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## Hot answers tagged operator-theory

2

Example, $$A = \frac{1}{i}\frac{d}{dx}$$ on the domain $\mathcal{D}(A)$ of absolutely continuous functions $f \in L^2[0,1]$ for which $f' \in L^2[0,1]$ and $f(0)=0$. Then $A^*$ is the same as $A$ except that the condition $f(0)=0$ is replaced by $f(1)=0$. Then $A^{\star\star}=A$ because $A$ is closed and densely-defined. However, ...

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Given $f$ and $\epsilon$, choose a polynomial $p$ with $\Vert f-p\Vert_{\infty,X}<\epsilon$ (where $\Vert\cdot\Vert_{\infty,X}$ is the supremum norm oin $X$). Now see the corresponding polynomial function in $\mathcal{A}$, $p:\mathcal{A}\to\mathcal{A}$. (Remember: the functional calculus respects this notation, i.e., $p(a)$, in the functional calculus, is ...

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The Spectral Theorem for $A$ is given in terms of a Borel Spectral measure $E$ $$Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x,$$ and $x \in \mathcal{D}(A)$ iff $$\int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty.$$ The operator $e^{iA^2}$ is defined through the functional calculus as $$e^{iA^2}x = ... 2 Unfortunately this is not so simple/immediate and does depend on the type of equation and perturbation. I suggest that before anything else you have a look at the Engel-Nagel book, more precisely at Chapter 3 (such as at their Theorem 3.14). They also include discussions of some specific types of equations, although I don't remind elliptic. 1 Answering to my question after a great help from Jake, through comments and chat : Let's say that (a_1,a_2,a_3,a_4) are the four vertices of our square. Then, the operator that assigns them exactly to the next one will give back the vector (a_2,a_3,a_4,a_1). (Note that the fourth one goes back to the first one). Then, a vector v of the space \mathbb ... 1 a) We have that U_sU_{-s}=Id=U_{-s}U_s and$$<U_sf,g>=\int_{\mathbb{R}} U_sf(x)\overline{g(x)}\mathrm{d}x=\int_{\mathbb{R}} f(x-s)\overline{g(x)}\mathrm{d}x=\int_{\mathbb{R}} f(x)\overline{g(x+s)}\mathrm{d}x=<f,U_{-s}g>.$$Hence U_s^*=U_{-s}. By the first identity U_s is unitary. Try V_s yourself. c) We have that$$(U_tV_sf-V_sU_tf)(x)= ...

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The norm of a bounded linear operator $A:\ell_1\to Y$, for any normed space $Y$, is simply $\sup_{n}\|Ae_n\|$. (This follows from the triangle inequality.) For $Y=\ell^1$, in terms of the coefficients $a_{m,n}$ this becomes $$\|A\| = \sup_{n}\sum_m|a_{mn}|\tag1$$ the supremum of $\ell^1$ norms of columns. For $A$ to be the limit of finite-rank operators, we ...

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Using the Spectral Theorem, $$Ax=\int_{0}^{\infty}\lambda dE(\lambda)x \\ \mathcal{D}(A) = \left\{ x : \int_{0}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty \right\}.$$ Then the positive square root $\sqrt{A}$ is $$\sqrt{A}x = \int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)x \\ \mathcal{D}(\sqrt{A}) = \left\{ x : ... 1 There is something off with that argument as it is written. Nothing prevents, for instance, that x-\Phi_i(x_i')\leq0 and nonzero. In that case,$$f(x-\Phi_i(x_i'))=0,$$and then$$ 0<\|x-\Phi_i(x_i)\|,\ \ \ \ \|f(x-\Phi_i(x_i'))\|=0. $$This is how I think that argument can be saved. If x\geq0, y=y^*, and \|x-y\|<\varepsilon, then ... 1 Well one place we could start would be to note that we could define 𝔾 on a larger set; namely any real sequence of coefficients for which the generating series is absolutely convergent or even Abel summable (namely the limit of the series exists as s→1 even if the series diverges when one substitutes s=1). If we are willing to consider quasiprobability ... 1 You are done. You have shown that$$ \sigma(A)=\{3-i\lambda:\ |\lambda|\leq1\}. $$But multiplying the unit disk by \pm i is still the unit disk, so$$ \sigma(A)=\{3+\lambda:\ |\lambda|\leq1\}. $$In other words, the spectrum of A is the disk of radius 1 centered at 3. 1 There is no hope for an upper bound. To spell out @Keith's comment, assume that \|L(\sigma)^{-1}\|\le g(\sigma) for all such families. Then g(\sigma)\ge\frac{e^{b/\sigma}}B. Now consider the family \begin{equation*}L(\sigma)=\begin{pmatrix}Be^{-b/\sigma}&0\\0&g(\sigma)^{-2}\end{pmatrix}\end{equation*}to arrive at a contradiction. 1 Let \{v_n\} be a dense countable subset of V. Then we can find:$$ \alpha_1,\ \ \ \text{ such that } \|T_{\alpha_1}(v_1)-v_1\|<1.  \alpha_2,\ \ \ \text{ such that } \|T_{\alpha_2}(v_j)-v_j\|<\frac12,\ \ j=1,2.  \alpha_3,\ \ \ \text{ such that } \|T_{\alpha_3}(v_j)-v_j\|<\frac13,\ \ j=1,2,3.  \vdots  \alpha_n,\ \ \ \text{ such ...

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