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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 = ...


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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.


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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 ...


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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 : ...


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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 ...


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Well one place we could start would be to note that we could define š”¾ 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 sā†’1 even if the series diverges when one substitutes s=1). If we are willing to consider quasiprobability ...


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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.


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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.


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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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