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## New answers tagged compact-operators

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Let $S=\{x\in\mathcal{H}:\|x\|=1\}$. Since $S$ is closed and bounded, then $S$ is weakly compact. But $T$ is compact. Then $T$ is weak-norm continous. Thus $f:H\to\mathbb{R}$ given by $f(x)=\|Tx\|$ is weak-continous. Since $H$ equipped with the weak topology is Hausdorff and $S$ is weakly compact, then $f$ attains a globlal maxima on $S$. That is, there is ...

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Regarding your comment about bounded functionals. Your answer is not correct. Hahn-Banach theorem works in other direction. Given $x\in X$, then there exists a bounded functional $f$ with $\|f\|=1$ and $f(x)=\|x\|.$ Your argument proves what I claimed. What you need to do on Hilbert spaces, just apply Riesz representation theorem to obtain a vector $y\in ... 3 We will prove that$\exists x_0\in S(0,1)=\{x\in H:\|x\|=1\}:\|Tx\|=\|T\|$. You know that$\|Tx\|\leq \|T\|\,\forall x\in B(0,1)\Rightarrow \exists$a sequence$\{x_n\}\subset B(0,1): \|Tx_n\|\to \sup\limits_{x\in B(0,1)} \|Tx\|=:\|T\|$. From this sequence you can chose a weakly convergent subsequence, still denoted by$\{x_n\}$, in$B(0,1)$, say ... 2 If you prove that the image of the closed unit ball of$\mathcal H$by$T$is compact in$\mathcal H$, then there exists$x\in \mathcal H$with$\|x\|=1$such that$\|Tx\|=\|T\|.$Indeed, since$T(B_\mathcal H)$is compact, the norm$\|\cdot\|$atains its maximum on$T(B_\mathcal H)$. Therefore, the set$\{\|Tx\|:\; x\in B_\mathcal H\}$is closed. This ... 0 No. Any operator on a finite dimensional space is compact and in particular, any non-trivial nilpotent operator$T$is a counterexample. 0 The sequence$x=\{ \cdots,0,0,-1,0,1,0,0,\cdots\}$where the middle$0$is$x_0$is mapped to$\{ \cdots,0,0,\frac{-2}{1^2+1^2},0,\frac{2}{1^2+1^2},0,0,\cdots\}$under the transformation$T$. That is$Tx=\frac{2}{1^2+1^2}x$. In this way you can see that$\frac{2}{1+n^2}$is an eigenvalue for$n=1,2,3,\cdots$. And$0$is an eigenvalue also because$Ty=0$... 2 Hint: if$[1-\lambda(n^2+1)]x_n=x_{-n}$for all$n$, then$[1-\lambda(n^2+1)]x_{-n}=x_{n}$for all$n$as well. 0 Let$A : X \rightarrow Y$be a compact operator and$A^* : Y^* \rightarrow X^*$its dual. The trick is to note that $$\Vert A^* y^* \Vert = \sup_{\Vert x \Vert \leq 1} \Vert y^* A x \Vert \leq \sup_{y \in \mathrm{cl}(A(B))} \Vert y^*(y) \Vert$$ where$B$is the closed unit ball in$X$. Let$(y_n^*)_{n \in \mathbf{N}} \subset Y^*$be a bounded sequence, say ... 0 You will find complete descriptions of all closed ideals in the papers M. Daws, Closed ideals in the Banach algebra of operators on classical non-separable spaces, Math. Proc. Cambridge Philos. Soc. 140 (2006), no. 2, 317–332. W.B. Johnson, T. Kania, G. Schechtman, Closed ideals of operators on and complemented subspaces of Banach spaces of ... 0 You can see the ideal of compact operators as the ideal generated by the finite-rank operators. You can play the same game and consider the ideal generated by those operators which have separable range. 0 Take any$0\neq x\in \mathbb{H} $and cosider ideal$\mathcal{M} =\{ T\in \mathcal{B} (\mathbb{H} ) : T(x) =0\}.$1 Yes. Simply note $$\|AB \phi - BA \phi \| \leq \|AB \phi - A_n B \phi\| + \|A_n B\phi - BA\phi\| = \|AB \phi- A_n B \phi\| + \|B A_n \phi - BA\phi\|\leq \|B\phi\|\cdot \|A-A_n\| + \|B\| \|A_n \phi - A\phi\|$$ as$n\to \infty$. Here, I assumed$A_n \to A$in operator norm. But actually, an easy modification of the argument shows that it suffices to have ... 0 So this is how I solved it: I defined$(f_n)$as: $$f_n(x)=\chi(2n,2n+1)$$ where$\chi$is the characteristic function. Follows $$Kf_n(t)=\lambda f_n(t)-f_n(t-1)=\lambda\chi(2n,2n+1)-\chi(2n-1,2n)$$ now lets calculate$||f_n-f_m||_2\$: $$||f_n-f_m||_2=\sqrt{\int_{-\infty}^{+\infty}|f_n(x)-f_m(x)|^2}=\sqrt{2\lambda^2+2}$$ which means, that ...

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