All Questions
14 questions
0
votes
0
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49
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Existence of sequence of regular projections
Reading the book :Krasnosel'skii, M.A.; Pustylnik, E.I.; Sobolevskii, P.E.; Zabreiko, P.P. (1976), Integral Operators in Spaces of Summable Functions, Leyden: Noordhoff International Publishing, 520 p....
5
votes
0
answers
94
views
When a compact subset of a TVS can be continuously projected on a closed linear subspace?
Let $V$ be a (Hausdorff) topological vector space, $W\subset V$ a closed linear subspace, $X\subset V $ a compact.
(Q):
When there is a continuous map $P:X\to W$ such that $P(x)=x$ for every $x\in X\...
- 60.5k
11
votes
1
answer
309
views
Which closed subsets $Y$ of a compact space $X$ admit a linear extensor $C(Y)\to C(X)$?
In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$.
The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), so it admits a continuous right ...
- 60.5k
1
vote
0
answers
292
views
Closure of finite rank operators on $L^p$
It well-known that, an operator $T:H\to H$ on a Hilbert space, is compact if and only if T is limit of finite rank operators.
Besides this, the results by Per Enflo 1973 shows that this results is ...
- 1,101
1
vote
0
answers
539
views
Is the set of compact operators closed with the strong topology?
It is well-known that the space of compact operators over Banach spaces is closed within the norm topology.
My question:
Let $X$ be a Banach space.
Considering the strong topology (defined by ...
- 299
4
votes
0
answers
202
views
Double commutant of compact operators
So my question is straightforward. Let $\mathfrak{X}$ be a (complex, if necessary) Banach space and $K\colon\mathfrak{X}\to\mathfrak{X}$ a nonzero compact operator. Denote by $\mathcal{C}(K)$ the ...
- 1,453
2
votes
1
answer
135
views
A non-condensing operator with a power condensing
Let $\alpha$ to be the Kuratowski measure of non-compactness, in a Banach space $E$.
It's very easy to prove that $\alpha (D_1\times D_2)\leq \alpha (D_1)+\alpha (D_2)$, where $D_1$ and $D_2$ are ...
- 291
-1
votes
1
answer
102
views
Compactness of a special kind of Integral operators
Let $(S(t))_{t>0}$ be a continuous operator from $L^2(0,1)$ to its self and Let $K$ be the operator $$\eqalign{
& K:{L^2}(0,1) \to {L^2}(0,1) \cr
& f: \to (Kf)(x) = \int\limits_0^1 {k(...
- 617
0
votes
0
answers
172
views
Compact operator
Let $k:[0,1]^2 \to [0,1]$ be a measurable function. Define $K:L^2([0,1])\to L^2([0,1])$ to be the operator:
$$
(Kf)(x) = \int_0^1\int_0^1 f(z) k(x,y) \mathbf{1}_{x\leq z\leq y} \ \mathrm{d}z \mathrm{d}...
- 31
4
votes
2
answers
244
views
Compact images of nowhere dense closed convex sets in a Hilbert space
Let $B=-B$ be a nowhere dense bounded closed convex set in the Hilbert space $\ell_2$ such that the linear hull of $B$ is dense in $\ell_2$.
Question. Is there a non-compact linear bounded operator ...
- 41.8k
1
vote
0
answers
196
views
Compact embedding result
Let $\tau$ and $\ell$ be positive numbers. We know that the space $H^2(0,\ell)\cap H^1_0(0,\ell)$ is compactly embedded into $L^6(0,\ell)$. Now, is the space $L^2(0,\tau;H^2(0,\ell)\cap H^1_0(0,\ell))$...
- 395
0
votes
2
answers
1k
views
Does point-wise weak convergence give weak convergence in $L^2(I;X)$?
Let $X$ be a separable reflexive Banach space, $F$ be a locally Lipschitz nonlinear operator on $X$ that is weakly continuous on $X$, and $u_n$ are $u_n$ weakly converges to $u$ on $L^2(0,T;X)$. Now, ...
- 395
4
votes
2
answers
467
views
When is a Nemytskii map between Sobolev spaces compact?
Let $f:\mathbb{R} \to \mathbb{R}$ be a smooth function with bounded derivative. Define the Nemytskii map $F:H^1(\Omega) \to H^1(\Omega)$ by $F(u)(x) := f(u(x))$. Here $\Omega$ is a bounded smooth ...
- 65
1
vote
0
answers
380
views
If an upper semicontinuous multivalued map is compact on a set, is it compact on the boundary as well?
I have stumbled upon the following problem during my research:
Let $X$ and $Y$ be Banach spaces, $K\subset X$ nonempty, $F:\overline{K}\rightarrow 2^{Y}$ an upper semicontinuous multivalued map with ...
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