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You can prove this using the following observation: Let $T$ be a bounded linear operator on a Hilbert space. Then $T$ is a partial isometry if and only if the spectrum of $T^{\ast} T$ consists only of $\{0,1\}$. Furthermore, let $A$ be a bounded, self-adjoint operator. Then $\sigma(A) \subset \{0,1\}$ if and only if $A = A^2$; this follows from the spectral ...

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Here is a proof from scratch, in particular, without the Lax-Milgram. Lemma. Let $\widetilde H$ be a real vector space with positive definite inner product (completeness not assumed). Suppose that $K$ is a subspace of $\widetilde H$ which is complete. Then any element $v\in \widetilde H$ can be written as $v_1+v_2$ where $v_1\in K$ and $v_2\in K^\perp$. ...

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Here is the answer to my own question. Please feel free to comment. Any suggestion is welcome: Let $\widetilde{H}$ be the closed subspace generated by $D$ and the one-dimensional subspace of $H$. Since $B$ is positive definite, there exists a one-to-one bounded symmetric operator $L$ on $\widetilde{H}$ such that $B(u,v)=\langle Lu,\,v\rangle$. Let $y\in ... 1 (A2) is correct. Note that it assumes that$A$be injective. And that$X$be a normed vector space suffices. For (Q1), a natural family of counterexamples is given by non surjective isometries on a Banach space, since their ranges are closed proper subspaces. For instance, the unilateral shift operator$S:\ell^2(\mathbb{N})\longrightarrow ...

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Yes. If $\varphi$ is a holomorphic map of unit disk into itself, the composition operator $f\mapsto f\circ \varphi$ is bounded on $H^p$. In fact, $$\|f\circ \varphi\|_{H^p}\le \left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^{1/p}\|f \|_{H^p} \tag1$$ Original source: John V. Ryff, Subordinate $H^p$ functions, Duke Math. J. Vol. 33, no. 2 (1966), ...

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Suppose $X = Y$ and $T$ be the Fourier transform on the Schwartz space of functions. If $X = Y$, it is trivially dense in Y and the Fourier transform is bounded on the Schwartz space with the $L^2$ norm (and so is its inverse).

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Counterexample. Let $\lambda\in c_0$ with non zero entries. Consider diagonal operators $$T:\ell_1\to\ell_1:x\mapsto\lambda x\\ S:\ell_\infty\to\ell_\infty:x\mapsto\lambda x$$ This is straightforward to check that $S=T^*$. Obviously $\operatorname{Ker}(T)^\perp=\{0\}^\perp=\ell_1^*=\ell_\infty$. Since $\lambda\in c_0$ then $\operatorname{Im}(S)\subset ... 4 We want$L^*$such that $$\langle f \vert Lg \rangle = \langle L^*f \vert g \rangle$$ We have $$S=\int_0^{\infty} e^{-x}f(x)\left(x+\dfrac{d}{dx}\right)g(x) dx = \int_0^{\infty}x e^{-x}f(x) g(x) dx + \int_0^{\infty} e^{-x} f(x) dg(x)$$ We have $$\int_0^{\infty} e^{-x} f(x) dg(x) = \int_0^{\infty} d\left(e^{-x} f(x)g(x)\right) - \int_0^{\infty} g(x) ... 1 A more general question has been asked and answered on here before. The bounded operators on a Hilbert space form a Banach Algebra. A beautiful argument of H. Wielandt shows that the equation AB-BA = I is not possible in any normed algebra. I won't repeat it here, but it follows by induction that if AB -BA = I, we have AB^{n}-B^{n}A = nB^{n-1} for all ... 0 There are a number of comments which might be appropriate here. Firstly, the general rule of thumb is that you require n boundary conditions to specify a self-adjoint operator for a differential operator of order n (in the sense of unbounded operators in Hilbert space), not the 2 n given here. In fact, if one considers the formal differential ... 3 Since \|T^{-k}\| \leq M for all k \geq 1, we have by definition \|T^{-k}(x)\| \leq M\|x\|. \|x\| = \|id(x)\| = \|T^{-n}(T^{n}x)\| \leq \|T^{-n}\|\|T^n x\| \leq M\|T^n x\|. It seems we are done. 1 You already know the answer: derivatives of odd orders are anti-symmetric, not symmetric. But if you want a concrete example, take something like f(x)=(x-a)^{2n+1}(b-x)^{2n+1} and g=f^{(n)}. (Check that these functions satisfy the boundary conditions). Then \int_a^b f^{(n)}g>0 and \int_a^b fg^{(n)}=-\int_a^b f^{(n)}g<0, so these integrals are ... 1 Dirac won a Nobel Prize for finding the square root of a pesky differential operator while working on the relativistic Schrodinger equation. http://www.youtube.com/watch?v=zM-Lc16nyho. <--- That video gives a clear overview of the hacks Dirac used. 1 It relies on several things. First rewriting the derivative as limit:$$ \int^b_a \lim_{\epsilon\to 0} \frac{\frac{\partial }{\partial x}u(x,t+\epsilon) - \frac{\partial }{\partial x}u(x,t) }{\epsilon} dx\tag{1} $$The question is: whether above is the same as$$ \lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(\int^b_a \frac{\partial }{\partial ... 3 I'm not in position to explain any$K$-theory. But it is not hard to see that$M_n(K(H))\simeq K(\bigoplus_1^nH)$canonically. So you can see matrices of compact operators as compact operators on a bigger Hilbert space, and you can calculate its trace if it is trace-class. More concretely, if you have a matrix of operators$T=(T_{kj})_{k,j}\in M_n(B(H))$, ... 2 1) There exist bijection between bounded bilinear operators and bounded opeartors. The proof of this fact requires Banach-Steinhaus theorem. If bilinear form is symmetric, then the respecitive opeartor is (obviously) symmetric too 2) Spectrum of any bounded operator is compact, and as the consequence closed 3) No, consider bilinear form $$... 1 You need to specify the domain of L more precisely, as the spectrum depends on it very much, e.g. the resolvent set is empty unless L is a closed operator. I assume henceforth that \text{dom}L = \{f \in L^{2}[0,\infty): f' \in L^2[0,\infty)\}. In that case, if I haven't made any mistake, \text{dom}L^{\ast}=\{f \in L^{2}[0,\infty): f(0)=0, f' \in ... 0 Your integral kernel is actually separable, since \cos(x+t)=\cos(x)\cos(t)-\sin(x)\sin(t). This should make the problem significantly easier to handle, i.e. you can solve it explicitly without the Neumann series. 1 Denote Y=\operatorname{T_\lambda}, then Y is a proper subspace of H. By projection theorem Y^\perp\neq \{0\}, so you can choose y\in Y^\perp\setminus\{0\} which is by definition of Y is orthogonal to Y, i.e. to \operatorname{Im}(T_\lambda) 3 From Gelfand's formula, which holds for every element of a unital complex Banach algebra more generally,$$ \lim_{n\rightarrow +\infty}\|A^n\|^\frac{1}{n}=0\quad\iff\quad \rho(A)=0. $$This characterizes the set of quasinilpotent operators, which extends the set of nilpotent operators. There is nothing you can say about \rho(A+A^*) without further ... 1 Your argument does not prove the result at this point. For a bounded linear operator, T is self-adjoint (T^*=T) iff (Tx,y)=(y,Tx) for every x,y\in H. So you need to do more. Using the textbook lemma is a good idea. It yields a straightforward argument. The proof of that lemma is no very difficult if you are familiar with polarization. It boils down ... 2 The first, and main, step is to show that T is bounded on L^q. This is done "by duality": the starting point is the identity$$\|f\|_{L^q}=\sup\left\{ \int fg : \|g\|_{L^p}\le 1\right\}$$Apply it to Tf :$$\|Tf\|_{L^q}=\sup\left\{ \int (Tf)g : \|g\|_{L^p}\le 1\right\}$$Now it's time to re-read the assumptions of the problem and decide what to do ... 2 In general, the integral triangle inequality holds for functions f from a measure space \Omega to a normed linear space X. It says that$$\left\|\int_\Omega f \right\|\le \int_\Omega \|f\| \tag1$$One can then follow up with the pointwise inequality \|f\|\le \sup_\Omega \|f\| and conclude that$$\left\|\int_\Omega f \right\|\le \sup_\Omega \|f\| ... 1 The subsequences you took must not have any index in commom: For example, let$K_1(x)=K_2(x)=x$in$\mathbb{C}$(or$\mathbb{R}$, if you prefer) and take$x_n=(-1)^n$. Then$K_1$and$K_2$are compact operators and$\left\{x_n\right\}_n$is a bounded sequence If$n_k=2k$and$n_l=2l+1$, then$\left\{K_1(x_{n_k})\right\}$and$\left\{K_2(x_{n_l})\right\}$... 1 Recall principle of uniform boundedness: If$\{T_n\}$are linear bounded operators from$X$to$X$where$X$is a Banach space, for which$\sup_n\lVert T_nx\rVert<\infty$for all$x$, then$\sup_n\lVert T_n\rVert<\infty$. Here, the sequence$(T_ny,n\geqslant 1)$is bounded for all$y$(as it converges weakly to$0$), which gives the result. 0 Foguel gives here a nice proof of the weak limit case: http://www.ams.org/journals/proc/1965-016-04/S0002-9939-1965-0180862-5/S0002-9939-1965-0180862-5.pdf 1 I'm assuming that$E,F$are Hilbert spaces with some scalar products$\langle \cdot , \cdot \rangle_{E}$and$\langle \cdot , \cdot \rangle_{F}$. Let$\begin{pmatrix} e \\ f \end{pmatrix} \in G^\perp$, i.e.$\left\langle \begin{pmatrix} e \\ f \end{pmatrix}, \begin{pmatrix} x \\ Ax \end{pmatrix}\right\rangle_{E\times F}=0$for all$x\in D(A)$. Using the ... 0 Let$A_n:=T_n-T$. Write $$T_n^p-T_n^p=(T+A_n)^p-T^p=\sum_{\substack{J\subset [p]\\J\neq \emptyset}}P_J,$$ where$P_J$is a product of$p$operators$B_1\dots B_p$, for which$B_j=A_n$if$j\in J$and$B_j=Totherwise. This gives $$\lVert T_n^p-T^p\rVert\leqslant \sum_{\substack{J\subset [p]\\J\neq \emptyset}}\lVert A_n\rVert^{|J|}\lVert ... 1 Yes. Operator multiplication is continuous in operator norm. To see this (if you haven't already), suppose S_n \to S and T_n \to T. In particular, \sup_n \|T_n\| < \infty. Then$$\|S_n T_n - ST\| \le \|S_n T_n - S_n T\| + \|S_n T - ST\| \le \|S_n\| \|T_n - T\| + \|S_n - S\| \|T\|$$and both terms go to 0. Now to show your claim, use induction ... 0 I think i have it. You just have to approximate the interval with Riemann sum where you take arbitrary value in the intervals of the partition... For example,$$ I_n = \int_0^{n}e^{-\lambda t}S(t)udt \approx \sum_{i=1}^k \frac{n}{k} e^{-\lambda t_i}S(t_i)u $$where t_i \in [(i-1)\frac{n}{k},i\frac{n}{k}]. Everything else should be ok. 3 Suppose that \{F_n\} is a sequence of finite-rank operators such that F_n\to I strongly. Note that by the uniform boundedness principle the sequence is bounded, i.e. there exists k>0 with \|F_n\|<k for all n (thanks julien for reminding me of this). I will assume that all F_n are selfadjoint (I need for my estimates, but didn't think if ... 2 Look at some of the hits from a google search for "exponentiation of operators". Among other things, you'll see that exponentials of various differentiation operators, e.g. e^{\frac{d}{dx}}, arise in quantum mechanics. More general than exponentiation of operators are operators of operators defined by various power series expansions of an operator. Also, ... 3 Hint:$$ \Vert T_1(T_2x)\Vert\leq\Vert T_1\Vert\Vert T_2(x)\Vert\leq\Vert T_1\Vert\Vert T_2\Vert\Vert x\Vert 0 if ||A||=0 then it means that ||A(1,0,0,000,0)||=0 so first column of A is zero and you also ||A(0,1,0,000,0)||=0 so scond column of A s zero ,,,,,,,,,,,,,,, ||A(0,0,0,000,1)||=0 so end column of A s zero 0 I have proven it. It is indeed a supremum of numbers, a subset of \Bbb R. We have to prove this: \forall A\ne 0, A\in M_n(\Bbb F), \exists v\in \Bbb F^n: w=Av\ne 0. We can chose vector v to be the first non-zero row of matrix A, then at least one component of w is non-zero, if the matrix is real. If the matrix is complex, we chose vector v to ... 3 This proof is (sort of) from Advanced Linear Algebra by Steven Roman. Let V be a complex inner product space and let A be a linear operator on V with the given property. For all v\in V and r\in\mathbb{C} we have \begin{align} 0 &= \langle A(rv+Av),rv+Av \rangle \\ &= |r|^2\langle Av,v\rangle + \langle A^2v,Av\rangle + r\langle Av, ... 4 Recall that the compact operators form a two-sided ideal in B(X) for any normed vector space X. Writing T=L^{-1}LT shows that every bounded operator is therefore compact. In particular your (L+B)^{-1}. Remark: actually, the existence of a compact invertible operator in B(X) is equivalent to X being finite-dimensional. Indeed, this is equivalent ... 1 Since B is linear, we need only show that B is weakly continuous at zero. Suppose V is a weak neighborhood of 0. Then V contains a set of the form W = \{ x \,|\, |\phi_k(x)| < \epsilon, \ k \in I\}, where I is a finite set of indices and \phi_k \in \mathbb{H}^*. Then we need to find a weak neighborhood U of 0 such that BU \subset ... 0 It would work in any normed space, since for any x, the sequence (F_n(x),n\geqslant 1) converges to both F(x) and 0. 1 Try U_1 = \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}, U_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}. Both are unitary, hence normal. However U_1+U_2 = \begin{bmatrix} 0 & 2 \\ 0 & 0\end{bmatrix}, and (U_1+U_2) e_1 = 0, but (U_1+U_2)^* e_1 = 2 e_2, hence U_1+U_2 is not normal. 0 with attention to ||T||=\inf \bigl\{ k\in\mathbb{R} \ :\ \forall x \in E, \ ||T(x)||_{F} \leq k||x||_{E}\bigr\} k>0 so0\le ||T|| it will be remain to show||T_1+T_2||\le ||T_1||+||T_2||$$||(T_1+T_2)(x)||_F\le ||T_1(x)||_F+||T_2(x)||_F\le( k_1+k_2)||x||_E (k_1 and k_2 are coresspond to inf of defintion of T_1and T_2) so k correspond to ... 0 Note that the assumption of continuity implies that \lVert\cdot\rVert is well-defined. We need to check the axioms. Homogeneity (\lVert aT\rVert=|a|\lVert T\rVert) is a consequence of \inf\{\lambda S\}=\lambda\inf S for \lambda\geqslant 0. Let T\in\mathcal C(E,F) be such that \lVert T\rVert=0. Then for all k>0 and all x\in E, \lVert ... 1 Yes, that’s exactly what it is. And you derive it in the same way, as a consequence of the product rule for the difference operator:$$\begin{align*} \Delta\big(f(x)g(x)\big)&=f(x+1)g(x+1)-f(x)g(x)\\ &=f(x+1)g(x+1)-f(x)g(x+1)+f(x)g(x+1)-f(x)g(x)\\ &=f(x)\Delta g(x)+Eg(x)\Delta f(x)\;. \end{align*}\$

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Take the spectral decomposition of the matrix and use the method stated in the answer of the above question.

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