The Invariant Subspace Problem for Separable Hilbert Spaces

Abstract

In this paper, we prove that every bounded linear operator on a separable Hilbert space has a non-trivial invariant subspace. This answers the well-known invariant subspace problem.

1. Introduction

In the field of operator theory, an important area of research is how linear operators behave when acting on a Hilbert space H. Moreover, the concept of the invariant subspace is essential to the subject of operator theory and functional analysis. It make operators more interesting and much more easy to deal with.

Let H be a complex Hilbert space and $L(H,H)$ be the space of all bounded linear operators acting on H. A closed subspace W of H is called a non-trivial invariant under T∈$L(H,H)$, if $T\left(W\right)\subseteq W$, where $W\ne \left\{0\right\}$ and $W\ne H$. This concept generalizes the idea of eigenspaces comprising $n\times n$ matrices [1,2,3].

One of the most renowned partially unresolved questions in mathematics is known as the “invariant subspace problem”. This enduring puzzle asks a seemingly straightforward question: “Does every bounded linear operator T in $L(H,H)$ have a non-trivial invariant subspace?” Despite being posed over fifty years ago, this problem remains open, with mathematicians employing a wide array of techniques to tackle it. The challenge and complexity of this problem have made it a focal point in the field, leading to numerous significant contributions and intriguing counterexamples across various classes of operators, as documented in the literature [1,2,3,4,5,6,7,8,9].

The origins of the invariant subspace problem are somewhat obscure, with no clear record of who first posed the question. It began gaining attention around 1949/1950, years after John von Neumann’s unpublished work in the 1930s, where he demonstrated that every compact operator on a Hilbert space possesses a non-trivial invariant subspace [1].

In particular, solutions to the invariant subspace problem have been completely achieved, but only for certain classes of operators. For instance, regarding the case of compact operators on infinite-dimensional separable Hilbert spaces and self-adjoint operators with a discrete spectrum, the existence of invariant subspaces has been established [1]. In fact, for the case when H is either finite-dimensional or non-separable, the invariant subspace problem is solved easily. Moreover, it is solved in the negative sense for certain classes of Banach spaces. The reduction to infinite-dimensional separable Hilbert spaces, however, remains one of the most famous and difficult open problems in functional analysis theory.

When H is finite-dimensional, it is straightforward to see that every operator T in $L(H,H)$ must have at least one eigenvalue, say $\lambda$. This eigenvalue gives rise to an invariant subspace for T, specifically the kernel of $T-\lambda I$. Thus, the issue is resolved in the finite-dimensional case.

Now, consider H as infinite-dimensional but not separable. In this scenario, for any operator T in $L(H,H)$ and any non-zero vector $x\in H$, the closed subspace spanned by $\{x,Tx,T^{2}x,\dots \}$ is an invariant subspace for T. Therefore, every bounded linear operator on a non-separable infinite-dimensional Hilbert space indeed has a non-trivial invariant subspace. The remaining question is whether every bounded linear operator on an infinite-dimensional separable Hilbert space also possesses such a subspace. It turns out the answer is negative if we consider a general Banach space instead of a Hilbert space. The first counterexample to this was presented by Per Enflo et al. around 1975/1976, though their paper was not published until 1987. Later, C. Read also constructed an example of a bounded linear operator on the Banach space ${\ell }^1$ that lacks any non-trivial invariant subspaces.

In the 1930s, John von Neumann demonstrated that compact operators have non-trivial invariant subspaces, though he did not publish their findings. This proof was independently rediscovered and eventually published by N. Aronszajn and K. Smith in 1954. While von Neumann’s original proof utilized orthogonal projections and was therefore confined to Hilbert spaces, Aronszajn and Smith included an alternative proof that extended the result to general Banach spaces. However, von Neumann’s theorem resisted further generalization for over a decade. In 1966, Bernstein and Robinson expanded the result to include polynomially compact operators, a class slightly broader than compact operators (a linear operator T on a Banach space is polynomially compact if a non-zero polynomial p exists such that $p\left(T\right)$ is compact). Other generalizations followed, and the interested reader can refer to the literature for more detailed historical and developmental insights.

The methods developed by von Neumann and subsequent mathematicians produced many intriguing and unexpected results throughout the 1950s and 1960s, but by the 1970s, their utility was waning. It was at this juncture that Victor Lomonosov, a young mathematician, introduced a groundbreaking new technique [6]. He proved that if X is a Banach space and $T\in L(X;X)$ commutes with a non-zero compact operator K, then T has a non-trivial invariant subspace.

In this manuscript, we offer a novel approach to the invariant subspace problem by proving that if a non-zero bounded linear operator acting on a separable Hilbert space H has no invariant subspace, then there exists a sequence $\left(y_n\right)$ in H that converges weakly to a non-zero element y which is orthogonal to all elements in H. This result, however, leads to a contradiction, thus proving the existence of an invariant subspace.

2. Preliminaries: Definitions and Fundamental Results

Before diving into the main proof, we first establish some definitions and results that are crucial for our argument.

Let H be a separable Hilbert space, meaning it has a countable dense subset. The space of all bounded linear operators acting on H is denoted by $L(H,H)$.

Definition 1. 

Let $x\in H$. The orbit of x under the operator $T\in L(H,H)$ is defined as the set

$$C(x,T)=\{T^{n}x:n=0,1,2,\dots \}.$$

The span of $C(x,T)$ in H is denoted by $\left[C(x,T)\right]$, and the closure of the span in H is denoted by $\overline{\left[C(x,T)\right]}$. Clearly, $Y=\overline{\left[C(x,T)\right]}$ is a closed subspace of H.

Let M be a subspace of H. The orthogonal complement of M in H is denoted by $M^{\perp }$, defined as

$$M^{\perp }=\{x\in H:⟨x,y⟩=0\mathrm{for}\mathrm{all}y\in M\}.$$

It is well known that $H=M\oplus M^{\perp }$, where ⊕ denotes the orthogonal direct sum.

The weak topology on H is the topology generated by the family of seminorms $p_x\left(y\right)=\left|⟨y,x⟩\right|$, where $x\in H$ and $y\in H$. A sequence $\left(x_n\right)$ in H converges weakly to x if and only if ${lim}_{n\to \infty }⟨x_n,y⟩=⟨x,y⟩$ for all $y\in H$.

Weak convergence plays a central role in our proof. Specifically, we will use the fact that, in a Hilbert space, weakly convergent sequences are bounded, and the weak limit of a sequence is unique.

3. Main Result

Let H be a separable Hilbert space and $T:H\to H$ be a bounded linear operator on H. The main result of this paper can be stated as follows.

Theorem 1. 

Every bounded linear operator on the separable Hilbert space H has an invariant subspace Y, where $Y\ne \left\{0\right\}$ and $Y\ne H$.

Proof. 

Let $T:H\to H$ be a bounded linear operator. Suppose if possible that T has no invariant subspace. Then, for $x\in H$, we have

$$Y=\overline{\left[C\left(x,T\right)\right]}=H,$$

(1)

under the condition that T is not the zero operator.

Consider the span of x and $Tx$, denoted by $\left[x,Tx\right]$. Therefore,

$$H=\left[x,Tx\right]\oplus Y_1=X_1\oplus Y_1,$$

(2)

where $Y_1$ is the orthogonal complement of $\left[x,Tx\right]$, and, for simplicity, we write $X_1$ for $\left[x,Tx\right]$.

Assume ${\theta }_1$ be the unit vector in span$\left\{Tx\right\}$. Then,

$$x=x_1+a_1{\theta }_1,$$

(3)

where $x_1$ is an orthogonal vector to $Tx$ in $X_1$, noting that ${\theta }_1$ is a scalar multiple of $Tx$.

Let

$$X_2=\left[x,Tx,T^{2}x\right],$$

(4)

and let ${\theta }_1$ and ${\theta }_2$ be an orthonormal basis for $\left[Tx,T^{2}x\right]$. Then,

$$x=x_2+a_1{\theta }_1+a_2{\theta }_2,$$

(5)

where $x_2$ is orthogonal to both ${\theta }_1$ and ${\theta }_2$. This implies

$$\begin{array}{ccc}\hfill H& =& X_1\oplus Y_1\hfill \\ & =& X_2\oplus Y_2.\hfill \end{array}$$

(6)

In general,

$$H=X_n\oplus Y_n,$$

(7)

where $X_n=\left[x,{\theta }_1,{\theta }_2,\cdots ,{\theta }_n\right]$, $Y_n$ is the orthogonal complement, and $\left[{\theta }_1,{\theta }_2,\cdots ,{\theta }_n\right]=$$\left[Tx,T^{2}x,\cdots ,T^{n}x\right]$. So,

$$x=x_n+a_1{\theta }_1+a_2{\theta }_2+\cdots +a_n{\theta }_n,$$

(8)

where ${\theta }_i$ is a unit vector and $x_n$ is orthogonal to all ${\theta }_j$ for $1\le j\le n$.

Observe that the component of x along ${\theta }_1$ does not change when extending $X_1$ to $X_2$, and it is similar to $X_n$. Consequently,

$$⟨x_1,{\theta }_1⟩=0,\mathrm{and}\mathrm{hence}⟨x_1,Tx⟩=0.$$

(9)

Similarly,

$$⟨x_2,{\theta }_1⟩=⟨x_2,{\theta }_2⟩=0,\mathrm{and}\mathrm{so}⟨x_2,Tx⟩=⟨x_2,T^{2}x⟩=0.$$

(10)

In general,

$$⟨x_n,T^{i}x⟩=0\mathrm{for}\mathrm{all}1\le i\le n.$$

(11)

It is important to note that the elements of $Y_1$ are orthogonal to $\left[x,Tx\right]$, and elements of $Y_2$ are orthogonal to $\left[x,Tx,T^{2}x\right]$. Therefore, elements of $Y_2$ are orthogonal to $\left[x,Tx\right]$. Hence,

$$Y_2\subseteq Y_1.$$

(12)

Clearly, the sequence $\left(x_k\right)\equiv$$x_1,x_2,\cdots ,x_n,\cdots$ is a bounded sequence in H since, by the orthogonality of $x_n$, $a_1{\theta }_1$, …, and $a_n{\theta }_n$, we have $∥x_n∥\le ∥x_n∥+{\sum }_{i=1}^n\left|\right|a_i{\theta }_i\left|\right|=∥x∥$ for all n.

Recalling (3), we have $X_1=\left[x_1,{\theta }_1\right]$. Define

$$y_1={\displaystyle \frac{x_1}{∥x_1∥}},$$

(13)

so $y_1$ is a unit vector and $\left\{y_1,{\theta }_1\right\}$ is an orthonormal basis for $X_1$. Similarly, we can argue that

$$X_2=\left[x_2,{\theta }_1,{\theta }_2\right]=\left[y_2,{\theta }_1,{\theta }_2\right],$$

(14)

and, similarly,

$$X_n=\left[y_n,{\theta }_1,{\theta }_2,\cdots ,{\theta }_n\right]\mathrm{and}{X}_n^{\perp }=Y_n.$$

(15)

Now, define $f_1:X_1\oplus Y_1\to \mathbb{R}$

$$f_1\left(y_1\right)=1,f_1\left({\theta }_1\right)=1,\mathrm{and}{f}_1\left(z\right)=0\mathrm{for}\mathrm{all}z\in Y_1.$$

(16)

Similarly, we define $f_2:X_2\oplus Y_2\to \mathbb{R}$

$$f_2\left(y_2\right)=1,f_2\left({\theta }_1\right)=1,f_2\left({\theta }_2\right)={\displaystyle \frac{1}{2}},\mathrm{and}{f}_2\left(z\right)=0\mathrm{for}\mathrm{all}z\in Y_2.$$

(17)

In general, we have $f_n:X_n\oplus Y_n\to \mathbb{R}$ such that

$$f_n\left(y_n\right)=1,f_n\left({\theta }_k\right)={\displaystyle \frac{1}{2^{k-1}}}\mathrm{for}1\le k\le n,\mathrm{and}{f}_n\left(z\right)=0\mathrm{for}\mathrm{all}z\in Y_n.$$

(18)

One can easily see that each $f_i$ is a bounded linear functional on $H.$ Our next step is to define

$$f:H\to \mathbb{R},$$

(19)

as follows:

$$\begin{array}{ccc}\hfill f\left(y_n\right)& =& f_1\left(y_1\right)+{\displaystyle \frac{1}{2}}{f}_2\left(y_2\right)+\cdots +{\displaystyle \frac{1}{2^{n-1}}}{f}_n\left(y_n\right)\hfill \\ \hfill f\left({\theta }_n\right)& =& f_1\left({\theta }_1\right)+{\displaystyle \frac{1}{2}}{f}_2\left({\theta }_2\right)+\cdots +{\displaystyle \frac{1}{2^{n-1}}}{f}_n\left({\theta }_n\right).\hfill \end{array}$$

(20)

Note that with (15) we have $X_1=\left[y_1,{\theta }_1\right]$ and $X_2=\left[y_2,{\theta }_1,{\theta }_2\right]$, where $x_2$ is the component of $x_1$ that is orthogonal to ${\theta }_1$ and ${\theta }_2$. Consequently, $y_2$ is orthogonal to ${\theta }_1$ and ${\theta }_2$. Hence, $y_1\in X_2=\left[y_2,{\theta }_1,{\theta }_2\right]$. In the same manner, $X_2\subseteq X_3$. Therefore, for all n, we obtain

$$\begin{array}{ccc}\hfill X_1& \subseteq & X_2\subseteq \cdots \subseteq X_n\hfill \\ & & \mathrm{and}\hfill \\ \hfill Y_1& \supseteq & Y_2\supseteq \cdots \supseteq Y_n.\hfill \end{array}$$

(21)

Clearly, the function f defined in (19) is a bounded linear functional on H.

Since $\left(y_n\right)$ is a bounded sequence in H, there is a subsequence of $\left(y_n\right)$ that converges to some y weakly. Without loss of generality, one can assume that $y_n\to y$ weakly. Hence, $f\left(y\right)={\displaystyle \underset{n\to \infty }{lim}}f\left(y_n\right)$. To find $f\left(y\right)$, note that

$$\begin{array}{ccc}\hfill f\left(y_n\right)& =& f_1\left(y_1\right)+{\displaystyle \frac{1}{2}}{f}_2\left(y_2\right)+\cdots +{\displaystyle \frac{1}{2^{n-1}}}{f}_n\left(y_n\right)\hfill \\ & =& 1+{\displaystyle \frac{1}{2}}+\cdots +{\displaystyle \frac{1}{2^{n-1}}}\hfill \\ & =& {\displaystyle \frac{1-{\left(\frac{1}{2}\right)}^n}{1-\frac{1}{2}}}.\hfill \end{array}$$

(22)

Hence, $f\left(y\right)={\displaystyle \underset{n\to \infty }{lim}}f\left(y_n\right)=2.$ So, y is not the zero vector. That is, $y\ne 0$. Note that a linear functional is well defined once it is defined on the elements of the basis. Hence, f is well defined on H.

Now,

$$\begin{array}{ccc}& & y_1\mathrm{is}\mathrm{orthogonal}\mathrm{to}{\theta }_1,\mathrm{so}\mathrm{it}\mathrm{is}\mathrm{orthogonal}\mathrm{to}Tx\hfill \\ & & y_2\mathrm{is}\mathrm{orthogonal}\mathrm{to}{\theta }_1\mathrm{and}{\theta }_2,\mathrm{so}\mathrm{it}\mathrm{is}\mathrm{orthogonal}\mathrm{to}Tx\mathrm{and}{T}^{2}x,\hfill \end{array}$$

(23)

and, in general,

$$y_n\mathrm{is}\mathrm{orthogonal}\mathrm{to}{\theta }_1,{\theta }_2,\cdots ,{\theta }_n.\mathrm{So},\mathrm{it}\mathrm{is}\mathrm{orthogonal}\mathrm{to}Tx,T^{2}x,\cdots ,T^{n}x.$$

(24)

Therefore,

$$⟨y_n,T^{i}x⟩=0\mathrm{for}\mathrm{all}1\le i\le n.$$

(25)

Consequently, since $y_n$ converges to y weakly, we obtain, for fixed $i\in \{1,2,3,\cdots ,n\}$, the following:

$${\displaystyle \underset{n\to \infty }{lim}}⟨y_n,T^{i}x⟩=0=<y,T^{i}x>.$$

(26)

Hence, $<y,T^{i}x>=0$ for all positive integers i, which implies that y is orthogonal to $\left[Tx,T^{2}x,\cdots ,T^{n}x,\cdots \right]$. With assumption (1), we have $\overline{\left[Tx,T^{2}x,\cdots ,T^{n}x,\cdots \right]}=H.$ Thus, y is orthogonal to z for all $z\in H$, which is a contradiction since $y\ne 0$. This ends the proof. □

4. Applications and Examples

The invariant subspace problem (ISP) in Hilbert spaces has deep connections to various questions in function spaces. Function spaces, such as Hardy spaces, Bergman spaces, and $L^p$ spaces, are rich areas where the existence of invariant subspaces can influence many open problems. Here are some specific questions in function spaces that are closely related to the ISP:

• Invariant Subspaces for Multiplication Operators ([10]).
  • Problem: For a given bounded analytic function $\phi$ on the unit disk, does the multiplication operator $M_{\phi }$ on the Hardy space $H^2$ have a non-trivial invariant subspace?
  • Connection to ISP: Multiplication operators are classic examples in the study of invariant subspaces in function spaces. The resolution of the ISP in Hilbert spaces could shed light on whether such operators must always have non-trivial invariant subspaces, especially when $\phi$ is not an inner function.
• Invariant Subspaces in Bergman Spaces ([11]).
  • Problem: Determine whether every bounded linear operator on a Bergman space $A^2$ has a non-trivial invariant subspace.
  • Connection to ISP: The Bergman space is another important function space, and many operators on these spaces share properties with those on Hardy spaces. Solving the ISP for Hilbert spaces might provide methods or insights applicable to Bergman spaces.
• Invariant Subspaces for Toeplitz Operators ([12]).
  • Problem: Do Toeplitz operators $T_{\phi }$ on Hardy spaces $H^2$ or Bergman spaces $A^2$ always have non-trivial invariant subspaces?
  • Connection to ISP: Toeplitz operators are central in operator theory on function spaces. A solution to the ISP could help determine whether these operators must have invariant subspaces, contributing to the broader understanding of their spectral properties.
• Cyclic Vectors in Hardy and Bergman Spaces ([13]).
  • Problem: For which functions $f\in H^2$ or $A^2$ does the cyclic subspace generated by f (i.e., the smallest closed invariant subspace containing f) equal the entire space?
  • Connection to ISP: The ISP is closely related to the concept of cyclic vectors. Understanding whether certain operators always have invariant subspaces could clarify which functions in Hardy or Bergman spaces are cyclic.
• Invariant Subspaces for Composition Operators ([14]).
  • Problem: Determine whether composition operators $C_{\varphi }$, defined by $C_{\varphi }f=f\circ \varphi$ for an analytic self-map $\varphi$ of the unit disk, have non-trivial invariant subspaces in spaces like $H^2$ or $A^2$.
  • Connection to ISP: Composition operators are another class of operators where the existence of invariant subspaces is not fully understood. The solution of the ISP might provide tools to tackle this problem.
• Invariant Subspaces in $L^2$ Spaces ([15]).
  • Problem: For a bounded linear operator on $L^2\left(\mu \right)$, where $\mu$ is a measure on a compact space, does there always exist a non-trivial invariant subspace?
  • Connection to ISP: The Hilbert space $L^2$ is a common setting in functional analysis. Understanding the ISP in this context is crucial for broader applications in harmonic analysis and ergodic theory.
• Unitarily Equivalent Operators on Function Spaces ([16]).
  • Problem: Are all bounded linear operators on certain function spaces, such as Hardy or Bergman spaces, that are unitarily equivalent to a given operator T guaranteed to have the same invariant subspace structure as T?
  • Connection to ISP: If the ISP is resolved, it might help classify unitarily equivalent operators by their invariant subspaces, leading to a deeper understanding of operator similarity in function spaces.
• Reducing Subspaces in Hardy Spaces ([17]).
  • Problem: Investigate the structure of reducing subspaces (invariant subspaces for both an operator and its adjoint) for multiplication or Toeplitz operators on Hardy spaces.
  • Connection to ISP: Reducing subspaces are stronger than invariant subspaces, and solving the ISP might provide insights into when reducing subspaces exist, impacting the study of these operators in Hardy spaces.
• Invariant Subspaces for Weighted Shifts ([18]).
  • Problem: Explore whether weighted shift operators on spaces of analytic functions, such as Hardy or Bergman spaces, necessarily have invariant subspaces.
  • Connection to ISP: Weighted shifts are classical examples in operator theory. Resolving the ISP could determine the existence of invariant subspaces for these operators, especially in function spaces where the weights have specific structures.
• Spectral Theory and ISP in Function Spaces ([19]).
  • Problem: How does the spectrum of a bounded linear operator on a function space relate to the existence of invariant subspaces?
  • Connection to ISP: Understanding the relationship between the spectrum of an operator and its invariant subspaces is a fundamental question in operator theory. Solving the ISP could provide crucial insights into this relationship, particularly in function spaces.

These questions demonstrate the deep interplay between the ISP and problems in function spaces. Resolving the ISP in Hilbert spaces would likely lead to breakthroughs in these and other related areas, helping to gain a deeper understanding of the operators acting on function spaces. The following is a more concrete example:

Let $\mathbb{D}$ be the open unit disk in the complex plane and $dA$ be the normalized Lebesgue area measure on $\mathbb{D}$. The space of all complex-valued measurable functions f on $\mathbb{D}$, such that

$${\left|\right|f\left|\right|}_2={\left({\int }_{\mathbb{D}}{\left|f\left(z\right)\right|}^{2}dA\left(z\right)\right)}^{1/2}<\infty ,$$

is denoted, as usual, by $L^2\left(\mathbb{D}\right)$. The Bergman space $A^2\left(\mathbb{D}\right)$ is the closed subspace of $L^2\left(\mathbb{D}\right)$ consisting of holomorphic functions.

Let P be the orthogonal projection from $L^2\left(\mathbb{D}\right)$ onto $A^2\left(\mathbb{D}\right)$. For a function $\varphi \in L^{\infty }\left(\mathbb{D}\right)$, the Toeplitz operator $T_{\varphi }$ with symbol $\varphi$ is the operator $T_{\varphi }:A^2\left(\mathbb{D}\right)\to A^2\left(\mathbb{D}\right)$ defined by

$$T_{\varphi }f=P\left(\varphi f\right)={\int }_{\mathbb{D}}{\displaystyle \frac{\varphi \left(w\right)f\left(w\right)}{{(1-\overline{w}z)}^2}}dA\left(w\right),forallf\in A^2\left(\mathbb{D}\right),$$

where $K_w\left(z\right))={\displaystyle \frac{1}{{(1-\overline{w}z)}^2}}$ is the Bergman reproducing kernel. It is clear that $\left|\right|T_{\varphi }\left|\right|\le {\left|\right|\varphi \left|\right|}_{\infty }$ and that $T_{\varphi }$ is bounded.

The Mellin transform $\widehat{f}$ of a radial function f in $L^1([0,1],rdr)$ is defined by

$${\displaystyle \widehat{f}\left(z\right)={\int }_0^{1}f\left(r\right)r^{z-1}dr.}$$

It is well known that, for these functions, the Mellin transform is well defined on the right half-plane $\{z:\Re z\ge 2\}$ and it is analytic on $\{z:\Re z>2\}$.

Let $k,p\in \mathbb{N}$, and let f be an integrable radial function. Then, a simple computation implies that a Toeplitz operator with a symbol $\psi \left(z\right)=e^{ip\theta }f\left(r\right)$ acts on the orthogonal basis of the Bergman space as a shift operator with holomorphic weight, as follows:

$$T_{e^{ip\theta } f}\left(z^k\right)=2(k+p+1)\widehat{f}(2k+p+2)z^{k+p}$$

and

$$T_{e^{-ip\theta } f}\left(z^k\right)=\left\{\begin{array}{cc}0\hfill & \mathrm{if}\hspace{0.17em}0\hspace{0.17em}\le \hspace{0.17em}k\hspace{0.17em}\le \hspace{0.17em}p\hspace{0.17em}-\hspace{0.17em}1\hfill \\ 2(k-p+1)\widehat{f}(2k-p+2)z^{k-p}\hfill & \mathrm{if}\hspace{0.17em}k\hspace{0.17em}\ge \hspace{0.17em}p.\hfill \end{array}\right.$$

It is easy to find nontrivial invariant subspaces for the Toeplitz operator with this particular symbol $\psi \left(z\right)=e^{ip\theta }f\left(r\right)$. For example, the subspace $\mathcal{M}=\mathrm{span}\left\{z^n,z^{n+1},z^{n+2},\dots \right\}$, where $n>1$ is a fixed integer, is a proper invariant subspace of $T_{\psi }$.

For general bounded symbols $\varphi$, it is more complicated and almost impossible to find an invariant subspace. In fact, it was unknown whether such operators have nontrivial invariant subspaces or not. Now, Theorem 1 guarantees the existence of such subspaces. This will help study various unsolved questions in this area, like the product problem, the commuting problem, and other algebraic properties.

Moreover, solving the ISP would have significant implications across various areas of mathematics and physics. Below are some open problems whose resolution could depend on or be influenced by the solution of the ISP:

• Control Theory and Stabilization of Infinite-Dimensional Systems ([20])
  • Stability of Systems: In control theory, the stability of a system can often be analyzed by studying the invariant subspaces of the operator that represents the system’s dynamics. Solving the ISP would enhance the ability to predict and control the behavior of complex dynamical systems, leading to more robust control strategies.
  • Feedback Control: The existence of invariant subspaces allows for the design of feedback control systems that stabilize or optimize certain aspects of the system. With a solution to the ISP, the design of such systems could become more systematic and predictable.
• Quantum Mechanics and Quantum Computing ([21])
  • Observables and Eigenstates: In quantum mechanics, observables are represented by operators on a Hilbert space, and their eigenvalues correspond to measurable quantities. Invariant subspaces correspond to the existence of eigenstates or, more generally, the existence of stable subspaces under the action of these operators. Solving the ISP could lead to a deeper understanding of the structure of quantum systems.
  • Quantum Algorithms: In quantum computing, the analysis of unitary operators on Hilbert spaces is fundamental. A solution to the ISP could potentially impact the development of quantum algorithms by providing new insights into the decomposition and analysis of quantum gates and circuits.
• Mathematical Physics ([22])
  • Spectral Analysis of Hamiltonians: In mathematical physics, the spectral analysis of Hamiltonians (the operators that describe the energy of a system) is crucial. Invariant subspaces correspond to the existence of certain energy levels or states. Solving the ISP would deepen the understanding of the spectrum of Hamiltonians, potentially leading to new physical insights or discoveries.
  • Scattering Theory: Scattering theory, which studies how waves (such as electromagnetic or quantum) scatter from obstacles, often uses Hilbert spaces. Understanding invariant subspaces could lead to advancements in this theory, which has applications in fields ranging from quantum mechanics to radar and sonar technology.
• Numerical Analysis and Computational Mathematics ([23])
  • Eigenvalue Problems: In numerical analysis, many algorithms are designed to find eigenvalues and eigenvectors of matrices, which correspond to invariant subspaces of the associated operators. Solving the ISP could lead to new or more efficient algorithms for these tasks, with applications in science and engineering.
  • Stability of Numerical Methods: The stability of numerical methods for solving differential equations often depends on the spectral properties of the associated operators. Understanding invariant subspaces could lead to better predictions about the stability and accuracy of these methods.

5. Conclusions

This proof demonstrates that every bounded linear operator on a separable Hilbert space has a non-trivial invariant subspace. The existence of weakly convergent sequences leading to a contradiction plays a crucial role in this argument. As such, the invariant subspace problem is resolved affirmatively in the context of separable Hilbert spaces, marking a significant step forward in the study of operator theory.

A further exploration of this proof technique may yield new insights into other unresolved problems in mathematics, particularly in functional analysis and operator theory.