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Definition [$C_0$ Semigroup]: Let $X$ be a Banach space, and let $T : [0,\infty)\rightarrow\mathcal{B}(X)$ be a function into the bounded linear operators on $X$. Then $T$ is a semigroup if $T(0)=I$ and $T(t)T(t')=T(t+t')$ for all $t,t' \ge 0$. $T$ is a $C_0$ semigroup if $\lim_{t\downarrow 0}T(t)x=x$ for all $x\in X$. Suppose $X$ is a Banach space, and ...

4

Consider the Laplacian matrix of the cycle on $n$ vertices, given by \begin{equation*} L = \begin{bmatrix} 2 & - 1 & 0 & \cdots & 0 & -1\\ -1 & 2 & -1 & \cdots & 0 & 0\\ 0 & -1 & 2 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \...

2

Let's assume we're on a bounded domain $\Omega\subset\mathbb{R}^N$ and we've fixed boundary conditions. Recall that the Laplacian $\Delta$ induces an orthogonal decomposition of $L^2(M)$ into eigenspaces $E_k$, where $E_k$ is associated to the $k^{th}$ eigenvalue $\lambda_k$, and $\Delta$ acts by scaling on $E_k$. Write $0\leq \lambda_1\leq\lambda_2\leq\... 2 The operator consists of two unrelated parts: reflection in the plane$(x_1,x_2)$, which contributes norm$1$and eigenvalues$\pm 1$, and scaled backward shift, which contributes the eigenvalue$0$. There are no other eigenvalues because if$\lambda \notin \{0,\pm 1\}$then$x_3,x_4,\dots $must be nonzero and satisfy$x_{n+1} = n\lambda x_n$for all$n. ... 2 You can write unravel this definition by writing \begin{align} Af(x) & = \int_{0}^{x}K(x,y)f(y)dy+\int_{x}^{1}K(x,y)f(y)dy \\ & =\sinh(1-x)\int_{0}^{x}\sinh(y)f(y)dy+\sinh(x)\int_{x}^{1}\sinh(1-y)f(y)dy. \end{align} This is a typical kind of Green function expression. First note that $$(Af)(0) = 0,\;\;\; (Af)(1)=0.$$ Then, \begin{... 2W_C = \{u + iv \in X_C \mid u, v \in W\}$is invariant under$T_C$, so we may as well forget about the real Banach spaces and consider an operator$T$on a complex Banach space$X$with a finite-codimensional invariant subspace$W$, such that$\sigma(T) \subseteq \mathbb R$. The claim is that$\sigma(T|_W) \subseteq \mathbb R$. If$\lambda \notin \sigma(T)...

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A simple example Let us look at a specific example in two dimensions (the main point would probably come forward better in three dimensions, but that is harder to draw, and is left as an exercise to you in the end). We study, as an example, the symmetric matrix $$A= \begin{bmatrix} 4 & -2\\ -2 & 7 \end{bmatrix}.$$ A standard calculation shows ...

2

As mickep observes in the comments, the spectra are not the same! The Dirichlet spectrum is $$\lambda_{kl} = \bigg(\frac{\pi k}{a}\bigg)^2 + \bigg(\frac{\pi l}{b}\bigg)^2,\ k,l = 1,2,3,\ldots$$ while the Neumann spectrum is $$\nu_{kl} = \bigg(\frac{\pi k}{a}\bigg)^2 + \bigg(\frac{\pi l}{b}\bigg)^2,\ k,l = 0,1,2,3,\ldots$$ The intuition is that the ...

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Seems like you're jumping the gun with this question... I'd like to take a few steps back then work up to whats going on here. If $A$ is a hermitian matrix ( or symmetric for real matrices) then all the eigenvalues of $A$ are real (why?) and the eigenvectors are orthogonal (also why?). This allows us to rewrite our basis in terms of these eigenvectors. i.e. ...

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You seem to be getting twisted in a knot. The actual problem is, as I'm sure you realize, pretty simple: you are trying to minimize $x^\top A x$ subject to the constraint $x^\top x = 1$. For that you can use any of several techniques, such as Lagrange multipliers. You seem to be doing fine for a while: you've written $A = P^\top D P$ where the diagonal ...

1

The connection with the spectral measure $P$ is $$P(E) = \chi_{E}(A).$$ So, for example, $$P[a,b] = \chi_{[a,b]}(A), \;\; P(a,b) = \chi_{(a,b)}(A) \\ P[a,b] = P(a,b) + P\{a\}+P\{b\}$$ The spectral measure $P$ is regular in the strong topology, which gives you \begin{align} P[a,b]x & = \lim_{\...

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Another way of writting Jeb's solution is the following. Let $v_1,v_2,v_3$ be an orthonormal basis of $\mathbb{R}^3$ of eigenvectors of $A$ associated to the eigenvalues $\lambda_1\leq\lambda_2\leq\lambda_3$. Let $V=\text{span}\{v_2,v_3\}$. If $X$ is a 2-dimensional subspace of $\mathbb{R}^3$ then $\dim(X\cap V)\geq 1$. Let $u\in X\cap V$ with norm 1. ...

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Notice that $\|Lx\|^2 = \langle x, L^2 x\rangle$, and by assumption, this is $\le \|Ax\|^2=\langle x, A^2 x\rangle$. In other words, $L^2\le A^2$, and since the square root function is operator monotone, this implies that $L\le A$ and thus also $L+t\le A+t$. Now we obtain in the same way that $$\|(L+t)^{1/2}x\|^2 = \langle x, (L+t) x\rangle\le \|(A+t)^{1/2}... 1 First, note that X_2 is an A-invariant subspace, so that A_2:X_2\to X_2. We can show formally that the adjoint of A_2 should be the restriction of A^* to X_2, which is again A^*. It then suffices to note that the image of the closed unit ball under A_2 is a closed subset of the image of the closed unit ball under A and is therefore ... 1 It follows from the definition of \sup. If$$ M=\sup\left( f(x)\ : x \in B\right), $$then there exists a sequence x_n\in B such that$$ M=\lim_{n\to \infty} f(x_n).Apply this observation with B=\text{unit sphere}, f(x)=|(Ax_n, x_n)| and M=\|A\|. 1 Yes. The necessary and sufficient condition is that A is bounded. Indeed, the domain of AE_I is the whole Hilbert space, and the domain of E_IA is the domain \mathcal D(A) of A. 1 If \lambda^n \ne 1, then you can directly verify that T-\lambda I is invertible by showing \begin{align} I&=(T-\lambda I)\left[\frac{1}{1-\lambda^n}(T^{n-1}+\lambda T^{n-2}+\cdots+\lambda^{n-2}T+\lambda^{n-1}I)\right] \\ &=\left[\frac{1}{1-\lambda^n}(T^{n-1}+\lambda T^{n-2}+\cdots+\lambda^{n-2}T+\lambda^{n-1}I)\right](T-\lambda I).... 1 Your first inequality is usually known as the Kadison-Schwarz inequality. It only requires 2-positivity. Claim. A=\begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\geq0 if and only if a^*a\leq b. Proof. If A\geq0, then for any \xi\in H, \langle (b-a^*a)\xi,\xi\rangle=\left\langle \begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\,\begin{...

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You're not going to find an explicit function for $N$. (For example, $N$ is not even continuous.) To prove the asymptotic expansion $$N(\lambda) = c_n|\Omega|\lambda^{n/2} + o(\lambda^{n/2})$$ for the Dirichlet counting function on a domain $\Omega$, the simplest way in bounded Euclidean domains --- and one that I assume you're to use, since you appear to ...

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Denote by $N_{D_p}(\lambda)$ the number of eigenvalues for the Dirichlet Laplacian on the rectangle $D_p$ that are less than $\lambda$. Then, and this is what you seem to be missing, since the sequence $\mu_1\leq\mu_2\leq\cdots$ enumerate all eigenvalues on the different rectangles, the quantity $M(\lambda)$ that you define in the question is just the sum ...

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You are looking in the wrong volume. You should look in Vol I, Chapter VI, Section 4, Subsection 4 starting on page 436. The title of that section is "Asymptotic Distribution of Eigenvalues for an Arbitrary Domain". Good luck!

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I think it's a bit confusing what they wrote, bordering on incorrect. I'm sure you can find a better reference for these things. The idea is to use the theorem that $\lambda_1(M) \leq E(u) / \langle u, u \rangle = \| \nabla u \|^2 / \| u \|^2$ for any sufficiently nice $u$ (it does not have to be $C^\infty$). Let $\varphi_1 \in C^\infty(M_1)$ be ...

1

The right inequality implies $$\frac{\lambda_n}{n}\geq \frac{4\pi}{ab}.$$ We use that in the left inequality (for the square root term only), $$n\geq \frac{\lambda_n ab}{4\pi}-C\sqrt{\lambda_n} \geq \frac{\lambda_n ab}{4\pi}-C\sqrt{\frac{4\pi n}{ab}}.$$ Rearranging, $$\frac{\lambda_n}{n}\leq\frac{4\pi}{ab}+C\Bigl(\frac{4\pi}{ab}\Bigr)^{3/2}\frac{1}{\... 1 The eigenvalues$$ \lambda_{ab} = \pi^2\bigg(\frac{1}{a^2} + \frac{1}{b^2}\bigg) $$for a,b \in \Bbb{N} are squared lengths of vectors in the lattice (\pi/a)\mathbb{N}\times(\pi/b)\mathbb{N}, where I let \Bbb{N} = \{1,2,3,\ldots\} be the positive integers. So the counting function N(\lambda) counts lattice points (in the upper quadrant) whose length ... 1 If T is invertible, then so is T': indeed, if ST=TS=I, then$$ S'T'(x+iy)=STx+iSTy=x+iy. $$Conversely, if KT'=T'K=I, then define an operator S on X by Sx=K(x+i0). Then$$ STx=K(Tx+iT0)=KT'(x+i0)=x+i0. $$In both cases the induced inverse is bounded by the Open Mapping Theorem. It follows that if \lambda\in\mathbb R, then T-\lambda I is ... 1 The notation can also be interpreted as follows: order the eigenvalues \lambda_1\le\lambda_2\le \ldots. Consider \lambda_N for a large N, and now follow Willie's outline to approximately determine N: The number of smaller eigenvalues is one fourth the number of lattice points in the ellipse$$ x^2/a^2 + y^2/b^2 = \lambda_N/\pi^2 , $$and this is ... 1 I suggest the book by Pazy on Semigroups of Linear Operators. It's one of the most elegant and readable books in Functional Analysis I've encountered. On page 14, there is a theorem: Theorem: A Linear operator is dissipative if and only if$$ \|(\lambda I-A)x\| \ge \lambda \|x\|,\;\; x\in\mathcal{D}(A),\; \lambda >0. $$His setting for ... 1 Your proof is fine. What you are missing to work at points other than zero is the following lemma: Lemma. Let f:X\to\mathbb C with X a compact subset of \mathbb R, \varepsilon>0 and R>0. Then there exists \delta=\delta(\varepsilon,R)>0 such that if a,b\in A^+, with \|a\|+\|b\|<R, with \sigma(a)\cup\sigma(b)\subset X, and such ... 1 The Laplacian is given by:$$ L = \begin{pmatrix} 2 & -1 & 0 & \cdots & -1 \\ -1 & 2 & -1 & \cdots & 0 \\ 0 & -1 & 2 & \cdots & 0 \\ \vdots & & & \vdots & \\ -1 & 0 & 0 & \cdots & 2 \end{...

1

Yes. In fact, if $A$ is $m \times n$ and $B$ is $n \times m$, $m \ge n$, the characteristic polynomial of $AB$ is $\lambda^{m-n}$ times the characteristic polynomial of $BA$.

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