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A compact operator has a closed range iff it has a finite dimensional range. Without loss of generality,we can assume that the range of A is closed,otherwise we can consider the restriction of A to E where E=T−1(F),In all cases the theorem assures that F is finite dimensional. to prove the theorem,consider the canonical map associated with A and the fact ...

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Let $F\subset T(X)$ be a closed (in $Y$) subspace. Then $E = T^{-1}(F)$ is a closed subspace of $X$, and $T\lvert_E \colon E \to F$ is a compact surjective operator. Since $F$ is closed in $Y$, the open mapping theorem implies that $F$ is finite-dimensional. So: the image of a compact operator cannot contain an infinite-dimensional Banach space.

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Indeed we also have $T_F \to T$ in the strong operator topology. The strong operator topology is defined by the seminorms $$p_F(S) = \sup \{\lVert S(x)\rVert : x \in F\},$$ where $F$ traverses the finite subsets of $X$. The construction immediately yields $$p_F(T_F - T) = 0$$ for any linearly independent finite $F\subset X$, and it is straightforward to ...

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I think I found the answer.It's a modification of the proof in "Introduction to Tensor Products of Banach spaces, R.A.Ryan" Prop4.6. Suppose $x^{**}$ is an element of the dual of $W^*$ endowed with the given topology. It's easy to see that there exsist a compact subset $K\subset X$, s.t. $$|x^{**}(f)|\leq \sup_{x\in K}|f(x)|,\forall f\in W^*.$$ Use that ...

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In what follows we prove that: If $\{\lambda_n\}$ are eigenvalues (corresponding to linearly independent eigenvectors) of the self-adjoint compact operator $K$, then $\lambda_n\to 0$. Suppose not, and in particular, that there exists an $\varepsilon>0$, such that $\lvert\lambda_{j_n}\rvert\ge\varepsilon$, for $\{\lambda_{j_n}\}$ a subsequence of ...

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Since dim $Y_\varepsilon < \infty$ and $K$ is bounded, $\overline{K}$ in $Y_\varepsilon$ is compact and we can take a finite subcover of $\cup_{y\in K} B(y,\delta)$, for all $\delta>0$.

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It suffices to prove that $B_X(0,1)$ is totally bounded in $Y$; that is, for every $\epsilon>0$ it admits a finite $\epsilon$-net (in the norm of $Y$). Pick linear functionals $\lambda_1,\dots,\lambda_n\in X^*$ such that $$\|x\|_Y\le \frac{\epsilon}{3}\|x\|_X+\max_{1\le i\le n} |\lambda_i(x)|\qquad \forall\ x\in X$$ The image of $B_X(0,1)$ under the map ...

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The commutator is compact if and only if $f$ is in VMO. Source: On the compactness of operators of Hankel type by A. Uchiyama (free access; note that $VMO$ is called $CMO$ there). Since $C_0(\mathbb R)\subset VMO$, the answer to your question is positive.

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Let's prove that every finite-dimensional subspace of a topological vector space is closed. Suppose $M$ is the linear span of linearly independent vectors $x_1,\dots,x_n$, and we have a convergent sequence of the form $v_k = \sum_{i=1}^n c_{ik} x_i$. Consider two cases: There is a constant $C$ such that $|c_{ik}|\le C$ for all $i,k$. Then we can choose a ...

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Since you requested a hint, here is a start: Our operator is $T(\{x_n\}) = \{x_n/n\}$, and we are applying it to $B=B(l_2(\mathbb{N}))$. We wish to show that for $\delta > 0$ the image $T(B)$ is coverable by a finite number of balls of radius $\delta$. Since every element in B is of norm less than or equal to 1, we have that if $\{x_n\} \in B$ then ...

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Let $x=(x_n)_{n\in\mathbb N^*}\in \mathrm{B}(\ell_2)=\{x=(x_n):|x|_2\leq 1\}$ so $$|T(x)|_2^2\leq \sum_{n=1}^{+\infty} \left|\frac{x_n}{n}\right|^2\leq \sum_{n=1}^{+\infty} \frac{1}{n^2}=\frac{\pi^2}{6}.$$this implies that : $$|T(x)|_2\leq \frac{\pi}{\sqrt6}\hspace{2mm}\forall x\in\mathrm B(\ell_2).$$ (I think that is true I am not specialise in ...

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Yes, you can use Ascoli-Arzelà. Most easily seen here if you factor the map through $L^\infty([0,1], \lVert\,\cdot\,\rVert_{L^\infty})$: C^1([0,1],\lVert\,\cdot\,\rVert) \underbrace{\hookrightarrow}_{\text{compact}} L^\infty([0,1], \lVert\,\cdot\,\rVert_{L^\infty}) \underbrace{\hookrightarrow}_{\text{continuous}} L^1([0,1], ...

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More is true, in fact. If $H$ is a Hilbert space, and $T\colon H \to H$ a compact normal operator, then $H$ is the closure of the direct sum of the eigenspaces of $T$. For $\lambda \in \sigma_P(T)$, let $E_\lambda$ be the eigenspace of $T$ corresponding to the eigenvalue $\lambda$, and let $S = \bigoplus\limits_{\lambda\in\sigma_P(T)} E_\lambda$. Then $S$ ...

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Suppose $X$ is a Banach space and suppose that $T : X\rightarrow X$ is linear. Then the following are true: 1. If $X$ is finite-dimensional, then $T$ is compact, regardless of whether or not $0 \in \sigma(T)$. 2. If $X$ is infinite-dimensional and $T$ is compact, then $0 \in \sigma(T)$.

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If $0$ is not in the spectrum of $A$, then $A$ is invertible with inverse $B$. Since the product of compact operators is compact, the identity $I = AB$ is compact. But this is impossible if $A$ acts on an infinite-dimensional Banach/Hilbert space, because the closed unit ball is not compact.

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This is indeed only true in infinite dimensions. Suppose your compact operator $A$ has finite spectrum. Then there are only finitely many eigenvectors with nonzero eigenvalue. Let $F$ be the subspace spanned by those eigenvectors and let $E$ be its orthogonal complement. Since $F$ is finite dimensional, $E$ is infinite dimensional, and in particular $E ... 0 It suffices to establish that any sequence$\,u_k=u_k(|x|)\to 0\,$weakly convergent in$\,H_{rad}^1(\mathbb{R}^n)\,$will be srongly convergent in$\,L^{2\sigma+2}(\mathbb{R}^n)$, which might be ruined by the unboundedness of$\,\mathbb{R}^n$if it were not for the remarkable inequality$\,(\ast\ast)\,\$ that provides cutting off the infinity.  Indeed, ...

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