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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
8
votes
Applications of functional analysis beyond analysis(towards algebra, geometry, number theory...
Besides Hodge and index theories, mentioned in Qiaochu Yuan's comment above as applications of functional analysis to (complex) algebraic geometry and algebraic topology respectively, I believe that a …
2
votes
Accepted
Fourier transform of function on compact set and Sobolev norm equivalence
This seems to be one special case of the so called Fourier restriction problem (see, for instance, Terry Tao's discussion: http://terrytao.wordpress.com/2010/12/28/the-bourgain-guth-argument-for-provi …
1
vote
1
answer
316
views
Uniformly continuous functions and Borel hierarchy in the compact-open topology
Let $\Omega\subset\mathbb{R}^n$ be open, $\mathscr{C}(\Omega,\mathbb{R})$ the Fréchet space of real-valued continuous functions on $\Omega$ endowed with the compact-open topology, and $\mathscr{C}_u(\ …
4
votes
What properties should $C(M,\mathbb{R})$ have when $M$ is a $n$-dimensional manifold?
This is not really an answer but rather a long-ish comment.
First of all, if $M$ is not compact, $\mathfrak{A}=C(M,\mathbb{R})$ is not really a C${}^*\!$-algebra but actually only a locally C${}^*\!$- …
7
votes
Accepted
For a tempered distribution $F$ on $\mathbb{R}^2$, what exactly does it mean by $\lvert F(x,...
I'll try to explain what Igor meant in his comments in a different way, maybe it helps.
Of course, any tempered distribution is a distribution in the broader sense - more precisely, any compactly sup …
10
votes
Accepted
Is the space of tempered distribution second countable?
Since $\mathscr{S}'$ is the dual of an infinite-dimensional Fréchet space, the weak-* topology in $\mathscr{S}'$ is not even first countable. What people usually do is to define probability measures i …
14
votes
Accepted
Abstract result on partitions of unity?
I will leave to Yemon Choi discussing the answer from Gelfand-Raikov-Shilov's book (Commutative Normed Rings, I suppose?), and restrict myself to more recent discussions on the matter...
There is an …
8
votes
2
answers
357
views
Can smoothness of curves into a convenient locally convex vector space be tested with just a...
Let $E$ be a (Hausdorff) locally convex vector space (from now on just "lcs" for short). We say that $E$ is convenient (also called locally complete, Mackey-complete or $c^\infty$-complete) if, given …
3
votes
Fourier Transforms restricted to mass shell
I doubt it. My reason is related to your comment: take your favorite function $f$ on $\mathscr{D}$, multiply it by a Gaussian with covariance $\sigma$ and centered around a point $x$ in the support of …
4
votes
Quantum fields and infinite tensor products
The "infinite tensor product" picture may be useful as a sort of concrete image of the state space of a quantum field theory, but in practice is rarely used because of the technical difficulties it br …
11
votes
4
answers
665
views
Is every non-negative test function the limit of a sequence of sums of squares of test funct...
Let $0\leq f\in\mathscr{D}(\mathbb{R}^n)$. As shown e.g. by J.-M. Bony, F. Broglia, F. Colombini and L. Pernazza, Nonnegative functions as squares or sums of squares, J. Funct. Anal. 232 (2006) 137-14 …
2
votes
1
answer
598
views
Stein's extension operator and wave front sets
Let $K\subset\mathbb{R}^d$ be a compact set with non-empty interior and Lipschitz boundary. In Section VI.3 of his book "Singular Integrals and Differentiability Properties of Functions", E. M. Stein …
5
votes
2
answers
667
views
Is every Montel locally convex vector space compactly generated?
Let $X$ be a Hausdorff locally convex vector space. Recall (my reference is the book of H. Jarchow, Locally Convex Spaces. B.G. Teubner, 1981) that we say that $X$ is a semi-Montel space if every boun …
8
votes
Separate continuity implies (joint) continuity
This is easy and comes from the fact that $V$, being a Fréchet space, is barrelled, that is, the locally convex topology of $V$ coincides with its strong topology $\beta(V,V')$ (where $V'$ = topologic …
2
votes
"Completeness" for weak convergence of unbounded closed operators on a separable Hilbert spa...
Not always. What you are talking about is also called "convergence in the sense of sesquilinear forms", because you are taking a pointwise limit in $D\times D$ of a sequence of sesquilinear forms $\al …