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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.

2 votes
1 answer
180 views

A question regarding $L^1(R)\cap L^2(R)$ [closed]

Does $L^1(R)\cap L^2(R)$ have finite or infinite corank in $L^2(R)$? I guess the latter is the case but I have never seen this discussed, and would like to see a simple proof.
Arnold Neumaier's user avatar
0 votes
1 answer
318 views

On the Riesz representation theorem II

I have a follow-up question to On the Riesz representation theorem . Let $V$ be a subspace of a Hilbert space, and let $V^\times$ be the space of all antilinear functionals on $V$, equipped with the w …
Arnold Neumaier's user avatar
11 votes
3 answers
647 views

norm inequalities

Let $p>2$. I'd like to know the best possible lower and upper bound for $\|x\|_p$ given that $x\in R^n$ and $\|x\|_1$, $\|x\|_2$, and $\|x\|_\infty$ have fixed values. It is well-known that $$\|x\|_ …
Arnold Neumaier's user avatar
2 votes
1 answer
422 views

Which topology for $C^\infty(X)$ works?

Let $X$ be a smooth manifold. What is the appropriate topology on $C^\infty(X)$ such that a linear functional $\lambda$ on $C^\infty(X)$ is continuous iff it can be represented as a limit of the form …
Arnold Neumaier's user avatar
0 votes
0 answers
129 views

homomorphisms into tensor product algebras

Given a decomposition $H=H_1\otimes H_2$ of a Hilbert space $H$ into the tensor product of the Hilbert spaces $H_1$ and $H_2$ and a *-isomorphism $U: B(H_0)\to B(H)$, where $H_0$ is another Hilbert sp …
Arnold Neumaier's user avatar
5 votes
1 answer
609 views

On the Riesz representation theorem

Let $V$ be a vector space with inner product $(\phi,\psi)$ antilinear in the second argument - not necessarily a Hilbert space. Let $\Phi$ be an antilinear functional on $V$. What are the precise (n …
Arnold Neumaier's user avatar
2 votes
0 answers
84 views

Implicit function theorem and selfadjoint operators

Is there a version of the implicit function theorem in some space of functions from $[0,1]$ to a Hilbert space $H$ that contains as a special case the unique solvability of the initial-value problem $ …
Arnold Neumaier's user avatar
2 votes
0 answers
147 views

On limits of positive linear functionals

I am looking for pointers to the literature on questions of the following kind. ($Y$ and $\Omega$ might be open subsets of some Euclidean space, but I am interested in the kind of conditions that need …
Arnold Neumaier's user avatar
27 votes
1 answer
1k views

Do Sobolev spaces contain nowhere differentiable functions?

Does the Sobolev space $H^1(R^n)$ of weakly differentiable functions on a bounded domain in $R^n$ (or a more general Sobolev space) contain a continuous but nowhere differentiable function?
Arnold Neumaier's user avatar
1 vote
1 answer
132 views

Embedding Hermitian forms into Hilbert spaces

Let $H$ be a nondegenerate, not positive definite, Hermitian form on a complex vector space $V$ such that $$|H(x,y)|\le u(x)u(y)\tag{B}$$ for some map $u:V\to R_+$ with $u(\lambda x)=|\lambda|u(x)$ fo …
Arnold Neumaier's user avatar
2 votes

Embedding Hermitian forms into Hilbert spaces

Since I posed the question I found that the book J. Bognár, Indefinite inner product spaces, Springer 1974 has a counterexample in infinite dimensions; see his Example 5.6, p.90.
Arnold Neumaier's user avatar
3 votes
0 answers
141 views

Automorphism group of Lie algebra of bounded operators

What is the automorphism group of the complex Lie algebra of bounded operators on a complex Hilbert space, with the commutator as Lie bracket? What for the real Lie algebra of bounded antihermitian op …
Arnold Neumaier's user avatar
4 votes
1 answer
380 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?
Arnold Neumaier's user avatar
4 votes
0 answers
2k views

Eigenvalues and spectrum of the adjoint

In a finite-dimensional Hilbert space, the eigenvalues of the adjoint $A^*$ of an operator $A$ are the complex conjugates of the eigenvalues of $A$. But in infinite dimensions this need no longer be …
Arnold Neumaier's user avatar
2 votes

About the quantum spectrum of a certain potential.

This question (and a partial converse with counterexamples) is answered in B. Simon, Some quantum operators with discrete spectrum but classically continuous spectrum, Annals of physics 46 (1983), 2 …
Arnold Neumaier's user avatar