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2d
revised If $T: X \to Y$ is norm-norm continuous then it is weak-weak continuous
fixed formulas
2d
answered If $T: X \to Y$ is norm-norm continuous then it is weak-weak continuous
Aug
18
answered Kernel of the Extension of a Bounded Linear Operator
Aug
15
awarded  functional-analysis
Aug
14
answered What does local space of a given Banach space says intuitively?
Aug
8
comment Completation of an n.v.s. and dimensions of subspaces.
Are you sure that $X\cap Y$ is different from $\lbrace 0 \rbrace$?
Aug
7
comment Prove this inclusion: $\bigcup_{k<p}\ell^k\subsetneq\ell^p$
look here math.stackexchange.com/questions/480807/…
Aug
6
comment Essential support vs. classical support for a continuous function
If a continuous function is a.e. zero on an open set it is in fact zero because open sets with measure zero are empty.
Aug
6
answered For holomorphic functions, if $\{f_n\}\to f$ uniformly on compact sets, then the same is true for the derivatives.
Jul
28
answered Inductive Limit of directed locally convex Frechet Spaces
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 The space of distributions endowed with the topology of uniform convergence on bounded sets is not Fréchet.
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.