# Tagged Questions

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### Show these operators converge to a particular limit

Let $H$ be a Hilbert space, and $T$ be a operator on $H$ of the form $T=\sum_{n=1}^{\infty}{\lambda}_{n}<x,e_{n}>e_{n}$ where $e_{n}$ are the eigenvectors of $T$ and an orthonormal basis of H ...
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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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### The eigenvalues of a compact and self-adjoint operator on Hilbert space

Show that if $K$ is a compact self-adjoint operator on Hilbert space then it has either finitely many eigenvalues or a sequence of eigenvalues $\lambda_n\to 0$ as $n\to \infty$.
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### Totally boundedness of a compact operator [closed]

Let $T:\ell_2(\mathbb N)\longrightarrow \ell_2(\mathbb N)$ bounded linear operator such that $$T(\{x_n\})=\{x_n/n\}.$$ I need to prove that $TB(\ell_2(\mathbb N))$, that is closed unit ball in ...
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### Compact embedding

Prove that the embedding $j\colon (C^1[0,1],\|\cdot\|)\to(L^1[0,1],\|\cdot\|_{L^1})$ where $\|f\|=\max\{\|f\|_\infty,\|f'\|_\infty\}$ and $\|f\|_\infty$ denotes the supremum norm, ...
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### Direct sum of eigenspaces of a compact operator has finite codimension

In an infinite dimensional Hilbert space the orthogonal complement of the (closure) of the direct sum of eigenspaces of a compact normal operator is finite dimensional. Why is this the case? thanks.
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### Inequalities on kernels of compact operators

Suppose we have a $\sigma$-finite positive measure $\mu(v)$ on $\Bbb R^d$ and we have two positive kernels on $\Bbb R^d\times \Bbb R^d$ $k_1(v,u)>0$, $k_2(v,u)>0$. We define integral operators ...
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### Pitt's theorem on automatic compactness of bounded operators between sequence spaces

Why is it called Pitt's theorem? I couldn't locate the origin of the statement.
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### Compactness of the solution operator

Let $\Omega$ be a smooth open bounded subset of $\mathbb{R}^n$. The bilinear form $$a(u,v)=\int_{\Omega}\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}dx$$ is elliptic on ...
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### compact operators and finite dimentional spaces

Let $Q_n$ a finite dimentional space. Since any finite rank operator is compact, it's true that any linear operator $K:Q_n\to Q_n$ is compact 'cause $\dim(R(K))<\infty$?
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### Operator from $\ell_{4}$ to $\ell_{1}$ is compact, if it's continuous.

Define $T: \ell_{4} \rightarrow \ell_{1}$ as $Tx=(a_1x_1, a_2x_2, \ldots)$. I showed that $T$ is continuous if and only if $\sum \left| a_i \right|^{\frac{4}{3}} < \infty$. How can I prove that if ...
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### Is operator from $X^{*}$ to $c_{0}$ compact

$X$ - Banach space. Let $X \ni (x_n)$ be such that $x_n \stackrel{\omega}{\rightarrow}0$ weakly. Define $T: X^{*} \rightarrow c_0$ by $T=\left(x^{*}(x_n) \right)_{n=1}^{\infty}$. Is operator $T$ ...
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### Compact operator with closed range has finite dimensional range

Let $X,Y$ be Banach Spaces, and let $T\in K(X,Y)$ be a compact operator from $X$ to $Y$. I have to prove that $T(X)$ is closed in Y if, and only if, $\dim(T(X))<\infty$. Can anybody help me with ...
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### Multiplication operator with a function non-vanishing on the cantor set

Let $M_f$ be the multiplication operator, which acts on bounded functions $g$ on the unit interval as $g\mapsto fg$, with $f:[0,1]\rightarrow \mathbb{C}$ such that $f$ is nonzero only on the Cantor ...
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### Closure of the range of a compact operator

Let $X$ be an infinite-dimensional Banach space, and let $Y$ be a banach. Let $T$ be a compact operator from $X$ to $Y$, ie. if $(x_n)$ is a sequence in $X$ then there is a subsequence s.t. ...
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### $L(\ell_{p})$ contains only one proper closed ideal

I am trying to solve the following problem: Show that if $1<p<\infty$ and $T:\ell_{p}\rightarrow\ell_{p}$ is not compact then there is a complemented infinite dimensional subspace $E$ of ...
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### Why $K(X,Y)$ is closed?

If $X$ and $Y$ are Banach space then $K(X Y)$ is a closed vector space of $B(X, Y)$. $B(X,Y)$ is the vector space of all bounded linear maps from $X$ to $Y$. $K(X,Y)$ is the set of all compact ...
We will say that ${T_i}\subset B(X,Y^*)$ converges to $T$ in W*-operator topology if $T_i(x)\rightarrow T(x)$ in W*-topology of $Y^*$( \$\forall y\in Y \langle T_i(x),y\rangle \rightarrow \langle ...