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Theorem: Suppose $A : \mathcal{D}(A)\subseteq X\rightarrow X$ is a surjective linear operator on a complex Hilbert space $X$ for which $(Ax,x) \ge \|x\|^{2}$ holds for all $x\in\mathcal{D}(A)$. Then $A$ is densely-defined and selfadjoint. Proof: To show that $A$ is densely-defined suppose otherwise. Then there exists $y \perp \mathcal{D}(A)$. However, ...

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We will show that $\sigma(A) = \{0\}$, and that $0$ belongs to the residual spectrum. As you have shown, $0$ is not an eigenvalue, and as $\operatorname{im} A \subseteq \{x \in C[0,1]: x(0) = 0\}$, $A$ does not have dense image. Hence $0 \in \sigma_r(A)$. To see that $A - \lambda$ is invertible for $\lambda \ne 0$, let $y \in C[0,1]$ be given. We have to ...

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Answer. $S=\{z\in\mathbb C: \lvert\,z\rvert=1\}$. Explanation. The space $\ell^2(\mathbb Z)$ is isometric to $L^2(\mathbb T)$, where $\mathbb T$ is the unit circle (equivalently, the domain of $2\pi-$periodic functions), and this is done by the transfomation $$\varPhi\big((a_k)_{k\in\mathbb Z}\big)=\sum_{k\in\mathbb Z}a_k\mathrm{e}^{ikx},$$ as ...

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Suppose $T$ is a bounded operator on a Banach space $X$. $\lambda\in\rho(T)$ iff $T-\lambda I$ is a linear bijection. In that case, the inverse $(T-\lambda I)^{-1}$ is automatically continuous by the closed graph theorem. There are three basic things that can stand in the way of $T-\lambda I$ being invertible. $T-\lambda I$ is not injective. Equivalently, ...

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More elementary way: a. If $\lvert\,\lambda\rvert>1$, then $\lambda\not\in\sigma(\mathcal L)$. This is clear as the series $(\lambda-\mathcal L)^{-1}=\lambda^{-1}\sum_{n=0}^\infty (\lambda^{-1}\mathcal L)^n$ converges. b. If $\lvert\,\lambda\rvert<1$, then $\lambda\not\in\sigma(\mathcal L)$. This is clear as the series $(\lambda-\mathcal ... 2 Some important tricks/theorems: The spectrum is non-empty for Banach algebras (over$\mathbb{C}$) The spectral radius formula $$r(a) = \lim \|a^n\|^{1/n}$$ tells you that if$a$is nilpotent, then$\sigma(a) = \{0\}$If$A$is commutative, then $$\sigma(a) = \{\tau(a) : \tau \in \Omega(A)\}$$ where$\Omega(A)$denotes the set of non-zero multiplicative ... 2 Actually convolution with$k$defines a bounded operator on$L^2(G)$if and only if$\hat k$is bounded. In this case the spectrum is exactly the essential range of$\hat k$. In fact if$m$is any bounded measurable function on$\hat G$then$m$defines a bounded operator on$L^2(G)$by $$\widehat{Tf}=m\hat f,$$whether or not there exists$k$with$m=\hat ...

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First, all functions in the span of $u$ and $v$ are critical points. Second, consider $u_0 = \sin(\pi x)\sin(\pi y)\ge 0$, $u=t\cdot u_0$, i.e., eigenfunctions to the smallest eigenvalue of the Laplacian. Then $$I[u] = \int_\Omega \frac12|Du|^2 - \frac{5\pi^2}2|u|^2 = \frac12\int_\Omega u(-\Delta u - 5\pi^2u) = \frac12t^2 \int_\Omega u_0(2\pi^2 - ... 1 Suppose \lambda is an isolated point of the spectrum of T. Let F be a continuous function that is identically 1 near \lambda, and is 0 on the remaining part of the spectrum \sigma(T)\setminus\{\lambda\}. Then F is either 1 or 0 on the spectrum, which makes P=F(T) an orthogonal projection. Clearly (T-\lambda I)P=0 because ... 1 Here is an alternative proof of the fact that r(T)=0 ( I add it because it does not seem to appear in the related posts and I find it quite nice). First show by induction that |A^n x(t)|\leq \frac{t^n}{n!}\|x\|_\infty for all t\in[0,1]: Basis n=0: |x(t)|\leq \|x\|_\infty for all t\in[0,1] by definition of the supremum norm. Inductive step: For ... 1 If the characteristic polynomial of f has no multiple roots, then there exist distinct complex numbers \lambda_1, \dots, \lambda_n and a basis v_1, \dots, v_n of \mathbb{C}^n such that$$ fv_k = \lambda_k v_k  for $k = 1, \dots, n$. Now \begin{align*} f(gv_k) &= g(fv_k) \\ &= \lambda_k (gv_k), \end{align*} which means that $gv_k$ is in the ...

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The operator $L$ maps $\{ a_n \}_{n=-\infty}^{\infty}$ to $\{ a_{n+1} \}_{n=-\infty}^{\infty}$. If $e_n$ is the standard basis element defined as a sequence of all $0$'s except for a $1$ in the $n$-th place, then $Le_n = e_{n-1}$, which is why $L$ would be considered to be a left shift. And this follows from the definition $L\{ a_n \} = \{ a_{n+1} \}$: the ...

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HINT.If $a_{n+1}-z a_n=b_n$ then $a_2=z a_1+b_1$, $a_3=z a_2+b_2=z^2 a_1+z b_1+b_2$ ,$a_4=z a_3+b_3=z^3a_1+z^2b_1+z b_2+b_3$, etc. If $|z|=1$ and $b_n= z^{n}/(1+|n|)$, what happens to $|a_n-z^{n-1}a_1|$ for large positive $n$?

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For $T$ to have $\lambda$ as an eigenvalue, $T-\lambda I$ must be non-injective. For $\lambda$ to be in the spectrum of $T$, it must only be non-invertible. These are equivalent when $T$ is an operator on a finite-dimensional space, but not in general! For example, let $T$ be the shift operator $(x_0,x_1,\dots) \mapsto (0,x_0,x_1,\dots)$ on your favorite ...

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