# Questions tagged [unbounded-operators]

Let $X$ and $Y$ be normed spaces and $T: D(T)\rightarrow Y$ a linear operator, where $D(T)\subset X$. The operator $T$ is said to be unbounded if there exists a sequence $\{x_n\}\subset D(T)$ s.t. $$\| Tx_n\| \geq n\| x_n\|$$

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### Is a dense subset in the domain of a closed, densely defined linear operator a core?

Let $X_0,X_1$ be Banach spaces. Let $A:D(A)\subseteq X_0\to X_1$ be a closable linear operator. Recall the definition of a core for such an operator: A set $\mathcal D\subseteq D(A)$ is called a core ...
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### Why the statement of one of the spectral theorems involves an integral rather than a sum?

My question is about this theorem regarding projection-valued measures. Suppose $H$ is a separable Hilbert space (in many cases we are just concerned about the separable case). Suppose $A$ is an ...
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### Example of discontinuity of position operator $X\colon f(x) \mapsto xf(x)$ for $f(x) \in L^2(\mathbb{R})$

It is well known that continuous linear operators are bounded and vice versa. It is also well known that the position operator (which I shall call $X$) causes many headaches in quantum mechanics due ...
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Suppose $A$ is the infinitesimal generator of a $C_0$ semigroup $S(t)$ on an Hilbert space $X$. If $$\langle Ax, x\rangle \leq \omega \|x \|^2 \ \ \ \forall x \in \mathfrak{D}(A)$$ then $$\|S(t)\... 1answer 54 views ### The unbounded antipode for Woronowicz's quantum group \operatorname{SU}_q(2) For non-zero q\in [-1,1], Woronowicz's quantum group \operatorname{SU}_q(2) is given as the universal unital \mathrm{C}^*-algebra generated by elements a,c\in C(\operatorname{SU}_q(2)) subject ... 1answer 32 views ### Couterexample to Littlewood-Paley theorem Let d\geq 2 and let P_j be the Fourier multiplier defined on L^2(\mathbb{R}^d) by \widehat{P_kf}(\xi)=\mathbf{1}_{2^k<|\xi|\leq2^{k+1}} \hat f(\xi), for any k \ge 0. It has been proven by ... 1answer 89 views ### Is Tf(x)=\frac{1}{x}\int_{0}^{x}f(y)dy bounded as operator on L^2((0,1);\mathbb{R} )? Given the operator T:L^2((0,1);\mathbb{R} )\rightarrow L^2((0,1);\mathbb{R} ) defined by Tf(x)=\dfrac{1}{x}\displaystyle\int_{0}^{x}f(y)\,\mathrm dy, say if it is well defined and discuss its ... 1answer 34 views ### Show that A is a bounded operator in ℓ2 and ∥A∥2≤∑∞j=1∑∞i=1|aij|2 I want to prove this: Let (a_{ij})_{i,j=1}^{\infty} be an infinite matrix such that \sum_{j=1}^{\infty }\sum_{i=1}^{\infty} \left | a_{ij} \right |<\infty  and A: \ell^2 \rightarrow \ell^2 ... 0answers 28 views ### Reference request for unbounded linear operators The following question is not as well formulated as I would like, but here goes. How should one think about unbounded linear operators on a Hilbert space H? Even though I have read a little on the ... 1answer 45 views ### Is this counterexample for T closed, symmetric \iff T self-adjoint valid? Lemma Let T: \text{dom}(T) \to \mathscr{H}, where \mathscr{H} is a Hilbert space, be densely defined. Then T is closed and symmetric if and only if T = T^{**} \subset T^*. To better get a ... 1answer 42 views ### Are all bounded linear operators including the ones that are Banach, also isometries? Are all bounded linear operators including Banach bounded linear operators, also isometries? An isometry is a homeomorphism that preserves distance, i.e. only reorders the points. May an unbounded ... 1answer 16 views ### Proving Boundness of Two Linear Operators I have that K:C[0,1] \rightarrow C[0,1]  and K_N:C[0,1] \rightarrow C[0,1] where:$$K \phi (x) = \int_0^1 k(x,t) \phi (t) dt K_N \phi (x) = \int_0^1 k_N(x,t) \phi (t) dt $$Where k(x,t):= ... 0answers 40 views ### Unbounded operators with dense domain in Banach space. Let X and Y two Banach spaces, and let T: X \rightarrow Y be a closed operator. we known that if T with dense domain in hilbert space X, then the domain of T^{*} is also dense in hilbert ... 0answers 33 views ### Description of the adjoint of this closed operator Consider the Hilbert space L^2(\mathbb R) and the multiplication operator A \colon D(A) \to L^2(\mathbb R) defined by$$(Af)(x) := (1+x^2)f(x)$$with domain$$D(A) := \left \{f \in L^2(\mathbb R): ...
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A is essentially self-adjoint if and only if the equation $A^ * x = -x$ has no non-trivial solutions. Where in addition, A is a densely defined, symmetric and positive operator. I considered a ...
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### Closure of a Hilbert subset with respect to a norm not defined on the entire Hilbert space

I have some problems to understand the definition of the closure of an operator. Let's $H$ be a Hilbert space with the norm $||\cdot||_H$, and let $A$ be a linear and symmetric operator defined on ...
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### Contradiction with Uniform Boundedness Principle using Hamel basis

Let $\{e_{\alpha}\}_{\alpha \in \mathcal{A}}$ be a Hamel basis of an infinite dimensional Banach space $(X, \| \cdot \|_X)$. Without loss of generality, we assume $\|e_{\alpha}\|_X = 1$. It is known ...
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### Definition of Inverse of Unbounded Operator

I am currently studying unbounded operators. However, in the text I am using, the definition for the inverse of an unbounded operator is given. I also went to several other books I own and the ...
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### Proof that any commutative family of unbounded self-adjoint operators has a “generating” operator

In Quantum Mechanics for Mathematicians by Leon A. Takhtajan, the author remarks on page 73: According to von Neumann's theorem on a generating operator, for every commutative family $\mathbf{A}$ ...
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### Integral with respect to spectral measure

Let $A:D(A)\subset H \to H$ be a self-adjoint unbounded operator on complex Hilbertspace $H$ with corresponding spectral measure $E:\mathcal{B}(\mathbb{R})\to\mathcal{L}(H)$. I want to show that an ...
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### Bounded superposition operator in the space of bounded variation functions

For $n\in\mathbb N$ put $A_n:=\{k/2^n\ |\ k\in\{1,\ldots,2^n-1\}\}$. Consider $g:[0,1]\times\mathbb R\to\mathbb R$, defined by \begin{align*} g(t,u)= \begin{cases} 2^{-n}\chi_{A_n}(t) & ...
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### Is it a mistake if I delete the condition in Banach - Steinhaus Theorem?

I'm studying the Banach-Steinhaus theorem recently. There is a question to me, I'm sure I have made a mistake, but I can't find out where the mistake is. The Banach - Steinhaus theorem says: Let X be ...
### If $T$ is self-adjoint then $T^2$ is also self-adjoint,
Let $T:D(T)\subset H \to H$ be selfadjoint unbounded linear operator on a complex Hilbertspace $H$. Show that $T^2$ is self-adjoint. Since $T$ is selfadjoint it's spectrum must be real, so $T\pm i Id$...