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bio website math.uni-trier.de/~wengenroth
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age 46
visits member for 1 year, 10 months
seen Jul 7 at 12:53

Jul
4
comment Find vector space $X$; so that vector space operations are not continuous
q egreg I noticed that. I didn't claim to answer the question.
Jul
4
comment Find vector space $X$; so that vector space operations are not continuous
@egreg Sure, addition $X \times X\to X$ is then continuous but multiplication $\mathbb C \times X \to X$ fails to be continuous.
Jul
2
awarded  Curious
Jun
24
answered Space of Distribution wrt to topology of uniformly convergence on bounded sets not Frechet-Space.
Jun
15
comment Characters only on commutative unital algebras?
This makes perfectly sense.
Jun
14
comment Find vector space $X$; so that vector space operations are not continuous
Try the discrete metric $d(x,y)=1$ for all $x\neq y$.
Jun
14
comment Check continuity of functional in weak topology
The first seminorm $\|f\|$ (which is the $l^2$-norm) dominates the others (substitute $t=ns$). Continuity would thus mean that $|f(0)|\le C \|f\|$ for some constant $C$ holds for all $f\in C[0,1]$. Now try functions with large $f(0)$ but very small integral. (This is apparently the same as Lukas' comment 15 seconds ago.)
Jun
3
comment Sequence of linear functionals
The closed graph also does the job: $\Phi$ has closed graph because $pi_n\circ \Phi =\phi_n$ are continuous (where $\pi_n: \ell^1\to \mathbb C$, $y\mapsto y_n$) and $y=0$ whenever all $\pi_n(y)=0$.
May
30
comment Bounded below operators
You are right, of course.
May
27
comment Do we have $C^\infty \cap \mathcal{O}_C' = \mathcal{S}$ and/or $C^\infty \cap \mathcal{S}' = \mathcal{O}_M$?
What about $f(x)=\exp(ie^x) \in C^\infty \cap L^\infty \subseteq C^\infty \cap \mathscr S'$? The derivative $f'(x)= ie^x \exp(ie^x)$ grows very fast.
May
9
comment Proving functions are uniformly continuous help!
$f'(x)= \frac 14 x^{-3/4}$ is not bounded. However, it is bounded on $[1,\infty)$ which gives uniform continuity there and $f$ is uniformly continuous on $[0,2]$ because continuous functions on compact sets are always uniformly continuous.
May
8
comment what are contraction(Lipschitz) maps on $\mathbb C$?
$\xi$ is somewhere on the segment joining $x$ and $y$. For the proof, take $z=f(x)-f(y)$ and apply the mean value theorem to $\varphi:[0,1]\to \mathbb R$, $t\mapsto \Re (\bar z f(x+t(y-x))$.
May
8
comment what are contraction(Lipschitz) maps on $\mathbb C$?
The mean value theorem fails in $\mathbb C$ but the mean value inequality $|f(x)-f(y)|\le |f'(\xi)| |x-y|$ is true. However, if $f$ is everywhere differentiable then so is $f'$ and if the derivative is bounded then it is constant by Liouville's theorem.
May
8
comment Lipschitz function and uniform boundedness principle
Implicitly, you are using Hahn-Banach which is a good idea. The problem with your argument is that the Lipschitz constant of $L \circ f$ may depend on $L$.
May
7
comment Separability of a certain space of continuous functions
If you want to have equality $C(I,V)=C(I)\tilde{\otimes}V$ for the completed tensor product you need the injective tensor product (I am not sure which assumptions you need for $I$, it is often stated for compact $I$). However, If you just want separability you don't need a topology: If the algebraic tensor product (that is the span of functions of the form $x\mapsto f(x) v$ for $f\in C(I)$ and $v\in V$) is dense it would be enough to take dense subsets of such $f$ and $v$.
May
6
comment An abstract integration problem from a mathematical finance calibration problem
As a toy model, think of $f\in \mathbb R^n$, a matrix $P$, and replace the integral by matrix multiplication. Even in this simplified case you get necessary conditions.
May
5
comment Gradient Estimate - Question about Inequality vs. Equality sign in one part
Since you want to prove an inequality the $\le$ is sufficient. If you have equality depends on whether $A$ is actually equal to the other expression.
May
2
comment Tridual-“Reflexive”
It is classical and (using the characterization of reflexivity by weak compactness of the unit ball) not too difficult to show that $X^*$ refexive implies $X$ refexive.
May
2
comment Tridual-“Reflexive”
What do you mean by $=$? As hatsoff points out the usually written $X=X^{**}$ should be more precisely $JX=X^{**}$ where $J$ is the canonical embedding. However, I do not know of a "canonical" embedding $X\hookrightarrow X^{***}$.
May
2
comment isometric embedding of l^2
What I wrote in the answer is an (isometric) embedding because $\|\sum_n \alpha_n x_{i(n)}\|_H^2 = \sum_n|\alpha_n|^2 = \|(\alpha_n)_n\|_{\ell^2}$.