + + + g = − if and only if ⇢ T which gives us that T . ( ) = − ( ) ( ) < +∞ The∈ ( other) claims follow from the( ) maximum⊆ D 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 ntn n 1 (3.2.1) 1 t 1 k 1 +∞ n 1 n! k 1 2 √ (− ) Definition 3.2.2. Alinearoperator,− = + = T X =X,hasa− + hyperinvariant subspace, W X if S W W for any operator, S,thatcommuteswithT . ∶ → ⊂ ( ) ⊆ 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 n 4 n 1 n S I k 1 T T 2 I K +∞ n 1 n! k 1 2 (− ) It follows that S (and= + therefore= K)iswelldefineddueto(3.2.1),andthatsincefor= − + − = + t 1wehave1 1 t 1thenweconclude K 1. Further reference to the Taylor Expansion gives:√ < − − < < (3 . 2 . 4 ) S2 I 4 T T 2 K 2T S I 2T 0(

= − ( − ) ⇒ ( + ) ○ ( + − ) = PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 9

Let F K 2T .WeobserveF is nonzero since F K 2T 2T K 0 = + We turn to the subspace U ker F .ToshowthatU 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 ,isgivenby ( ) 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 n 1 n ( ) ( ) defined, lim supn T .Itremainstobeshownthatthislimitiswell 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 T − T − T − ( − 1) = ( − 2) ○ ( − n 1 + ) n n where 1,2,...,n 1 C.ObservethatkerT − ker T .Wehavethat 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 n n ⇢ T .Moreover,if ⇢ T then there exists some i such that i ⇢ T . n n n n However, we deduce that i and so ⇢ T . ∈ ( ) ⇤ ∈ ( ) ∉ ( ) n n ∉ ( ) Corollary 3.3.3. It follows= that T ∉T ( . ) Theorem 3.3.4 (Gelfand). The spectral( ) = radius( ) of an operator, r T , is given by n 1 n n 1 n r T lim supn T limn T ( ) →∞ ( ) ≤ →∞ = 10 BEN GOLDMAN

n 1 n Proof. [B11] It remains to be shown that limn T exists. Deﬁne a sequence, a ,by n n N →∞ n ∈ an log T { } an Since T is bounded n is non-increasing and bounded. Moreover, we use the inequal- ities below to deduce some important= properties n m n m T T T an m an am + nm n m + T ≤ T ⇒ anm ≤man+ By the Euclidean Algorithm, we deduce that for m n we have that ≤ ⇒m ≤ a qa a a ↵ m n r n n >

m ar an where q n ,0 r n,and↵≤ max+ r n≤ r .If+⌘ infn N n it suces to show lim a ⌘.Foreach" 0selectm 0suchthat am ⌘ ".Forn m,we n n ≤ m ∈ have that= an am ≤ ↵ ≤by using the= above decomposition.= Select N ↵ so that for →∞ n m m " n max N,m= we have that> > < + > ≤ + = an ↵ an n 1 n an ⌘ > { } ⌘ " lim ⌘, lim T 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 X is a compact operator. Then T necessarily has a hyperinvariant subspace. ( ) Our∈ L 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(andx )andletB 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∉ aB nonzero vector= and( define) My X by ( ) M Sy S is a bounded operator that commutes with T y ∈ ⊂ = { ∶ } PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 11 so that My is hyperinvariant in T ,asisMy.SinceeachMy contains a nonzero vector (for nonzero y), it suces 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.ForeachS X let S y Sy B X.SinceMy is dense in B we have = ∈ L( ) S X 0 U( ) = { ∶ ∈ } ⊆ ST TS Observe that the S sets form an= openU( ) cover= ∼ of{T} B ,andsothereexistsafinite n n subcollection of operators, Si i 1 such that T B i 1 Si . Finally, we are ready to present Hilden’sU( method.) ( ) { } = ( ) ⊆ = U( ) 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 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 TSi1 T x B and thus extract∗ an indefinite product( ) ∈ forU( any) j, ( ( )) ∈ ∗ ( ( ( ))) ∈ ( ) j 1 ( ) ∈ − Sij k T x B k 0 − ∗ Moreover, letting c max1 i n Si= we get○ ( ) ∈ j 1 ≤ ≤ Si = j k cjT j x B − − k 0 c ∗ j j but we know that c T 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 ,thesmallestconvexsubsetcontainingY . ⊂ ( ) 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 " aﬁnite 2 -net with indices i I so that > ∈ " K B yi, i I,yi K i I 2

⊂ ∈ ∈ ∈ Set qi x max " F x yi , 0 .Wedefinethecontinuousmap,q i I qi,to be the sum of these distances. ∈ ( ) = − ( ) − = ∑ Picking some arbitrary, x C,setz 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,andsooneapprox- ∈ ( ) qi − ≤ imates z by the yi by taking a weighted average z q yi.Wehavethatyi I i↵ − − ≥ z yi " by our definition. Finally, we can show that this approximation is " fine: ≈ ∑ ∈ qi qi qi − < z yi z yi " " i I q i I q i I q

We can now deﬁne a− function,∈ F"≤,thatwillapproximate∈ ( − ) ≤ ∈ F in general≤

i I qi x yi F" x , F F" sup F F" " ∈ q x x C ∑ ( ) ∞ where F is continuous.( ) = − = ∈ − ≤ " ( )

Let Q be the convex hull of yi i I .WeknowthatQ 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,andthusF" 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 { } asequenceinK, and thus we can extract a convergent subsequence Fmk xmk x. For any= ⌘ 0wehavethat= → ∞ → ( ) ( ) → F x x F x F x F x F x x x ⌘ > mk mk mk mk mk ( ) − ≤ ( ) − ( ) + ( ) − ( ) + − < PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 13 for k suciently large. Thus, F x x by continuity of F . ⇤

( ) = 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,ispolynomially 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 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 L( ) that it does not have an eigenvector. Once again, we look at the finite covering given n by T B i 1 Si .Foreachy T B let qi y max 0, 1 Si y x and n q y qi y so that the weighted average ∗ i 1 = ( ) ⊆ U( ) ∈ n( ) ( ) = { − ( ) − } = qi y ( ) = ∑ ( ) y Si y i 1 q y ( ) is a map into the convex subset,( B) =.Likewise,letting ( ) be the convex hull of = ( ) T B gives that T maps into (by the same logic we used in 4.1.4). Finally, Schauder’s Fixed Point Theorem gives a vector, x ,suchthatC T x x(and○ )( ) n ( ○ )(C) qi Cy thus i 1 ↵iSi T x x where ↵i q y . The subspace we look for is given by ( ) ∈ C ( ○ )( ) = n = ( ( )) = = ( ) ∑ W ker ↵iSi T I i 1

= = ○ − 14 BEN GOLDMAN and is hyperinvariant, which suces 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. Apoint,x X,issaidtobecyclic in T if ( ) ∶ → span x, T x, T 2x, . . . , T nx, . . . span O X ∈ 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,thenW must contain Ox for any x W .Foranon-cyclicpoint,x X,thenW 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.

2 Example 3. Consider the right shift map, T l ,givenbyT x1,x2,x3,x4,... 2 0,x1,x2,x3,... .Itisfairlyeasytoseethate2 0, 1, 0, 0,... l is a non-cyclic vector, and thus T has a closed, nontrivial invariant∈ L( ) subspace. To( verify this claim,) = one( can show that) the closed unit ball, B x X= ( x 1 ,issuchasubspace.) ∈

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 suces 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

n Proposition 5.1.4. If x is a cyclic point then the nth order orbit, Ox , is a linearly independent set.

n i i Proof. Suppose that T i n ↵iT for some n N.Foranyi n,onecanexpressT in terms of these ﬁrst n iterates. We prove this assertion by induction. For i n 1 < we have = ∑ ∈ >

n 1 i i 1 = + T T ↵iT ↵iT + i n i n + which suces. For i n 2, we= have○ < = <

n 2 i 1 n i 1 i 1 i 2 T T ↵iT = +T ↵n 1T ↵iT ↵n 1 ↵iT ↵iT + i n + i n 1 + i n + i n 1 + − − = ○ = ○ + n j = i + In general, if it< holds for i n j i.e. that<− T i n iT

We next need to think about the space that we will be using for our counterexample.

Deﬁnition 5.1.5. Our traditional norm, . 1,forapolynomial,p,isgivenby

i p z 1 aiz 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 ↵ifi ↵i i 0 i 0

We deﬁne to be the completion ( ) = of the≥ space = of≥ polynomials equipped with the norm, . . + P 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 back- ground on the space of polynomials. Definition 5.2.1. Let , . be (as before) the Banach space of polynomials with complex coecients. The subset, ,isdefinedasthecollectionofpolynomialswith + n degree not exceeding n(P. ) P i Let p z i 0 aiz be a polynomial in .Wenametheindexofthefirstnonzero coecient to be the valuation of the polynomial i.e. val p min j aj 0 ≥ + ( ) = ∑ P Lemma 5.2.2. Let n 0 and ",,M 0 and g n 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 m then there exists q n≥ such that Pn>gq z ∈1 P " and q 1 K. ≤ > ≤ ≤ ( ) ≥ Proof. [B88] Define∈ theP compact subset ( (for) − each < 0 m n) ≤

⌦m p n p 1 M and P

If g ⌦m,thenpolynomialdivisiongivesaquotientandremainder,= ∈ P ≤ ( ) ≥ q,r,suchthat m m m z gq r and val r n.Thusgq z r and Pn gq z .Wenowlookat the open∈ cover of ⌦m given by neighborhoods of g, N g,⌘ ,thataresuciently m small= (⌘ + 1) so that( ) if>p N g,r then= P−n pq z (1 )"=. One extracts a ﬁnite subcover, Ug1 ,Ug2 ,...,Ugj ,composedofpolynomialswheretheremaindertheorem( ) gives respective<< quotients, ∈ qi (i I ) q1,q2,q3,...,q( ) −j .Oneboundsthesequotients < by K {max q .Repeatingthisprocessforeach} m n allows us to take m i I i 1 ∈ K max K which completes{ } the= { proof. } m n m∈ ⇤ = ≤ ≤ = 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 b0 1 a1 b1 a2 b2 an bn ... and let vn n 1 be given= = by

2 Let f0 1, f1 z 1, and define the function, k,r m, n R R by r 1 m k = = − m, n 2 ( ) ∶ → k,r n − − We begin with the case where n ( 2and1) = r n 1. We require that an,bn satisfy r 1 an vn r k ran and that k ≥ k a ≤ ≤ − (5.2.3) − f a z z n if ra k ra v ( − ) + < ≤ k n r n n n r 1 − − − − k,r an,bn 1 k (5.2.4) fk =2 ( − z if) r 1 a≤n ≤vn r +k ran ( − ) This allows us, for each n 1, to define f for k v −up to k a .Onecanusethe = (k − ) +n 1 < < n same logic to move upward to k a v and then to k 2a .Onecancomplete n n 2 − n this process inductively to≥ define f for k up to k= n 1 a .Fromhere,weneedto= − n expand our definitions. = + = = ( − )

If n 2and1 r n 1thenwerequirethat n 1 an r 1 bn r an bn and let

≥ ≤ ≤ − k k bn ( − ) + ( − ) < ( + ) (5.2.5) fk z bnz if r an bn k n 1 an rbn − (5.2.6) f 2k,r bn,an zk if n 1 a r 1 b k r a b k = − ( n + ) ≤ n≤ ( − ) n + n and if n 1 then use (5.2.5)( ) for k a b and (5.2.6) for a k a b . Now, one = ( −1 ) 1 + ( − ) < < 1 ( + 1) 1 can deﬁne fk for k n 1 an up to k an bn,allthewaytok vn n 1 an bn . = = + < < + = ( − ) = + = = ( − )( + ) 5.3. The Remaining Lemmas.

Remark. If (5.2.3) holds, then vn r an and n 2 an 1 bn 1 an so that j 1 2 .Puttingthisalltogether,onearrivesattheidentities i m a am − − − i < ( − )( + ) < = k k an 1 ∑ ≤ z z fk − an r 1 k an k 2−an = − z z fk an − − an r 1 − − = − + r 1 1 r 1 1 2 zk zk ran f zk zk ran ⋮ − k ian − − i 0 an r i − i 0 an r i an r − One can show− almost= = identically−( − ) that when⇒ (5.2.5)− holds, = = −( − ) ≤ − r 1 zk br zk rbn bi 2br 1 n − n − i 0 − − = = ≤ 18 BEN GOLDMAN

Theorem 5.3.1. The operator, T , given by

+ i i 1 T ∈ LaizP aiz i 0 i 0 + does not have an invariant subspace. ≥ = ≥

Lemma 5.3.2. If an , bn grow suciently rapidly then T 2 and the three identities below hold for n, r N { } { } 1 ≤ (5.3.3) ∈ Tfran vn r 1 an r + − − 1 ≤ − (5.3.4) Tfran 1 an − ≤ 1 (5.3.5) Tfr an bn 1 bn ( + )− These bounds follow fairly easily from the deﬁnitions ≤ so long as we require that an , bn grow suciently rapidly. An exact proof is found in Beauzamy’s paper.

{Deﬁnition} { } 5.3.6. We deﬁne Om R for each m N by Om k N n m, n m an k n m an vm 1 n m an, n m an vm 1 ⊆ ∈ n m − − We are= { now∈ prepared∶∃ > ( to− present) ≤ the≤ next( − key) lemma.+ } = I will> [( refrain− ) from( − proving) + this ] statement as the proof is very technical and does not shed much insight. It can be found in Beauzamy’s paper.

Lemma 5.3.7. If an , bn grows suciently rapidly for m 2 and k m 1 am,bm am s bm m 1 am then { } { } > > ( − (5.3.8) T sf 4 if k O ) + ≤ ≤ + ( − ) k m (5.3.9) T sf a zj s 1 if k O and we let k n m a j k m m ≤ ∉n + Although I am omitting+ the justiﬁcation,≤ ∈ I am going to= ﬂesh( − out) the+ utility of this statement.

Deﬁnition 5.3.10. Let Qm m 1 am be deﬁned by + f ( − ) 0 k m 1 a ∶ P →k P m Qmfk 0 k m 1 am,k Om j amz∶ else, where≤ ≤ (j is− as) above ∶ > ( − ) ∉ be a linear map. = − ∶ s s Proposition 5.3.11. If m 2 and bm am s bm m 1 am then T fk T Qmfk 4 for k N. > + ≤ ≤ +( − ) − ≤ ∈ PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 19

Proof. [B88] Before delving into the more dicult cases, we have that if k m 1 am s s then Qmfk fk gives T fk T Qmfk 0 ≤ ( − )

If k m 1= am and k Om−then Qmfk= 0sothatLemma5.3.7gives T sf T sQ f T sf 4 ≥ ( − ) ∉ k m= k k Otherwise, Q f a zj s (where j is the same as in 5.3.7) which gives T sf T sQ f m k m − = ≤ k m k 1duetoLemma5.3.7. + ⇤ = − − ≤ s s Corollary 5.3.12. If g then T g T Qmg 4 g .

+ Proof. [B88] It suces to∈ P prove that Qm− is bounded/continuous. ≤ Recall that vm 1 m 1 a v and that for each k v one can ﬁnd f in terms of a ,b ,a ,b ,...,a , m m m 1 k 1 1 2 2 − m 1 b ,a .If0 j m 1 a there exists a constant, C C a ,b ,a ,b ,...,a ,b< , m 1 m m − m m 1 1 2 2 m 1 m− 1 depending( − ) < on these sequence entries≤ such that f C gives a valid upper bound. − j 1 m − − Moreover, we≤ have≤ ( that− ) Qmfk amCm and Qmg =amC(m g which completes ) the proof. ≤ ⇤ ≤ ≤ Remark. It is useful to study the relationship between the two norms we have ↵ established. An important notion for the . 1 norm is that z p 1 p 1 for any polynomial, p ,andanypositiveinteger,↵ N.Forj m 1 am,onecan express zj in terms of f ,f ,...,f . Depending on whether or not =j v ,this + 1 2 j m 1 linear combination∈ P will involve the elements of the∈ sequence, ≤a (,b ,...,a− ) ,b , 1 1 m 1− m 1 or if larger, it will go up{ to a ,b ,...,a} ,b ,a .Thiswillallowustodeﬁnea≤ 1 1 m 1 m 1 m − − constant, C C a ,b ,...,a ,b ,a ,suchthat zj {C .Moreover, } 1 1 m 1 m 1 m− − ∗ ∗ { } ∗ − − (5.3.13) Qmfk 1 max fk 1 ,am Cm Qmg 1 Cm g = ( k m 1 am ) ≤

Corollary 2.2.4 gives that ≤ the≤( norm,− ) (. ,givenby) ≤ g ⇒ Q mg is≤ equivalent to . .

∗ ∗ = Lemma 5.3.14. If g (the unit sphere) and there exists m 2, 1 r m 2 for which ∈ B 1 > ≤ ≤ − (5.3.15) Pram Qm g 1 am then there exists some ' such( that○ ) ≥

+ 3 (5.3.16) ∈ P ' T g 1 am r 1

Proof. [B88] Fix m and r as above.( ○ ) Let− h ≤ Qm−g −so that the above remark gives h 1 Cm so that Lemma 5.2.2 gives a positive constant, K K Cm,am ,anda = ≤ = ( ) 20 BEN GOLDMAN

polynomial q such that 1 P qh zram , q K m 2 am 1 1 amCm

am ( − ) Set ' z q so that val ' a(m and) − ' m< 1 am and ≤

( − ) 1 P 'h z r 1 am = ( )m≥ 2 am ∈ P 1 ( + ) amCm ( − ) and (5.3.13) gives that ( ) − <

r 1 am 1 P m 2 am 'h z ( + ) am ( − ) Set ' 1 zbm ' 1 zam bm q so that( if )r −a b

1 1 ≥ 1 A j = ∑ 'h 1 ' 1 h 1 am j 0 bm bm

Cm≤ ≥ C m = Cm ≤ KC m 1 ' 1 ' 1 q 1 bm bm bm bm am where the ﬁnal inequality≤ emerges = ifbm grows= su ciently≤ larger≤ than am and (5.2.4) gives (for bm suciently large)

KCm 2 m 1 am 1 2bm ' h P m 1 a b ' h 2 'h 2 , m m ∗ 1 ∗ bm mam ∗ ( − ) + ∗ ∗ ( − ) − where b −KC a gives ' ≤h P ≤ ' h 1 =as an upper bound. m m m m 1 am bm am

∗ ( − ) + ∗ m 1 a b Since ' ≥ 1 zam bm q we can expand− in the( standard) ≤ basis ' m m zs bm s am bm s K K ( − ) + and note that ' + and thus s for each (relevant) s.Lemma5.3.2gives ∗ 1 bm bm ∗ = + s = s = ∑ that T g T Qm∗g 4 g and whenever am bm s m 1 am bm it follows that ≤ ≤ T sg T sQ g 4 g − ≤ m + ≤ ≤ ( − ) + and therefore that − ≤ m 1 am bm m 1 am bm 4K 1 ' T g ' h zsg zsh g ( − ) + s ( − ) + s b a s am bm s am bm m m ∗ ∗ ( ○ ) − = = + − = + ≤ < PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 21 where the ﬁnal inequality holds for bm 4kam.

Finally, we can complete the proof. Recall> at the beginning of this section where we discussed that for 1 r n 1andra k ra v we have zk zk an 2 n n n r 1 an r and that if k ra we get zran 1 1 .Onearrivesattheequivalentformulation,− n an r − − − z r 1 am 1 2 ≤. Applying≤ − the triangle≤ ≤ inequality+ gives our desired − identity: ≤ am r 1 − ( + ) = − ≤ − − ' T g 1 ' T g ' h ' h P ' h − ≤ m 1am bm r 1 a r 1 a P ' h ∗P 'h ∗ P ∗'h z ∗ m ( − z + m ∗ 1 m 1am bm ( ○ m)1 a−m ≤ ( ○ m) 1 −am + − ( ) ( + ) ( + ) ( − + ∗ ( − ) ( − ) 4 2 3 + ( ) − ( ) + ( ) − + − am am r 1 am r 1

≤ + − − ≤ − − ⇤

5.4. The Proof. Remark. Recall our discussion at the beginning of this section. As long as 1 is in the orbit of each element of (i.e. not just finite polynomials) then each vector, + p ,willbycyclic.ThisholdssinceO and 1 O implies that O O∈.P + 1 p 1 p P + The∈ P convenience of polynomials is that the= P elements∈ formed by the span⊆ of any given nth orbit can be equivalently given by multiplying the polynomial by another polynomial. Of course, for the elements which are not finite degree polynomials it is more practical to use the completeness of the space. Thus, in order to finally prove Theorem 5.3.1, it suces to show that for each g and " 0thereexistssome ' ' " such that ' Tg 1 ". 1 " ∈ B > Proof.∗ ∗[B88] Fix " 0andselect∗ k such that .ByLemma5.3.14,itsucesto = ( ) − < ak 3 find m, r ,m r 1 k such that P Q g 1 .Weproceedbycontradiction, N ram m 1 am and assume that no> such r, m exist such that

g ( − ) ↵jfj ↵jfj n n, n,n j 0 j n 1 an + = = + >( − ) = + ∈ P 22 BEN GOLDMAN

In terms of the standard basis, we have that

n 1 an vn 1 j n ( − ) n,j ,Qn n − n,jz j 0 j 0

= = ( ) = = where n,j an m n ↵j m n am . This follows directly from (5.3.10). Note that Qn n n,andso > +( − ) = − ∑ n 1 an vn 1 ( ) = j j Qn g ( − ) n,jz − n,jz j 0 j 0 If (5.2.3) holds then a( ) ↵= and= our assumption+ = gives that 1 m,j m r j j ram n,j am or likewise that ram vm r 1 ↵ 1 .Moreover,forn k 2somefurther ram vm r 2 −j amam r ≤ manipulation gives that+ −=− ∑ < + − − − ∑ < > + vn 1 vn 1 vn 1 1 2 − n,j − an ↵j an − ↵j m n am j vn r 2 1 j vn 2 1 m n m n j vn 2 1 m n am an 1 +( − ) − − = − = − < < Applying= (5.4.1)+ gives= v+n 1 > 1 which> = gives+ immediately that> vn 1 + j 0 n,j n,j an j vn 2 1 n,j 2 − − a . − n ∑ = + < ∑ = + < From our original deﬁnitions, we know that if j satisﬁes (5.2.3) or (5.2.4) i.e. that r 1 an vn r j ran then fj will not contain a term of degree i vn 2,vn 1 . Applying this to our situation gives, − − − ( − ) + < ≤ ∈ ( ] vn 1

− ↵jfj n,jzj vn 2 vn 2 1 n 2 1 vj n 1 − and thus − + − − + ≤ ≤ − ∈ P vn 1 vn 1 2Dn 1 1 − ↵j Dn 1 − n,j j vn 2 1 j vn 2 1 an− an − − − where the ﬁnal inequality= + holds ≤ for an= su+ciently < large. Equivalently,< √ we have that vm ↵ 1 .Returningtooneofourexpansionsfromabovei.e. j vm 1 1 j am 1 √ − v + v ∑ = + < n 1 n 1 1 1 − n,j − an ↵j an j vn r 2 1 j vn 2 1 m n m n am 1 an 1

− − = − > ≤ > √ + ≤ + vn= 1 + = n+ 1 an Using that n,j Pv ↵jfj 2 an we can conclude that j 0 n 1 j 0 1 − ( − ) − = = ∑ = n∑1 an 1 < 1 ( − ) ↵j j vn 1 1 an 1 an

= − + ≤ √ + < PROGRESS ON THE INVARIANT SUBSPACE PROBLEM 23

If vn 1 j n 1 an then our deﬁnitions yield

− n 1 an < ≤ ( − ) an 1 Pvn 1 fj 1 an 1 Pvn 1 ( − ) ↵jfj k vn 1 1 an− − − 1 − which combines to give( ) ≤ ⇒ = − + < vn 1 vn 1 an 1 2 an 1 2 − ↵jfj − ↵j Dn 1 j 0 −an j 0 −an + − + which gives that j=0 ↵j 0as< n ⇒which= contradicts ≤ our assumption. ⇤ Corollary 5.4.2 (C.≥ Read). There exists an operator on l1 with no invariant sub- ∑ = → ∞ space. [R10] It is not an immediate corollary, but it emerges due to the fact l1 . An actual construction can be found in Reed’s paper. + ≅ P 24 BEN GOLDMAN

6. Acknowledgements It has been a pleasure to work on and write about this topic during this otherwise dicult time. I want to issue a special thank you to my advisor, Dolores Walton, for being a constant source of support and helping me reﬁne several of the arguments that I presented here. I also express my gratitude to Peter May and the other vol- unteers who continued to run this program and transitioned to a remote setting.

Finally, I want to thank the professors and mentors who have gotten me to this point. Cheers to Andre Neves who introduced me to this area of study in the fall, as well as the remainder of the amazing 2019-2020 H. Analysis professors. And a ﬁnal thank you to Adam Wagner for encouraging me to pursue mathematics, and for being a constant source of support.

References [A54] N. Aronszajn and K. Smith, “Invariant subspaces of completely continuous operators,” Ann. Math.,60, 345–350 (1954). [A16] Axler, Sheldon J. Linear Algebra Done Right. New York: Springer, 2016. Print. [B88] Beauzamy, B. (1988). Introduction to operator theory and invariant subspaces. Amster- dam: North-Holland. [B11] Br´ezis, H. (2011). Functional analysis, sobolev spaces and partial di↵erential equations. New York, NY: Springer. [L73] Lomonosov, V.I. Invariant subspaces for the family of operators which commute with a completely continuous operator. Funct Anal Its Appl 7, 213–214 (1973). https://doi.org/10.1007/BF01080698 [M77] Michaels, A. J.. “Hilden’s simple proof of Lomonosov’s invariant subspace theorem.” Ad- vances in Mathematics 25 (1977): 56-58. [R85] Read, C. “A Solution to the Invariant Subspace Problem on the Space l1” The Bulletin of the London Mathematical Society, Vol 25 Issue 4 (1985): 305-317. [R10] Royden, H. L., & Fitzpatrick, P. (2010). Section II. In Real Analysis (pp. 183-334). Boston, Mass.: Prentice Hall. [T11] Tao, T. (2011, April 17). A proof of the Fredholm alternative. Retrieved August 10, 2020, from https://terrytao.wordpress.com/2011/04/10/a-proof-of-the-fredholm-alternative/