# Tagged Questions

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### Invariant convex subset

Let X be a banach space and $T$ a bounded operator on X with $||T||$ less than or equal to 1. If $T$ is an isometry and $r(T)$<1 show that there is a closed convex non-zero proper subset $C$ of the ...
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### Proving an operator $D: L^2[0,1]\rightarrow C'$, $Df(t)=\int^t_0 f(s) ds$ is unitary

Let $C'\subseteq C[0,1]$ be the space of all absolutely continuous function such that $f(0)=0$ and $f' \in L^2[0,1]$. Define an inner product on $C'$ as $\langle f,g \rangle = \int^1_0 f'(t)g'(t)dt$. ...
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### Abstract Wiener space and integration related to a trace class operator

Suppose I have a trace class operator $A$ of a Hilbert space $H$. Also suppose I have an abstract Wiener space $(H,B)$. Then, $\langle Ax, x \rangle$ is defined almost everywhere in $B$ with respect ...
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### Subspaces in the image of compact operator

Let $X$ and $Y$ be some infinite dimensional Banach spaces. Let $T:X\longrightarrow Y$ be some compact linear operator. It is easy to understand that $T$ cannot be surjective: the Open Mapping Theorem ...
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### renorm a Banach space to make an operator have spectral radius equal to norm

Let $X$ be an infinite-dimensional complex Banach space equipped with the norm $\lVert\cdot\rVert$, and let $T\in\mathcal{L}(X)$ a bounded linear operator on $X$. Let $r(T)$ denote the spectral ...
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### operator on separable banach space whose spectrum and point spectrum is prescribed compact set

I am interested in obtaining the following paper: G. K. Kalisch, "On operators with large point spectrum," Scripta Math. 29 No. 3-4, (1973), 371-378. According to Ben Mathes, "Strictly Cyclic ...
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### The trace class operators are the dual of the compact operators

I know that the map from the trace class operators $L_1(H)$ to the dual of the compact operators $K'(H)$ given by $A \mapsto tr( \cdot A)$ is an isometric isomorphism. Linearity is obvious by the ...
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### Prove that the Set of Bounded Linear Operators is Banach

Let $B(V,V')$ be the vector space formed by set of linear operators $T:V\rightarrow V'$. where $V,V'$ are normed vector spaces. Equip $B(V,V')$ with the norm $$\|T\|=\sup\frac{\|T(x)\|}{\|x\|}$$ ...
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### Pulling Operator Inside Integral

Say $Y$ is a Banach space and you have a family of continuous/bounded operators $L_{x}: Y \rightarrow Y$ for $x\in \mathbb{R}$ and say you have an bounded, smooth map $f(x):\mathbb{R}\rightarrow Y$. ...
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### Predual of $\mathcal{B}(K, H)$

Is there a predual of $\mathcal{B}(K, H)$? So, what does the space $X$ look like, such that $X^*=\mathcal{B}(K, H)$.
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### If space of bounded operators L(V,W) is Banach, V nonzero, then W is Banach (note direction of implication)

Let $V,W$ be normed vector spaces, and $L(V,W)$ be the space of bounded linear operators. Usually I would only see the statement "If $W$ is Banach, then $L(V,W)$ is Banach.". But Wikipedia writes that ...
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### $\gamma-$radonifying operators.

I am reading about $\gamma$-radonifying operators, and came across 2 similar definitions. And I hoped to understand why they are equivalent. Let $H$ be a seperable real Hilbertspace, $E$ banach ...
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### completely continuous implies compact

I'm searching for a proof of the fact that if: $T$ is a bounded operator in a reflexive Banach space that maps weakly convergent sequences onto convergent sequences then $T$ is compact. If we let ...
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### approximation property

In I. Namioka and R. R. Phelps's your paper "Tensor products of compact convex sets" Pacific Journal of Mathematics, Vol. 31, No. 2, 1969), they gave the following definition of approximation ...
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Let $X,Y,W,Z$ be Banach Spaces. Let $X\otimes_{\pi}Y$ denote the tensor product endowed with the projective norm. If I have $S\in B(X,Z)$, $T\in B(Y,W)$, it is straightforward to show that $S\otimes ... 1answer 51 views ### application of the theorem of the open application Let$ X, Y $be Banach spaces. Suppose that$ T: X \to Y $is a compact operator. show that if$ \dim Y $is infinite, then$ T $is not surjective. idea: Using the theorem of the open application 1answer 198 views ### A relatively bounded perturbation of a closed operator is a closed operator. Please I need help with an example I cant figure out and which will hopefully help me to get the theory: Let$X$be Banach space and$A, B$general operators. Furthermore$A$is closed, ... 2answers 217 views ### Isometry on a dense sub-space of a Banach space? Let$X$be a Banach space and let$D$be a dense sub-space of$X$. I don't know if the following fact is true: Fact: For every (linear) isometry$T\in\operatorname{Iso}(X)$and for every ... 1answer 92 views ### Pitt's theorem and reflexivity Does it follow from Pitt's theorem that the space of bounded operators from$\ell_2$to$\ell_p$($p<2$) is actually reflexive? We have$$\mathcal{B}(\ell_2, \ell_p) = \mathcal{K}(\ell_2, \ell_p) ... 2answers 142 views ### Closed operator I've got a very straightforward question : if$T : B \rightarrow B$is a linear continuous operator and$B$is a Banach space, is$T$a closed operator? This is obviously true in finite dimension, ... 1answer 64 views ### Is the adjoint operation WOT-WOT continuous? This is a well-known property of the Hilbert-space adjoint operator that it is WOT continuous. Is a similar true for Banach spaces? That is, for a given Banach space$X$is the operation ... 1answer 150 views ### Conditions for a kernel of a bounded operator to be complemented I am well aware of the problem of complementing subspaces in Banach spaces as it was discussed here and here . Nevertheless, I wonder whether there are conditions for existence of a complement$M$... 0answers 111 views ### Bounded operator on dense subspaces Give an operator like this or show it doesn't exist: Operator$T: X\rightarrow Y$is bijective.$X,Y$are dense subspaces of a Banach space$Z$, and$X$is proper subset of$Y$. Both$T$and$T^{-1}$... 1answer 98 views ### A problem on bounded invertible linear operator in Banach space Let$X$be a Banach space. Let$T : X \to X$be a invertible linear operator and$M > 0$be such that$\|T^{-k}\| \le M$for all$k \ge 1$. Prove that$\inf_ {n\ge1} \|T^n(x)\| > 0$for all$x ...
I'm looking for a counterexample in a Banach space. I've seen the counterexample at Sum of Closed Operators Closable?, but I don't understand why $A$ and $B$ are closed. Could someone expand on this ...