Reputation
2,329
Next privilege 2,500 Rep.
Create tag synonyms
Badges
2 14
Impact
~14k people reached

  • 0 posts edited
  • 0 helpful flags
  • 52 votes cast
9h
comment Existence stochastic integral
The covariation is the unique continuous process of finite variation starting at $0$ such that $LN-[L,N]$ is a (local) martingale.
14h
answered Existence stochastic integral
16h
comment Hyperplane in a complex vector space
A linear functional on a complex vector space is uniquely determined by its real part (because $Im(x'(x))=Re(-i x'(x)) = Re( x'(-ix))$).
Jul
28
answered Topology on compactly supported smooth functions
Jul
28
comment Topology on compactly supported smooth functions
If you declare an absolutely convex set $V \subseteq C_c^\infty(U)$ a $0$-neighbourhood if, for all $K$ there are $n\in\mathbb N$ and $\varepsilon$ such that $V\cap C^\infty(K) \supset \lbrace f\in C^\infty(K): |\partial^\alpha f(x) |\le \varepsilon$ for all $|\alpha|\le n\rbrace$ -- then the (locally convex) colimit topology is the limit topology.
Jul
27
comment $Hom(V,W)$ remains unchanged when norms of $V$ and $W$ are replaced with equivalent norms.
The notation $Hom(V,W)$ for the set of morphisms of the category of normed linear spaces and continuous linear mappings is not appropriate: This is NOT an abelian categroy. Here, homomorphisms are continuous linear maps $T:V\to W$ with closed range such that $T:X\to Range(T)$ is open. (This is, of course, not the fault of the OP.)
Jul
23
comment Fixed-point analysis similar to Banach Fixed Point Thm
Yes, as Robert Israel points out, it does.
Jul
23
answered Fixed-point analysis similar to Banach Fixed Point Thm
Jul
22
comment Finding a maximal complete subspace of Riemann Integrable functions on $[0,1]$
The sequence $(P_n)_{n\in\mathbb N}$ of spaces of polynomials of degree $\le n$ is an increasing sequence of closed subspace (e.g., because they are finite dimensional) whose union is not closed.
Jul
20
answered Norm of the integral operator in $L^2(\mathbb{R})$.
Jul
17
answered Need example of: Algebraic sum of closed vector subspaces need not be closed
Jul
13
comment Is $C(\Omega) \cong\prod_{n \in \mathbb{N}} C(K_n)$?
Every complete locally convex space is a closed subspace of a product of Banach spaces. For Frechet spaces like $C(\Omega)$ (where $\Omega$ is locally compact and $\sigma$-compact) you can have a countable product.
Jul
7
comment $T:X \to Y$ bounded linear map and $X$ separable implies $Y$ is separable?
Concerning reflexivity you can also argue with the weak* compactness of the closed unit balls. If $T:X\to Y$ is onto then the unit ball of $Y$ is contained in a multiple of $T(B_X)$ (by the open mapping theorem) which is weakly compact. Hence $B_Y$ is relatively weakly compact and weakly closed, hence weakly compact.
Jul
2
comment Is it possible to approximate $cos(x)$ with a linear combination of Gaussians $e^{-x^2}$?
Look at "convolution with Gaussian kernels". This, by the way, is the main trick how Weierstrass proved his famous approximation theorem.
Jun
29
comment Functional Analysis (Topological and Isometric Isomorphisms)
I think the OP meant linearly isomorphic. However, David's example works, of course. To see that $c$ and $c_0$ are not isometric one could observe that the unit ball of $c$ has extreme points but that of $c_0$ does not.
Jun
11
answered Matrix-like Representation of any linear map using Hamel Basis
Jun
5
awarded  Popular Question
Jun
5
comment No surjective bounded linear map from $\ell^2(\mathbf{N})$ to $\ell^1(\mathbf{N})$
A similar argument. $\ell^2$ is reflexive, quotients of reflexive spaces are reflexive and $\ell^1$ is not reflexive.
Jun
3
accepted Compact subsets of the plane with connected complement
Jun
3
comment Compact subsets of the plane with connected complement
@MikeMiller If I understand your comment well this answers the question. If you formulate it as an answer I would like to accept it.