All Questions
Tagged with fa.functional-analysis operator-theory
69 questions
8
votes
3
answers
1k
views
Ramanujan's Master Formula: A proof and relation to umbral calculus
The Ramanujan's master theorem states that:
$$
\int_0^{\infty}x^{s-1}\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}a_nx^ndx=\Gamma(s)a_{-s}
$$
I found a really strange proof recently on a personal blog:
Define
$...
- 302
10
votes
2
answers
1k
views
Harmonic oscillator discrete spectrum
Let us act intentionally stupid and assume we do not know that we can solve for the spectrum of the harmonic oscillator
$$-\frac{d^2}{dx^2}+x^2$$
explicitly.
Is there an abstract argument why the ...
- 501
6
votes
1
answer
2k
views
Comparing norms on tensor products of matrices
Given a Hilbert space $H$, let $S_1(H)$ denote the space of trace-class operators on $H$, with the trace-class norm or Schatten 1-norm. That is
$$ \Vert T \Vert_1 = \sum_{j\geq 1} |s_j| $$
where $(s_1,...
- 209
39
votes
3
answers
14k
views
Is the Invariant Subspace Problem interesting?
There's an amusing comment in Peter Lax's Functional Analysis book. After a brief description of the Invariant Subspace Problem, he says (paraphrasing) "...this question is still open. It is also an ...
- 1,241
34
votes
1
answer
3k
views
tr(ab)=tr(ba), part 2.
This is a Banach space version of Andre Henriques' question
Trace Question
for Hilbert spaces. Let $a:X\to Y$ and $b:Y\to X$ be bounded linear operators between Banach spaces s.t. $ba$ and $ab$ ...
- 31.5k
18
votes
4
answers
1k
views
Who first used the multiplication operator version of spectral theory
This is another history question.
Hilbert phrased the spectral theorem in terms of resolutions of the identity.
While this remained the form of Stone and von Neumann, they did also have the ...
- 827
13
votes
3
answers
1k
views
Is the set of separable quantum states closed?
Let $\mathcal H,\mathcal H'$ be Hilbert spaces (not necessarily separable).
A "separable state" is a trace-class operator of the form $\sum_i \rho_i\otimes\rho_i'$ where $\rho_i,\rho_i'$ are positive ...
- 896
11
votes
2
answers
714
views
A neat evaluation of an infinite matrix?
Let $M_n$ be an $n\times n$ matrix defined as
$$M_n
=\left[\frac{2i+1}{2(i+j+1)}\binom{i-1/2}i\binom{j-1/2}jx^{i+j+1}\right]_{i,j=0}^n.$$
With $I_n$ the identity matrix, consider $A_n:=I_n-M_n^2$. ...
- 43.2k
10
votes
2
answers
926
views
Continuity of the product map
Let $A$ be a $C^*$-algebra.
Is it possible to characterize $A$ for which the product map defined by
$$\sum\limits_{i=1}^n a_i\otimes b_i \mapsto \sum\limits_{i=1}^n a_i b_i$$
is continuous with ...
- 4,705
9
votes
1
answer
956
views
A problem in functional calculus
This is embarrassing, I think it must work, but I can't see how to prove it works. If anyone knows enough functional calculus of operators on a Hilbert space to tell me how to do it, I would be very ...
- 1,143
6
votes
1
answer
680
views
Is there an operator algebraic reformulation of the invariant subspace problem?
Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant ...
6
votes
1
answer
765
views
An equivalence relation on the space of polynomials in one complex variable
Let $P(z)$ be a polynomial with complex variable $z$. We consider the following distribution for the roots of $P(z)=0$: the distribution is a triple $(n_{1},n_{2},n_{3})$
where these integers are ...
- 356
5
votes
2
answers
459
views
Backward heat equation and forward perturbed heat equation well posed?
I consider the following scenario. Let $I$ be a compact interval in space and $f$ a nice function in the space $C^{\infty}(I)$. In the following we consider a self-adjoint realization of our operators ...
- 536
4
votes
1
answer
213
views
Mapping properties of backward and forward heat equation
In a previous question on mathoverflow, I asked about the following:
Let $\Delta$ be the Laplacian on some compact interval $I$ of the real line with let's say Dirichlet boundary conditions.
The ...
- 536
2
votes
1
answer
755
views
Existence of a solution to an infinite dimensional Stratonovich SDE
Let
$U,H$ be separable $\mathbb R$-Hilbert spaces
$Q\in\mathfrak L(U)$ be nonnegative and self-adjoint with finite trace
$U_0:=Q^{1/2}U$
$(\Omega,\mathcal A,(\mathcal F_t)_{t\ge 0},\operatorname P)$ ...
- 167
1
vote
1
answer
654
views
Properties of the trace term in the Itō formula
Let's consider the SDE $${\rm d}X_t=u_t(X_t){\rm d}t+\xi_t(X_t){\rm d}W_t\;\;\;\text{for all }t\ge 0\tag 1$$ where
$U,H$ are separable $\mathbb R$-Hilbert spaces
$Q\in\mathfrak L(U)$ is nonnegative ...
- 167
31
votes
3
answers
5k
views
When is an integral transform trace class?
Given a measure space $(X, \mu)$ and a measurable integral kernel $k : X \times X \rightarrow \mathbb{C}$, the operator
$$ K f(\xi) =\int_{X} f(x) k(x,\xi) d \mu(x),$$
the operator $K$ is Hilbert ...
- 11.2k
27
votes
0
answers
1k
views
Unital $C^{*}$ algebras whose all elements have path connected spectrum
A unital $C^{*}$ algebra is called a "Path connected algebra" if the spectrum of all its elements is a path connected subset of $\mathbb{C}$.
What is an example of a non commutative ...
- 356
18
votes
1
answer
564
views
Is the space of Hankel operators complemented in B(H)?
Let $H$ be $\ell^2({\mathbb N})$ and let $S:H\to H$ be the unilateral forward shift, so that $S^*S=I\neq SS^*$. Then a bounded operator $T:H\to H$ is Hankel if and only if it satisfies $TS=S^*T$.
Let ...
- 25.8k
17
votes
4
answers
4k
views
How much does the absolute value of an operator behave like an absolute value?
Recall that the absolute value of a bounded operator $T$ on a Hilbert space $H$ is the unique positive operator $|T|$ such that $$\||T|x\|=\|Tx\|$$ for all $x\in H$. It can be defined using the ...
- 3,115
13
votes
2
answers
1k
views
Homotopy groups of Fredholm operators
If $X$ is separable complex Hilbert space and $\mathcal{F}$ the topological space of Fredholm operators on $X$, then it is well-known, that
$$ \pi_0(\mathcal{F}) = \mathbb{Z}\, , $$
i.e. the connected ...
- 173
12
votes
1
answer
901
views
Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem?
This post is an appendix of this one.
Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.
Invariant subspace problem: Let $T \in B(H)$. Is ...
11
votes
1
answer
309
views
Which closed subsets $Y$ of a compact space $X$ admit a linear extensor $C(Y)\to C(X)$?
In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$.
The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), so it admits a continuous right ...
- 60.5k
11
votes
1
answer
451
views
Comparison of the absolute value of an operator with its positive parts
It is well known that the absolute value on operators does not satisfy the triangle inequality.
My question is whether for all positive operators $P,Q \in B(\mathcal H)$ is there a universal ...
- 3,984
11
votes
2
answers
545
views
Is $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$ a commutative algebra?
Consider $l^\infty(\mathbb{Z})$ the Banach space of bounded complex valued functions on the abelian group $\mathbb{Z}$ with the supremum norm. It has a natural action by $\mathbb{Z}$ given by $(zf)(g):...
- 1,550
11
votes
3
answers
445
views
Does the generator of a 1-parameter group of Banach space isometries know which elements are entire?
Let $X$ be a complex Banach space. Let $(\sigma_t)_{t \in \mathbb{R}}$ be a 1-parameter group of linear isometries of $X$ which is strongly continuous i.e. $t \mapsto \sigma_t(x)$ is continuous for ...
- 662
10
votes
1
answer
598
views
What happens if we rotate the kernel of an integral operator?
Given an integral operator $K$ on $L^2(\mathbb R)$ with kernel $k(x, y)$, consider the integral operator $L$ on $L^2(\mathbb R)$, whose kernel has the form $k(\alpha x+\beta y, \gamma x+\delta y)$, ...
- 452
9
votes
2
answers
848
views
$\zeta$-function regularized determinants
In (mathematical) physics in order to compute path integrals one often makes an infinite dimensional change of variables and uses infinite Jacobian as a purely formal expression. This step is done in ...
- 21.8k
9
votes
1
answer
359
views
Relaxation of notion of positive definite function
A function $f:\mathbb{R}\to\mathbb{R}$ is called positive definite (in the semigroup sense) if for all $n\geq 1$ and $x_1,\ldots,x_n\in\mathbb{R}$ pairwise different the matrix $(f(x_i+x_j))_{i,j=1}^n$...
- 3,031
8
votes
3
answers
691
views
Commutant of the conjugations by unitary matrices
Let $\mathcal{L}(\mathbb{C}^{n \times n})$ denote the algebra of all linear mappings from $\mathbb{C}^{n \times n}$ to $\mathbb{C}^{n \times n}$ and let $\mathcal{C} \subseteq \mathcal{L}(\mathbb{C}^{...
- 12.5k
8
votes
2
answers
488
views
If the cardinality of $B(X)$, the space of operators on $X$, is continuum, must $X$ be separable?
Does there exits any non-separable Banach space $X$ such that the size (cardinal number) of $B(X)$, bdd linear operators on $X$, is just of the continuum?
- 4,058
7
votes
2
answers
682
views
Hölder continuity for operators
Let $x,y$ be positive real numbers then
$$|\sqrt{x}-\sqrt{y}|=\dfrac{|x-y|}{\sqrt{x}+\sqrt{y}}=\sqrt{|x-y|}\cdot \dfrac{\sqrt{|x-y|}}{\sqrt{x}+\sqrt{y}}\leq 1\cdot |x-y|^{\frac{1}{2}}$$
we obtain $1/...
user127612
7
votes
1
answer
813
views
An equivalent condition for separability of $X^*$
Let $X$ be a Banach space. By the weak operator topology on $B(X)$, we mean the locally convex topology implemented by the following semi-norms:
$$B(X)\to[0,\infty) : T\to|\langle Tx,x^*\rangle|$$
...
- 4,058
7
votes
1
answer
414
views
Criteria for operators to have infinitely many eigenvalues
Normal compact linear operators on Hilbert spaces have infinitely many (counting multiplicities) eigenvalues by the spectral theorem.
For non-normal operators this no longer has to be true.
There ...
- 536
6
votes
1
answer
422
views
Absolutely 2-summable operator on a Hilbert space
An bouneded linear operator $A \in L(X, Y)$ (here $X$, $Y$ are Banach spaces) is called absolutely $2$-summable if there exists a $C>0$ such that
$$ \left( \sum_{j=1}^N \| A x_j\|_X^2 \right)^{1/2} ...
- 9,853
6
votes
3
answers
852
views
Are nuclear operators closed under extensions?
Given $X_i, Y_i$ Banach spaces, $f_j, g_j, T_i$ bounded linear operators for $i=1,2,3$ and $j=1,2$. We have the following diagram
$\require{AMScd}$
\begin{CD}
0 @>>> X_1 @>f_1>> X_2 ...
- 1,338
6
votes
2
answers
1k
views
The contractivity of the heat semigroup in $L^p$ spaces
Let $M$ be a Riemannian manifold. By functional calculus, it is immediate to show that the heat semigroup is a contraction in $L^2(M)$. I can also show that it is a contraction in any $L^p(M)$ with $p ...
- 5,407
6
votes
3
answers
3k
views
Non-empty resolvent set, then operator closed?
On Hilbert spaces, the following is true:
Let $T$ be a densely-defined linear operator with non-empty resolvent set, then $T$ is closed.
The obvious proof I see to show this uses explicitly the ...
- 115
5
votes
1
answer
552
views
Scattering theory for Coulomb potential
Both physical and mathematical theories of quantum scattering seem to be well developed in the case when the potential (or a more general perturbation of the Laplacian) decays fast enough at infinity ...
- 21.8k
5
votes
1
answer
456
views
The Bochner integral about a semigroup of bounded linear operators on a Banach space
Let $T(t)$ be a semigroup of bounded linear operators on a Banach space $X$. When does the following hold
$$
\int_0^t T(s)x ds = \Big(\int_0^t T(s) ds\Big)x, \quad x \in X \, ,
$$
where $ t \in (0,1)$?...
- 51
5
votes
1
answer
2k
views
Commuting with self-adjoint operator
Let $T$ be an (unbounded) self-adjoint operator. Assume that there is a bounded operator $S$ such that $TS=ST.$ For which kind of $f$ do we have that $f(T)S=Sf(T)?$
My thought was that using a ...
- 501
5
votes
1
answer
245
views
Convergence of functionals on compact projections on a separable Hilbert space
Let $H$ be a separable Hilbert space over $\mathbb{C}$, say $\ell_2$ for simplicity. Let $\mathcal{K}(H)$ denote the space of all compact operators on $H$ and $\mathcal{P}(H)$ the set of all finite ...
- 1,151
4
votes
1
answer
279
views
Schroedinger operator in 2 dimensions with singular potential
Consider the Schroedinger operator
$$H = -\Delta + \frac{c}{\vert x \vert^2}$$
in two dimensions with $c >0$
This operator has a self-adjoint realization, since it is a positive symmetric operator ...
4
votes
1
answer
308
views
Adjoint of the multiplication operator on a Sobolev space
Let $f\colon\mathbb{R}^n\rightarrow\mathbb{C}$ be a bounded function with a bounded first derivative. Then the multiplication operator $H^1(\mathbb{R}^n)\ni x\mapsto A_f x:=fx\in H^1(\mathbb{R}^n)$ is ...
- 128k
4
votes
2
answers
195
views
Regarding unital positive operators
Let $\Omega$ be a domain in $\mathbb{C}^n$. Let $\mathbb{D}$ denote the open unit disc in $\mathbb{C}$. Let $C_b(\Omega)$ and $C_b(\mathbb{D})$ denote the space of all bounded continuous complex ...
- 41
4
votes
0
answers
149
views
Cyclic vectors for the translation operator
Let $b\in \mathbb{R}\neq 0$, and consider the translation operators:
$$
\begin{align}
T_b:C(\mathbb{R}) & \rightarrow C(\mathbb{R})\\
f &\mapsto f(\cdot + b).
\end{align}
$$
*Are there known ...
- 5,405
4
votes
1
answer
386
views
Invertible unbounded linear maps defined on a Hilbert space
It is well-known that, assuming the axiom of choice, there are unbounded linear maps defined not only on a dense subset but on all of Hilbert space. Is it possible that such a map is invertible?
- 3,873
4
votes
1
answer
878
views
Commuting with an unbounded operator
Let $H$ be a Hilbert space. Let $A$ be a closed unbounded operator, and let $B\in B(H)$ be a bounded operator.
Definition:
$A$ and $B$ strong-commute if the partial isometry in the polar ...
- 43.2k
4
votes
2
answers
730
views
Finite dimensional approximations of operators on Hilbert spaces
Let $e_1,e_2,\dots$ be a Schauder basis for a Hilbert space $(V , \langle \cdot , \cdot \rangle)$. Let $A:V \to V$ be an operator. Finally, let $V_n = {\rm span}( e_1, \dots, e_n)$. Let $i_n : V_n \...
- 614
4
votes
1
answer
279
views
Reference request: Baire's theorem for operator ranges
Let $F$ be a Banach space. A vector subspace $U \subseteq F$ is called an operator range if there exists a Banach space $E$ and a bounded linear mapping $T: E \to F$ such that $TE=U$. By a quotient ...
- 12.5k