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Questions tagged [hilbert-spaces]

For questions involving Hilbert spaces, that is, complete normed spaces whose norm comes from an inner product.

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In an finite-dimensional Hilbert space, the sum of the diags give the identity only if it is an orthonormal basis [closed]

Let $\mathcal H$ be a Hilbert space of finite dimension $n$ and $e_1,\dots,e_n$ be vectors in $\mathcal H$. Show that $$(e_1,\dots,e_n)\text{ is an ONB in }\mathcal H\iff\sum_{i=1}^n|e_i\rangle\langle ...
Onur Aktan's user avatar
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24 views

Prove that: $x = \sum_{i, j\in I} x_{i, j} \otimes e_{i,j}$ [duplicate]

Let $\mathcal{H}, \mathcal{K}$ are Hilbert spaces and $(\epsilon_i)_{i\in I}$ be an ONB for $\mathcal{K}$. Suppose for each $i\in I, U_i:\mathcal{H}\to \mathcal{H} \otimes\mathcal{K}$ is the isometry ...
sigma's user avatar
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Norm of fractional Sobolev spaces

Let $\Omega$ a bounded open subset of $\mathbb{R}^N$ with a decomposition $\bar{\Omega} = \bar{\Omega}_1 \cup \bar{\Omega}_2$ (with $\Omega_1\cap\Omega_2=\emptyset$), and let $s\in(0,1)$. Consider the ...
Mathslover's user avatar
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33 views

How to show that the functor defined by, $ F ( H ) = H \otimes_{\mathbb{R}} \mathbb{C} $ is an equivalence of categories?

Let $ \mathcal{C}_{ \mathbb{R} } $ the category of real Hilbert spaces of infinite dimension, and morphisms between real Hilbert spaces of infinite dimension. Let $ \mathcal{C}_{ \mathbb{C} } $ the ...
YoYo12's user avatar
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1 answer
72 views

If an operator is defined over a Schauder basis then it's well defined

We know that in the finite dimensional case, given a linear operator $T:V\to W$ such that $V$ is finite dimensional, then defining $T$ over the basis of $V$ determines $T$ completely so $T$ is well ...
Math Admiral's user avatar
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1 answer
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Criterion for self-adjointness of non-negative symmetric operators

In a paper by Strichartz, the author claims the following: A densely-defined, closed, symmetric, non-negative operator $A$ (i.e. $\langle Ax,y\rangle\geq 0$ for all $x\in\mathcal{D}(A)$) is self-...
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strongly continuous one-parameter unitary group

Let $H = L^2(\mathbb{R}^2)$ and $C_0^\infty(\mathbb{R}^2) \subseteq H$. Show that $U_1(t) $ and $ U_2(t)$ are strongly continuous unitary groups. Here , i have shown the isometry , Do i have to show ...
User1's user avatar
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3 votes
1 answer
137 views

Sum of self-adjoint operators

Let $\mathcal{H}$ be a Hilbert space. I am very well aware of the fact that the sum of two (possibly unbounded) self-adjoint operators is in general not self-adjoint, in fact not even densely-defined ...
B.Hueber's user avatar
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Estimating Bessel Inequality for a Sequence of Functions Derived from a Frame in $L^2(\mathbb{R})$

Let $ \{f_k(x)\} $ be a frame for $ L^2(\mathbb{R}) $. For those unfamiliar, a frame in a Hilbert space $ \mathcal{H} $ is a family of functions $ \{f_k\} \subset \mathcal{H} $ for which there exist ...
Mark's user avatar
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1 vote
1 answer
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Proof of Zarantonello's Inequality

This question is from Brezis's Functional Analysis, exercise 5.23. Let $H$ be a Hilbert space, and let $T: H \rightarrow H$ be a (nonlinear) contraction. Assume that $\alpha_1, \alpha_2, \ldots, \...
camurphy's user avatar
1 vote
1 answer
19 views

Do positive functionals on a self-dual cone extend to the whole space?

Let $\mathcal H$ be a Hilbert space with self-dual cone $\mathcal H^+ = \{\,\xi\in\mathcal H \mid \langle\xi|\mathcal H^+\rangle \ge 0\,\}$. Suppose we have a function $f\colon \mathcal H^+ \to [0,+\...
Olius's user avatar
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Schauder bases and embeddings in Hilbert spaces

I would like to know whether continously and densely embedded orthonormal bases are always Schauder bases of the larger space. More precisely: Let $(H, \langle \cdot,\cdot\rangle_H)$ and $(\tilde{H}, \...
Dasi's user avatar
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1 answer
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I don't know how take weak derivative of norms in hilbert spaces

Let the following be hilbert spaces with dens inclusions $V ↪H=H^* ↪V^*$. Where $H^*$ and $V^*$ are the duals. $H$ has the product $(*,*)$ and $V×V^*$ has the product $⟨*,*⟩$. Let $u∈L^2 ([0,T];V); ...
Alucard-o Ming's user avatar
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1 answer
31 views

Topology for the resolution of identity equality

Let $H$ be a Hilbert space with orthonormal basis $\{x_{n}\}_{n\in \mathbb{N}}$. Given $x \in H$, one can write: $$x = \sum_{n\in \mathbb{N}}\langle x_{n},x\rangle x_{n}. \tag{1}\label{1}$$ Roughly ...
InMathweTrust's user avatar
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1 answer
37 views

If $C $ is a closed and convex subset of a Hilbert space $H$ then $y= P_Cx$ is the unique point characterized by

Just to understand answer OP gave, can someone please provide a link (or even a book) to the theorem (and of course it's prove) OP have pointed out in the answer of this link: Show that $C$ is a ...
user avatar
4 votes
1 answer
79 views

$\|T^*T\|=\|T\| \cdot \|T^*\|$ [duplicate]

Suppose we have a linear operator T between hilbert spaces and its adjoint T*. We know that $\|T^*T\| \leq \|T\|\cdot \|T^*\|$. Under which hypotheses can we conclude $\|T^*T\|=\|T\| \cdot \|T^*\|$? I ...
EdoRoundTheWorld's user avatar
2 votes
1 answer
48 views

Representation and the corresponding spectral measure

I'm reading a chapter about the relation between Hilbert space and quantum mechanics and got stuck at unclear correspondence. Let $\mathcal H$ be a Hilbert space, $V$ be $\mathbb{R^{1,d-1}}$ and $U$ ...
particle-not good at english's user avatar
8 votes
3 answers
390 views

Does strong convergence of operators imply pointwise convergence of their unbounded inverses?

Let $\mathcal{H}$ be a separable Hilbert space and let $A_n:\mathcal{H}\rightarrow \mathcal{H}$ be a sequence of bounded linear operators converging strongly to the identity operator, that is, one ...
S.Z.'s user avatar
  • 515
5 votes
1 answer
166 views

Understanding a proof that a bounded sequence in a separable Hilbert space contains a weakly convergent subsequence

I am trying to understand a proof that a bounded sequence in a separable Hilbert space $\mathcal H$ contains a weakly convergent subsequence, as in Theorem 5.12 of Gilbarg & Trudinger's "...
rosecabbage's user avatar
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5 votes
1 answer
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Are rank-1 operators $e_i e_j^*:\mathcal H\to\mathcal H$ dense, for a separable Hilbert space $\mathcal H$?

Let $\mathcal H$ be a separable Hilbert space with orthonormal basis $\{e_i\}_{i\in \mathbb{N} }$, and consider the linear operators $E_{ij}\equiv e_i e_j^*$, defined as $$E_{ij}v=e_i \langle e_j,v\...
glS's user avatar
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1 vote
1 answer
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Every closed convex subset of a finite-dimensional subspace of real inner product space is Chebyshev.

In my post I was now able to prove the following theorem : If $K$ is a closed convex subset of an real inner product space $X$ that is contained in a complete subset of $X$, then $K$ is Chebyshev.. ...
user avatar
0 votes
1 answer
22 views

Spectrum of Compact Hermitian Operator Accumulating at 0 From Both Directions

I'm getting into spectral theory and I'm wondering if someone could give an example of a Hilbert space $H$ and a compact hermitian operator $T$ on $H$ whose spectrum $\sigma(T)$ has both a positive ...
Miles Gould's user avatar
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1 answer
36 views

Separable Hilbert spaces, total orthonormal sets and Schauder basis in Banach spaces

A Schauder basis for a Banach space $X$ is called the sequece $\{e_n\}$ where $x\in X$ can be expanded as: $$x=\sum^\infty_{k=1} a_ke_k$$ A total orthonormal basis in a Hilbert space $H$ is a Schauder ...
Krum Kutsarov's user avatar
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0 answers
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Differentiability (and other properties) of functions mapping into countable product of Hilbert spaces

I have Hilbert spaces $\{H_i\}$ indexed by the natural numbers, so a countable set. Let me define a function $$f\colon X \to H_1\times H_2 \times H_3 \times ...$$ by $$f(x) = (f_1(x), f_2(x), ...)$$ ...
BBB's user avatar
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1 vote
1 answer
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I would like to prove that the corollary also implies the theorem

Theorem: Every closed convex set in a Hilbert space is approximatively Chebyshev compact. Proof: Let $K$ be a closed convex set in a Hilbert space $H$. Choose an arbitrary Cauchy sequence $(y_n)$ in $...
user avatar
6 votes
2 answers
162 views

Hilbert space question: if $M=\{f\in L^2(0,+\infty)\ |\ \int_0^\infty f^2(x)e^xdx<+\infty\}$, what is $M^\perp$?

This was from a Real Analysis exam: In the Hilbert space $H=L^2(0,+\infty)$. Consider $$M=\left\{f\in L^2(0,+\infty)\ \bigg|\ \int_0^\infty f^2(x)e^xdx<+\infty\right\}$$ Find $M^\perp$. Since the ...
Zima's user avatar
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0 votes
1 answer
42 views

Is having the closure of the infimum limit of a sequence of sets to be equal to the supremum limit enough to deduce convergence?

Let $H$ be an infinite-dimensional Hilbert space, and $(H_n)$ a sequence of finite-dimensional subspaces. I would like to show that $(H_n)$ converges. For this, I managed to show that $A \subseteq \...
Spida's user avatar
  • 373
0 votes
1 answer
50 views

Does the spectrum of $T(1+T^*T)^{-1}$ contain a compactification of that of $T$?

Let $T$ be a densely defined closed operator in the Hilbert space $\mathcal H$. Then $T^*T\ge0$, and $Q = T(1+T^*T)^{-1}$ has norm $\le1$ (Theorem 13.13 of Rudin's Functional Analysis). Define the ...
Olius's user avatar
  • 736
1 vote
0 answers
89 views

There exists a set of complex numbers $\alpha_n:n\in\mathbb{Z^*}$ such that $x=\sum_{n\in\mathbb{Z^*}}\alpha_ne^{nx}$?

Note this is an infinite series! I have the same question for $1$, is it possible to have $1=\sum_{n\in\mathbb{Z^*}}\beta_ne^{nx}$ for some complex coeficients $\beta_n\in\mathbb{C}$? I am considering ...
Armand's user avatar
  • 141
0 votes
0 answers
51 views

Limits of sequences of nonlinear "bounded" operators

I have a sequence of nonlinear operators $T_n\colon X \to Y$ between two separable Hilbert spaces. The operators are bounded uniformly in that there is a constant $C$ with $$|T_n(x)| \leq C|x|.$$ I ...
C_Al's user avatar
  • 672
3 votes
1 answer
73 views

Lifting matrix units from Calkin algebra

My question is about Examples 15.4.2(b) of Blackadar “K-Theory for Operator Algebras”. Let $H$ be a Hilbert space and $Q(H) = B(H)/K(H)$ be the Calkin algebra. Given a *-homomorphism $\tau: M_n(\...
Query600's user avatar
7 votes
1 answer
168 views

Is the weak topology on the infinite-dimensional separable Hilbert space $T_5$?

I was looking around in pi-base and apparently they don't know whether the weak topology on the infinite-dimensional separable Hilbert space is Completely Normal or not (although they know it is $T_4$)...
Carla_'s user avatar
  • 1,063
1 vote
1 answer
49 views

What is the relation between $A$ and $e^A$ for unbounded self-adjoint $A$?

Is there a generally accepted definition of the exponential of certain unbounded operators on a Hilbert space, which extends that for self-adjoint operators given by the functional calculus? I would ...
Olius's user avatar
  • 736
0 votes
0 answers
55 views

basis of direct sum Hilbert space

Let $H_i$ be a collection of Hilbert spaces indexed by $I$, define $$H = \bigoplus_{i\in I} H_i := \{x\colon I\to \bigcup_{i\in I} H_i: x(i)\in H_i\text{ and }\sum_{i\in I} \|x(i)\|^2<\infty\}.$$ ...
Zoro's user avatar
  • 33
0 votes
0 answers
40 views

Self-adjointness for real Hilbert space

For a complex Hilbert space $H$, a bounded operator $A$ is self-adjoint iff $\langle Ah,h\rangle\in \mathbb{R}$ for all $h\in H$. But is there a simple analogous result to determine when $A$ is self-...
user760's user avatar
  • 1,732
3 votes
0 answers
70 views

Continuity of optimal value of a functional on a Hilbert space

Let $f:X\times Y\to \mathbb{R}$ be a bounded continuous function on the product topological space $X\times Y$. It is well known that if $Y$ is compact, then the infimum function $g(x) = \inf_{y\in Y}f(...
zt wang's user avatar
  • 399
5 votes
3 answers
212 views

Limits of Compact Operators: a simple question about the diagonal trick.

Theorem. Let $H$ and $K$ be Hilbert spaces, and assume that $T_n\colon H \to K$ is compact for each $n \in \mathbb{N}$. If $T \in B(H,K)$ is such that $\lVert T − T_n\rVert \to 0$, then $T$ is compact....
Jack J.'s user avatar
  • 1,070
0 votes
0 answers
42 views

Prove that a space is a Hilbert Space with inner product different to the usual inner product used

I'm working with following academic paper : Stability of the solutions of differential equations whose author is Bernard Beauzamy. I'm trying prove that the two next spaces $\mathcal{B}_2$ y $P_2$ are ...
Richard's user avatar
  • 123
0 votes
1 answer
75 views

"Every linear subspace of a closed space is closed"

I am wondering how to show something that my book writes about $L^2$ and Hilbert spaces. First I will talk about what my book does: It defines what a random variable in $L^2$ is. It shows that $L^2$ ...
user394334's user avatar
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0 votes
1 answer
23 views

For quadratic functions on arbitrary Hilbert spaces, does strict convexity still imply strongly convexity?

Let $\mathcal{H}$ be a separable real Hilbert space. Let $q:\mathcal{H} \to \mathbb{R}$ be a bounded quadratic function. By this, I mean $q(x)=(Qx,x)+(b,x)+c$ for some self-adjoint bounded linear ...
Spencer Kraisler's user avatar
3 votes
1 answer
78 views

On the preimages under injective operators of a sequence of subspaces of a Hilbert space tending to 0

Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two separable Hilbert spaces and let $A_n:\mathcal{H}_1 \rightarrow \mathcal{H}_2$ be a sequence of uniformly bounded injective linear operators with a ...
S.Z.'s user avatar
  • 515
1 vote
1 answer
42 views

Why weak convergence doesn't Imply strong convergence on $\infty$- dimensional Hilbert spaces.

To start off, I know this is wrong. Im hoping someone can explain to me where Im going wrong. I know that on a finite dimensional Hilbert space that weak convergence implies strong convergence. But I ...
Alexander Sampson's user avatar
0 votes
0 answers
18 views

On strong convexity of the squared norm of an affine function.

Let $V,W$ be Hilbert spaces. Let $f:V \to W$ be a bounded affine function. That is, $f(x)=A(x)+b$ for some bounded linear function $A:V \to W$ and fixed $b \in W$. Since $V$ is a Hilbert space, the ...
Spencer Kraisler's user avatar
3 votes
2 answers
216 views

Operator norm of sum of tensor products

Let $A,B,C,D$ be bounded operators on a Hilbert space $\mathcal{H}$. I know that $$ \|AB\| \leq \|A\otimes B\| $$ where $\otimes$ is the tensor product and $\|\cdot\|$ is the operator norm. I wonder ...
felipeh's user avatar
  • 3,874
2 votes
2 answers
90 views

A is self-adjoint and $\|Ax\|=\|A\|$. Show x is an eigenvector for $A^2$.

Let $H$ be a Hilbert space and let $A$ be a bounded self-adjoint operator on $H$. Suppose that there exists a unit vector $x \in H$ such that $||Ax||=||A||$. i) Show that $x$ is an eigenvector of $A^2$...
Aester's user avatar
  • 107
1 vote
1 answer
48 views

Existence of Orthonormal Basis of Eigenvectors of Closed Operator Implies Diagonalizability?

Proposition: Let $H$ be a Hilbert space and $A$ a closed (possibly unbounded) linear operator on $H$. Let $(\lambda_n)_{n \in \mathbb{N}}$ be a sequence of complex numbers and $(\varphi_n)_{n \in \...
jd27's user avatar
  • 2,630
0 votes
0 answers
38 views

The map $A\colon H^* \to H$ defined as $Af=z$ is bijective and isometric

Let $H$ be a Hilbert space. Let $f\in H^*$ then for the Riesz representation theorem exists a unique $z\in H$ such that $f(x)=\langle x, z \rangle$ for all $x\in H$. This defines an operator $$A\colon ...
Jack J.'s user avatar
  • 1,070
0 votes
0 answers
25 views

Why does $D(A)^\perp\subseteq\{0\}$ imply $D(A)$ dense in $H$? [duplicate]

The problem comes from Functional Analysis Sobolev Spaces and PDE (H.Brezis) page 181, Chapter 7 (The Hille-Yosida. The) proposition 7.1, where A is the maximal monotone operator. It needs to prove ...
Furina de Fontain's user avatar
0 votes
3 answers
114 views

Example of a Banach space isomorphic to a Hilbert space in the category of Banach space

I'm new to the field of functional analysis and have been trying to understand the properties of Banach spaces that really distinguish them from a Hilbert space. In particular, I have been trying to ...
supernova's user avatar
  • 127
0 votes
0 answers
35 views

Is a matrix that is diagonalized by two different orthonormal bases determined by its spectrum in these common subspaces?

I would appreciate it if you could help me with a question. Suppose a $n \times n$ matrix $A$ is diagonalized by different orthonormal bases $\{|e_i\rangle\}_{i=1}^n$ and $\{|f_j\rangle\}_{j=1}^n$. ...
R. J. Ernest's user avatar

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