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### Question regarding conjugate operators and the harmonic operator.

Let us consider the operator $\hat{n}=\hat{a}^\dagger\hat{a}$ as the number operator of the harmonic oscillator. Let $|n\rangle$ be the eigenstates. Then we can say : $$\hat{n}|n\rangle=n|n\rangle$$ I'...
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### Spectrum analysis

I have a problem: Given the Hamiltonian $H_n= -\frac{d|}{d^2}+V_n$ where $V_n= a n e^{-n|x|}; a\in\Bbb R, n=1,2,....$ 1)Domain in which the operators are self adjoint; 2)study and list the ...
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### Column space of $A^*$ is row space of $A$?

I have seen that range $A^*$ = $(\text{null }A)^\perp$, for example here. And I have seen that $(\text{null A})^\perp$ = row $A$, for example here. Does it therefore follow that range $A^*$ = row $A$? ...
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### Is it possible to obtain an expression like $\langle T^*Tx,x\rangle_H\ge c\|x\|_H^2$ for $c>0$?

Let $H$ be a Hilbert space and let $S\in B(H)$ be a selfadjoint linear operator. We say that $S$ is positive if for all $x\in H$, $$\langle Sx,x\rangle_H\ge0.$$ If $V\in B(H)$ a not necessarily ...
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Suppose the vector space is a complex finite-dimensional inner product space $V$, and the operator $TT^*-T^*T$ is a positive operator on it. Can this information be sufficient to conclude that $TT^*-T^... • 756 0 votes 1 answer 44 views ### Fourier Transform definitions for operator I am following the convention of FT for operators, normalized to a length scale$L, whereby $$a_{k}=\frac{1}{\sqrt{L}}\int dx\,e^{-ikx}a(x)\,\,\,\,\,\,\,\,\,\,\,(1) \\ a^{\dagger}_{k}=\frac{1}{\sqrt{... • 145 4 votes 0 answers 83 views ### Adjoint of unbounded integral operator Consider a Borel-measurable function a:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{C}, and let T_a be the linear operator on L^2(\mathbb{R}) with domain \mathcal{D}(T_a)=\left\{... • 277 1 vote 1 answer 65 views ### About the proof of invertible quadratic expressions in Linear Algebra Done Right The theorem is that if T\in L(V) (and V is a real finite-dimensional vector space) is self-adjoint and b,c \in \mathbb R such that b^2<4c. Then the operator$$ T^2 + bT +cI$$is invertible.... • 756 1 vote 0 answers 26 views ### Geometric interpretation of adjoint Given the inner product \langle x, y \rangle = y^HMx, \, x,y\in\mathbb{C}^n where y^H = \bar{y}^T is the conjugate transpose and M^H=M positive definite, I find the adjoint A^* of A\in\mathbb{... • 1,748 0 votes 0 answers 29 views ### Construct the Green's function for the following boundary value problem and use it to find the solution y''-y'=t^2 Construct the Green's function for the following boundary value problem and use it to find the solution$$y''-y'=t^2,\quad y(0)=0,\: y(1)=0$$I know that,$$ \begin{aligned} &\langle L u, G\... 0 votes 1 answer 52 views ### Prove that in a complex Hilbert space\|Tx\|\le\|T\|^\frac{1}{2}\langle Tx,x\rangle^\frac{1}{2}$Let$H$be complex Hilbert space. Prove that$\|Tx\|\le\|T\|^\frac{1}{2}\langle Tx,x\rangle^\frac{1}{2}$. I am trying to prove this result by using the following result. Let$T:H\to H$be a bounded ... • 2,279 1 vote 1 answer 56 views ### Proving 2 properties of the adjoint map$T^*$on a Hilbert space Let$(\mathcal{H}, \langle \cdot, \cdot \rangle)$denote a seperable Hilbert space with orthonormal basis$(e_n)_{n \in \mathbb{N}}$with induced norm$\Vert \cdot\Vert$. Let$T:\mathcal{H} \...
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I know that a system with $\mathfrak{L}\equiv\mathfrak{L}$* and the boundary conditions also hold then the problem is known as fully self-adjoint, and if only the first statement holds it is formally ...
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### How to prove T is a closable operator

Let $T:H\to H$ a densely defined operator, with $H$ a Hilbert space such that: $$Re(x,Tx)\geq 0, \forall x\in Dom(T)$$ I want to prove that $T$ is a closable operator, that means... that there exists ...
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I'm trying to prove Theorem 2.21 in Brezis' book of Functional Analysis. The author leaves the proof as an exercise. Could you have a check on my attempt? Let $E, F$ be Banach spaces. Let $A: D(A) \... • 9,986 0 votes 0 answers 37 views ### Let$A^\star:\operatorname{dom}A^\star\subseteq F^\star\to E^\star$be the adjoint of$A$. Then$\operatorname{ker}A=(\operatorname{im}A^\star)^\perp$I'm trying to do exercise 2.18 in Brezis' book of Functional Analysis. Could you have a check on my attempt? Let$(E, |\cdot|_E), (F, |\cdot|_F)$be Banach spaces and$A: \operatorname{dom} A \...
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Let $(E, | \cdot|_E)$ and $(F, | \cdot|_F)$ be Banach spaces. An unbounded linear operator $A$ from $E$ to $F$ is a linear map of the form $A: D(A) \to F$ where $D(A)$ is a subspace of $E$. Let $A$ be ...