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Here is a sketch of the first part. For each $f\in C[0,1]$, \begin{eqnarray*} \|Tf\| &=& \max_{t\in[0,1]}\left|\int_{0}^{1}K(t,s)f(s)ds\right|\\ &\leq& \max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)f(s)\right|ds\\ &\leq& \|f\|_{\infty}\max_{t\in[0,1]}\int_{0}^{1}\left|K(t,s)\right|ds\\ \end{eqnarray*} Therefore $\|T\|$ is ...

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The definition of a compact operator $u \in B(X,Y)$ is that $u$ maps bounded sets of $X$ into relatively compact sets of Y. Therefore, (1) and (2) are equivalent by definition and no proof is required. This definition is equivalent to saying if $x_n$ is a bounded sequence in $X$, then $u(x_n)$ has a Cauchy subsequence (not necessarily convergent) in $Y$. To ...

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Let $H$ be a complex Hilbert space and let $T \in B(H)$ be an isometry. We claim that $T^\ast T = I$. To this end let $h \in H$ and note that by the Hilbert space Riesz representation theorem a linear functional in $H^\ast$ corresponds to an element $h \in H$ ($h \mapsto \langle \cdot, h \rangle$). Also note that if $\varphi (x) = 0$ for all $\varphi \in ... 0 You're right that Cauchy-Schwarz is important. It comes in here: $$\langle x, Ax \rangle \leq ||x||_1 ||Ax||_1$$ Can you fill in the rest? 0 The operator norm of an operator is $$\|A\| = \sup\left\{ \frac {\|Ab\|}{\|b\|} : b \neq 0 \right\}$$ Explanation: The term$\frac {\|Ab\|}{\|b\|}$is the factor how much the vector$b$is stretched by$A$. The operator is bounded, iff there is an upper bound for this stretching factor and the supremum of all possible stretching factors is called operator ... 0 Case 1: If the inverse is also continuous then the subset is necessarily all of space: $$A\cong Y\Rightarrow A\text{ complete }\Rightarrow A\text{ closed}\Rightarrow A=\overline{A}=X$$ Case 2: If the inverse is not continuous then the extension will be still surjective but no more injective: $$Y=\mathcal{R}(T)\subseteq\mathcal{R}(T_E)\subseteq ... 0 The new elements I believe will be going to the same places as before, since if A is dense in X we have sequences u_n\in A such that u_n\to u\in \overline {A}\setminus A=X\setminus A So originally we have Tu_n\to Tu\in Y , now we have T_Eu_n\to T_Eu=Tu except now u\in X 0 T must be densely defined for T^\ast to be defined. That T^\ast is a left inverse of T implies that \mathcal{R}(T) \subset \mathcal{D}(T^\ast). And hence for every x \in \mathcal{D}(T) we have$$\langle x,x\rangle = \langle x, T^\ast T x\rangle = \langle Tx,Tx\rangle,$$or in other words \lVert Tx\rVert = \lVert x\rVert, i.e. T is an ... 0 Yes. Let S_i be a set of cardinality \operatorname{hilb.dim}(H_i), then H_i=\ell_2(S_i). From this post we know that$$ H=\bigoplus_{i\in I} H_i=\bigoplus_{i\in I}\ell_2(S_i)=\ell_2\left(\bigsqcup_{i\in I}S_i\right) $$Hence$$ \operatorname{hilb.dim}(H)=\operatorname{Card}\left(\bigsqcup_{i\in I}S_i\right)=\sum_{i\in ... 0 I think not. Split your Hilbert space into a direct sum of two dimensional spaces. On the$n$th two dimensional space, find a matrix whose eigenvalues are$1/(2n-1)$and$1/(2n)$, whose eigenvectors are$(\cos(1/n),\sin(1/n))$and$(-\sin(1/n),\cos(1/n))$. Since the eigenvalues are all distinct, the eigenvectors are uniquely determined. The matrix is ... 0 Given two normed vector spaces$V$and$W$a linear map$A:V\rightarrow W$is continuous if and only if it is bounded. In other words if: $$||Av||\leq c||v||$$ The definition of the operator norm is given as: $$||A||_{op} = \inf\{c\geq 0: ||Av||\leq c||v|| \text{for all} v\in V\}.$$ The operator norm is defined by the smallest$c$so that this is true. From ... 0 It's a standard procedure which is usually deemed as obvious. Let's give a name to this set: $$C=\{c\ge 0\ :\ \lVert Av\rVert \le c\lVert v \rVert\}\subset [0, \infty)$$ To say that$C$is nonempty is the same as to say that$A$is bounded. In this case, we can take a sequence$c_n \in C$such that$c_n\to \inf C$. For every$n$one has, by definition, ... 1 Your operator$B$is not compact. 2 The only homomorphic images of a field$F$(when it is considered as a ring) are the field itself and the zero ring$\{0\}$. If we consider the ring$(\mathbb{Q},+,\cdot,-,0,1)$we see that$\mathsf{SH}(\mathbb{Q})$will just contain$\{0\}$and all the subrings of$\mathbb{Q}$. Note that every subring of$\mathbb{Q}will have characteristic 0. If we ... 1 It's easy to construct a counterexample. Consider, e.g. $$A=\pmatrix{1&1-\varepsilon&0\\ 1-\varepsilon&1&1-\varepsilon\\ 0&1-\varepsilon&1} =\text{entrywise maximum of } B \text{ and } C,$$ where $$B=\pmatrix{1&\varepsilon&0\\ \varepsilon&1&1-\varepsilon\\ 0&1-\varepsilon&1}, \ ... 1 You can just see it as an identity: the shift operator can be expressed in terms of a Taylor series, and then we just compute its closed form. There are other visualizations for this, though. You can think of 1 + \frac{d}{dx} as an infinitesimal shift operator, and exponentiation accumulates all of the infinitesimal shifts up into an actual shift. In ... 2 So you have suggested e^{t \frac d{dx}} f(x) = f(x+t). This is to be expected, because you would formally expect (i) e^{0 \frac d{dx}} to be the identity operator, (ii) \frac d{dt} [e^{t \frac d{dx}} f(x)] \big|_{t=0} = f'(x), and (iii) e^{t \frac d{dx}} e^{s \frac d{dx}}e^{(s+t) \frac d{dx}}. And look, the formula you propose works. Look here for ... 0 Note that$$(\lambda I - A)^{-1} = \dfrac{1}{\lambda} I + \lambda^{-1} A (\lambda I - A)^{-1}$$The obvious bound \|\int_C g(\lambda) \; d\lambda\| \le \text{length}(C) \max_C \|g(\lambda)\| will do for the second term. For the first, use Cauchy's theorem (either the "outside-the-circle" version, or transform z = 1/\lambda and use the usual version). 2 Your idea I thought about constructing some \hat{T}\colon X\times Z\to Y such that is linear, continuous and surjective is the right one. To make things a little easier to see, we assume that T is injective (by considering \tilde{T}\colon X/\ker T \to Y, we lose no generality). Since Y/\mathcal{R}(T) is finite-dimensional, we can choose Z to ... 2 Projections are characterized by being bounded and idempotent. The dual of P is bounded and also idempotent, as can be seen by dualizing the identity P^2=P. 1 By construction, every \alpha_k is an eigenvalue of T, hence we have$$K = \overline{\{\alpha_k : k\in\mathbb{N}\}} \subset \sigma(T).$$On the other hand, if \lambda\notin K, then there is a \delta > 0 such that \lvert\lambda - \alpha_k\rvert > \delta for all k, and then we find the inverse of \lambda I - T explicitly:$$\begin{align} ... 1 General helpful little fact IfA$is a bounded operator and there is an infinite dimensional subspace$M$on which$A$has a lower bound$c>0$, i.e., $$\|Ax\|\ge c\|x\|\quad \text{ for all }x\in M \tag{1}$$ then$A$is not compact. (Proof: pick a bounded uniformly separated sequence in$M$; its image under$M$has the same properties.) In your ... 1 What you have proved is $$\langle \psi_n | U^\dagger U -I | \psi_m\rangle = \delta_{m,n}.$$ Since the$\psi_n$are a basis this shows$U^\dagger U -I=0$, or equivalently$U^\dagger U =I$. 0 Obviously, $$\lim_{\|z\|\to\infty}\sum_i \alpha_i \exp(-||x_i-z||^2)x_i=0$$ (zero vector) and for$R$large enough $$f:B(0,R)\longrightarrow B(0,R).$$ By Brouwer, your equation will have always some solution (a fixed point of$f$). 1 Yes it does. As$M_n$is simple, both homorphisms are monomorphisms. If we replace$A$,$B$with$\pi_1(M_n)$,$\pi_2(M_n)$respectively, we may assume that$\pi_1$and$\pi_2$are isomorphisms. In particular, from$\pi_1(U)\in A_+$we get that$\sigma(U)=\sigma(\pi_1(U))\subset[0,\infty)$. But then$\sigma(U)=\{1\}$, and$U=I$. With a similar reasoning ... 0 The definition of the essential spectrum does not ask for any specific property of the operator. It is just the spectrum of the image of the operator in the Calkin algebra. More explicitly,$\sigma_{\rm ess}(T)$consists of those$\lambda\in\sigma(T)$such that there exists a sequence$\{x_n\}\subset H$with no convergent subsequence and such that ... 2 Instead of translation of a function$f(x) \to f(x+a)$, let us consider scaling. This means that we are going to make intervals of the independent variable smaller, or larger, with a factor$\lambda > 0$. The transformed function is then defined by: $$f_\lambda(x) = f(\lambda\,x)$$ Like with translations, it would be nice to develop the function ... 2 In a Banach space (or indeed in any normed space), the topology has a base consisting of open balls, i.e., sets of the form$\{x:\Vert x-a\Vert<\epsilon\}$, and these are convex because of the triangle inequality. 2 It is equivalent to define a space$X$to be locally convex if there is a collection$\{p_{\alpha}\}$of seminorms on$X$so that the topology on$X$coincides with the initial topology of the seminorms; in particular, the sets $$\bigcap_{i = 1}^n \{p_{\alpha_i}(x) < \epsilon\}$$ form a base for the neighborhoods of zero. In the case of a normed space, ... 2 The relevant consequence of the Hahn-Banach theorem(s) here is that in a normed space$X$, for every convex subset$M$, the weak closure and the norm-closure coincide. Since the norm topology is finer than the weak topology, we always have$\operatorname{cl}_{\lVert\cdot\rVert}(M) \subset \operatorname{cl}_w(M)$, and the Hahn-Banach theorem tells us that we ... 1 Hint: For a projection$P$, we have $$\ker P = \operatorname{im} (I - P).$$ 0 Please let me know if this (doesn't) makes sense. I could stand to think about it a little more carefully. Say you have a Cauchy sequence,$T_k$, of Hilbert-Schmidt operators. Then the the sequence of sequences $$(T_1(e_1), T_1(e_2), T_1(e_3),...) , (T_2 (e_1), T_2(e_2), T_2(e_3),...), ...$$ is a Cauchy sequence in the Banach space $$\big(H^\mathbb{N}, ... 0 we can take q_2=p_1\vee p_2-p_1, and so on, then you check this works. Add: Set q_1=p_1, q_i=p_1\vee \cdots\vee p_{i}-p_1\vee \cdots\vee p_{i-1}, 1<i<m, p_m=1-p_1\vee \cdots\vee p_{m-1}. Then use the fact p\vee q-p\sim q-p\wedge q so they have the same trace to get an upper bound for ||p_i-q_i||_2. 0 For your question about projections, I think it is true even in an arbitrary C^*-algebra. Such results are often called "perturbations". You can find a good account on Section III.3 in Davidson's "C^*-algebras by example". I cannot make sense of your second question: u=I and f(c)=c will always work. 1 You can take a look at this book by Doran and Belfi. The proof you are looking for takes most of chapter 3, some 15+ pages. Unfortunately, not all of it is shown in Google Books. On a brief historical note, the question you are asking was asked by Gelfand and Naimark in 1943. Glimm and Kadison proved it for unital algebras in 1960, using deep results. The ... 1 Let me outline the proof that the Daugavet property of X^* implies the Daugavet property of X. First note that if T:X\to X has rank 1, its adjoint T^*:X^*\to X^* also has rank 1. Indeed, since T has rank 1, there are f \in X^* and z \in X such that Tx=f(x)z. A short calculation shows that then T^*y^*=g(y^*)f, where g(y^*)=\iota ... 1 I don't believe your proof is complete. You have shown that the subspace \mathcal{M} consisting of all absolutely continuous functions f\in L^{2} with f(0)=f(1)=0 is contained in \mathcal{D}(T^{\star}), and that T^{\star}=-\frac{d}{dt} on \mathcal{M}. In other words, if T_{\mathcal{M}}=-\frac{d}{dt} on \mathcal{D}(T_{\mathcal{M}}), then you ... 0 If f and g are polynomials in z, \overline{z}, then the result is fairly clear. See if you can generalize from there. You're going to have to be precise about what you mean by f\circ g for Borel functions. 0 It depends on the concept of operator with which you operate. :) If an operator is required to be defined on the entire space L_p then yes, the identity isn't an operator at all. But it is also common to consider densely defined operators, which may or may not be bounded. The domain of such an operator acting on L^p is a dense linear subspace of ... 0 Take for instance f(x)= \min(x, 1-x). Then f has the properties you are asking for. 0 For one direction, if we start by assuming \|xx^*\| = \|x\|^2 then as C^*-algebras are Banach algebras we get \|x\|^2 = \|xx^*\| \le \|x\|\|x^*\| so that \|x\| \le \|x^*\|. Then \|x\| \le \|x^*\| \le \| (x^*)^*\| = \|x\| so that \|x\| = \|x^*\|. It then follows that \|xx^*\| = \|x\|^2 = \|x\|\|x^*\|. The other direction is a fair bit tricker ... 2 For every \varphi \in D(T^\ast), we have T^\ast\varphi \in L^2(\mathbb{R}), by definition of the adjoint. Now, we saw that$$T^\ast\varphi = (\psi_0,\varphi)\cdot f$$is a constant multiple of f - this part, "Thus T^\ast\varphi = (\psi_0,\varphi)f", needs a little justification, I'll do that below. But a constant multiple of f is in ... 1 I misinterpreted your question. I thought you wanted to see the standard construction of the V where V^{\star}V=P is the orthogonal projection onto \mathcal{R}(|T|)^{c}. This V is unique in all cases, even where \mathcal{N}(T)\ne \{0\}, whereas a general partial isometry V is not uniquely determined if T has a non-trivial null space. V is ... 1 The notation |T| means the unique positive square root of T^{\star}T. Notice that$$ \||T|x\|^{2}=(|T|x,|T|x)=(|T|^{2}x,x)=(T^{\star}Tx,x)=(Tx,Tx)=\|Tx\|^{2}. $$So \mathcal{N}(T)=\mathcal{N}(|T|). Let Y=\mathcal{R}(|T|)^{c}. Then |T| : Y\rightarrow Y is invertible because \mathcal{N}(|T|)\cap ... 1 We have T=V|T|. Left multiply both sides by V^{*} to get V^{*}T=V^{*}V|T|. Since V is a partial isometry on the closure of the range of |T|, V^{*}V=I on Im |T| whence V^{*}T=|T|. 1 I think I have answered this for you before, but I'll explain it again in more detail. The proof hinges on the following lemma, whose proof is given at the end. Perhaps you'll accept this answer. Lemma Let X be a complex Hilbert space and let T\in\mathcal{L}(X) satisfy (Tx,x) \ge 0 for all x \in X. Then (Tx,x)=0 iff Tx=0. That is, ... 0 Your computation is correct Yes, A is self-adjoint. More generally, if (\mu_n) is any bounded sequence of real numbers, then the multiplication operator (x_n) \mapsto (\mu_n x_n) is self-adjoint. This is the infinite-dimensional analogue of diagonal matrix with real entries. In your example, \mu_n alternates between 0 and 1. Put more ... 1 At some places you seem to be forgetting the coefficients. You have, using Cauchy-Schwarz,$$ \left|\sum_{k=1}^n\frac{\xi_k}k\right|\leq\left(\sum_{k=1}^n|\xi_k|^2\right)^{1/2}\left(\sum_{k=1}^n\frac1{k^2}\right)^{1/2}\leq\left(\sum_{k=1}^\infty|\xi_k|^2\right)^{1/2}\left(\sum_{k=1}^\infty\frac1{k^2}\right)^{1/2}=\|u\|\,\frac\pi{\sqrt 6}, $$so \|Tu\|\leq ... 0 Also, strong continuity of \pi is equivalent to continuity of (g,v)\mapsto \pi(g)v (see for example page 23 here: http://www.math.harvard.edu/~jbland/ma222_notes.pdf). 1 This statement is not true in general. Consider the following elliptic pde:$$ -\Delta u + \epsilon \chi_{K_\epsilon} u = f$$plus homogeneous Dirichlet-boundary conditions. Here,$K_\epsilon =\Omega \cap B(x_0, \epsilon^{-3/2d})$,$x_0\in\Omega$. Set$A_\epsilon:=-\Delta + \epsilon \chi_{K_\epsilon}I$. The inverses are uniformly bounded from$H^{-1}\$ to ...

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