All Questions
8 questions with no upvoted or accepted answers
5
votes
0
answers
94
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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
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
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
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
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 ...
- 11
0
votes
0
answers
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....
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