Tag Info

1

Fourier's Treatise on heat conduction was first submitted in 1807. In that work he introduced the method of separation of variables, and various expansions in orthogonal functions. Fourier died in 1830. Gauss looked at Potential Theory at roughly the same time. By 1830, Gauss was heavily involved in the study of Electromagnetism. He gave a minimization ...

4

Dominated convergence and related theorems don't apply here anyway. The norm here is almost certainly the supremum norm since this is the norm that makes $C([0,1])$ a complete space. Also, you'll need to to know something about $K$ in order to do this. For example, if $K$ is continuous the following argument works: what you need to do is find a $C > 0$ ...

1

Any one-to-one quasinilpotent operator will do (quasinilpotent are the operators $T$ with $\sigma(T)=\{0\}$); the Volterra operator from user3808066's answer is one example. Any finite set $K\subset\mathbb C$ can be realized this way: if $K=\{k_1,\ldots,k_n\}$ and $T$ is the Volterra operator, then $$\bigoplus_{j=1}^nT+k_jI$$ has spectrum ...

1

For hermitian $T$ you have $\ker(T) \cap \rm{im}(T) = \{0\}$, as $\langle x,x \rangle = 0 \iff x=0$, so if $x = Tz$ and $x \in \ker(T)$ you have $\langle x, x\rangle=\langle x, Tz\rangle=\langle Tx,z\rangle = 0$. Now for a positive bounded operator $T$ on a Hilbert space you have a root $T^{1/2}$ that is also positive. Then $\langle x, Tx\rangle=\langle ... 1 A counter example is the Volterra operator, it is a bounded linear operator between Hilbert spaces, with no eigenvalues while the spectrum is$\{0\}$. It is defined as $$V:L^2(0,1)\to L^2(0,1), f\mapsto \left(t\mapsto \int_0^t f(x)\mathrm{d}x\right).$$ 1 Martin gave a beautiful answer to your question but let me take this opportunity to advertise a solution to a more general question of which C*-algebras may be written as tensor products of two infinite-dimensional C*-algebras. In particular, it is proved that$B(\ell_2)$as well as the Calkin algebra cannot be decomposed like that no matter which tensor ... 1 This depends on the domain and range of the the operator$A$. For a linear operator$A : X \to Y$, where$X$and$Y$are normed spaces, one can show that if$A$is bounded and of finite rank, i.e.,$\dim A(X) < \infty$, then the operator$A$is compact. So, if you consider normed spaces, then your reasoning is fine. 0 Your proof is correct. Here is a slightly shorter way to do the last step (once you have$\langle Tx,Ty\rangle=\langle x,y\rangle$). Suppose$T$is surjective. Then it is bijective, since any isometry is injective. Now note that for any$x,y\in H$, $$\langle Tx,y\rangle=\langle Tx,T(T^{-1}y)\rangle=\langle x,T^{-1}y\rangle.$$ By definition of adjoints, ... 4 Suppose you have a non-trivial solution of $$T^*Tf = \int_{t}^{1}\int_{0}^{s}f(y)dy ds = \lambda f(t)$$ Then$\lambda \ne 0$because the above would give$f=0$after differentiating a couple of times. For$\lambda \ne 0$, any solution of the above must satisfy $$\lambda f'' = -f \\ f(1)=0,\;\; f'(0)=0.$$ Any ... 0 I'm not entirely sure what you are asking, maybe what I am writing does not give you what you are looking for. But there is essentially no difference from the finite dimensional case. If you write$v=\sum_i v_i e_i$then $$L_n(e_i)=\langle e_i, v\rangle = v_i$$ And you have$v=\sum_i L_n(e_i)\ e_i$, which is the same as your equation$\nabla f_n(x) = v$. ... 0 If$\lambda=q(s)$for some$s\in[0,1]$, then$T_q 1_s=q 1_s=q(s) 1_s=\lambda 1_s$. Thus,$\lambda\in \sigma(T_q)$. As the spectrum of an operator is closed, it follows that$\overline{\{q(t)\mid t\in[0,1]\}}\subset \sigma(T_q)$. If$\lambda\notin\overline{\{q(t)\mid t\in[0,1]\}}$, let$\phi=\frac 1{q-\lambda}$. Since$\lambda\notin\overline{\{q(t)\mid ...

6

The proof has nothing to do with the Schwartz space per se; nor with $i$ or $t$, or that $P$ and $Q$ are symmetric. If $P,Q$ are operators on a Hilbert space $H$ with domain $D$ and such that $PD\subset D$, $QD\subset D$, and $QP-PQ=\mathbb I$, then at least one of $P$ and $Q$ is unbounded. This applies to the case in the question because if we have ...

1

You have already done all the necessary work: $$u=\begin{bmatrix}\begin{matrix}u_1&&&\\&\ddots&&\\&&u_j&\\ &&&I_d\end{matrix} \end{bmatrix}.$$

1

Let's just try to find an eigenfunction for a given $|\lambda|< 1$: we want an $f$ that solves $$f\left( \frac{x+1}{2}\right) = 2\lambda f(x)- f(x/2) . \quad\quad\quad\quad (1)$$ Start out with an arbitrary (integrable) function on $(0,1/2)$, then use (1) for $0<x<1/2$ to define $f(t)$ for $1/2<t<3/4$, then reenter (1), with ...

1

A priori you cannot guarantee that $S$ will map $X_1$ into $X_2$. The necessary condition is also sufficient : the restriction of $S$ to $X_1$ will be invertible if and only if $X_2=SX_1$.

1

Let $f: \mathbb{R}^n \times l_2 (\mathbb{R} ) \to l_2$ $$f((x_1 ,x_2 , ..., x_n ) , (y_i )_{n\geq 1} ) = (x_1 ,x_2 , ..., x_n , y_1 , y_2 , ...)$$ then $f$ is injective.

4

When dealing with conjectures about unbounded operators, it's always good to test conjectures with a differential operator. John von Neumann defined closed unbounded operators to study differential operators. Differential operators are still the best examples. For example, let $X=C[0,1]$, and let $A=\frac{d}{dx}$ on the domain $\mathcal{D}(A)$ of ...

1

Every norm on $R^d$ are equivalent, thus you can suppose that the norm is $\|x\|=\sup_{i=1,..,n} \mid x_i\mid$ where $x_i$ are the coordinates of $x$ in the basis $(e_1,...,e_n)$. You have $\|A(x)\|\leq \mid x_1\mid \|A(e_1)\|+...+\mid x_n\mid \|A(e_n)\|\leq sup \mid x_i\mid_{i=1,..n} sup_{j=1,..n}\|A(e_j)\|\leq Sup_{j=1,..,n}\|A(e_j)\|\|x\|$.

0

To see that the linear map $\iota$ defined by (1) is injective, first notice that for any representation $u=\sum^n_{j=1}\phi_j\otimes y_j$ of $u\in X'\otimes Y$ one may assume that the $y_j$'s are linearly independent by simply redefining the $\phi_j$'s. Once this is done, we have that $\iota(u)=0$ implies that $\phi_j=0$ for each $j=1,\ldots,n$ and ...

2

Well, we have that the bilinear functional $\langle K,T\rangle_\mathrm{HS}$ defined via $$\langle K,T\rangle_\mathrm{HS}:=\int_0^1\int_0^1k(x,y)\overline{t(x,y)}\mathrm dx\mathrm dy,$$ is a scalar product on the space of Hilbert-Schmidt integral operators. It further induces a (the mentioned) norm by acknowledging that $z\cdot \overline z=|z|^2$ $$\langle ... 1 You have$$ (\lambda I-T)(x_1,x_2,\ldots)=((\lambda-1)x_1, (\lambda-\frac12)x_2,\ldots). $$Define an operator S  by$$S (x_1,x_2,\ldots)=(\frac{x_1} {\lambda -1},\frac {x_2}{\lambda -\frac12},\ldots). $$By the choice of \lambda  the linear operator S  is well-defined and bounded: by construction, S (\lambda I-T)=(\lambda I-T)S=I . So S=(\lambda ... 0 Let H be the subspace spanned by the orthonormal set \{1,e^{ix},e^{2ix},\cdots\} in L^2[0,2\pi] with the inner product$$ (f,g)=\frac{1}{2\pi}\int_{0}^{2\pi}f(t)\overline{g(t)}dt $$Consider the operator$$ T = -i\frac{d}{dx}(e^{ix}f(x)). $$Then T(e^{inx})=(n+1)e^{i(n+1)x} for n=0,1,2,3,\cdots. So this is the same as ... 1 The elements of X'\otimes Y \subseteq \mathcal L(X,Y) all have finite rank and, in particular, are compact operators. There is thus no chance for equality if X and Y are infinite dimensional. If you take the closure of the tensor product in \mathcal L(X,Y) you get (by definition) the approximable operators. This closure often coincides with all ... 0 Let R be the projection onto the range of PQ. Then, by page 116 of Kadison Ringrose' book.$$R:=P-P\wedge (I-Q). $$Then, R \sim P\vee (I-Q) - (I-Q). Then, \tau(R) =\tau(P\vee (I-Q) - (I-Q)) =\tau(Q) - \tau(I- P\vee (I-Q) ) \le \tau(Q)<\tau(P). Then, (P-R)(H) is not empty and it is easy to check P-R is Q^\perp \wedge P . 2 For 1, start with any bounded sequence \{ f_n \} in C[0,1] that has no convergent subsequence. Then F_n\int_{0}^{x}f_n(t)dt gives you a bounded sequence \{ F_n \} \subset C^1[0,1] whose image under T is \{ f_n \}. For 2, the same technique holds. Start with a bounded sequence in C^1[0,1] with no convergent subsequence, and integrate. For ... 0 Edit: First we recall the definition: T is a trace-class operator if$$||T||_1:=tr((T^*T)^{1/2})<\infty.$$We show that H\otimes H is dense in the trace class, with respect to that norm. Note that the definition is really all we're going to use about the trace class. In particular: It's not clear to the OP why a trace-class operator actually has a ... 0 Let E\subseteq H be the closure of Q(P(H)). Then Q is a bounded linear map from P(H) to E with dense image, and it follows that \dim E\leq \dim P(H) (if P(H) is finite-dimensional this is just linear algebra; if P(H) is infinite-dimensional, note that its dimension is the same as the minimal cardinality of a dense subset, and then the image ... 3 If you're working strictly within the context of an inner product space or Hilbert space, the statement you have learned is false. Even if you could identify certain "states" that you would call "eigenstates," those objects are not vectors in the space. They are some unknown, unspecified object in some unknown and extended space. For example, start with the ... 0 What you seem to be looking for is a proof of the first isomorphism theorem. Let \{k_i\}_{i \in I} be a basis of \ker T. This extends to a basis \{k_i\}_{i \in I} \cup \{x_j\}_{j \in J} of X (where \operatorname{span} \{k_i\}_{i \in I} \cap \operatorname{span} \{x_j\}_{j \in J} = \{0\}). Hence \{ x_j + \ker T\}_{j \in J} is a basis of X / (\ker ... 1 In a previous problem of yours, I considered L : L^2[0,1]\rightarrow L^2[0,1] given by$$ Lf = \int_{0}^{x}f(t)dt. $$This is a quasinilpotent operator, i.e., \sigma(L)=\{0\}, and \mathcal{N}(L)=\{0\}. The range of L is dense in L^2[0,1]. With this operator, let$$ H = L^2[0,1]\times L^2[0,1]. $$and define T : H \rightarrow ... 0 If you remember the Theorem of dimensions for linear transformation between finite dimension vector spaces, then the proof is quite similar. Maybe the equality$$\dim X = \dim \ker{(T)} + \dim \operatorname{Im}{(T)}$$does not make sense since the dimension of X could be infinite. But the quotient$$\frac{X}{\ker{(T)}}\cong \operatorname{Im}{(T)}$$... 3 If Tf = \lambda f for some unit vector f\in L^2, then \|Tf\|=\|f\| implies |\lambda|=1. Therefore,$$ f(x+n)=\lambda^n f(x),\;\;\; n=1,2,3,\cdots. $$But that's a problem because it means that$$ \int_{k}^{k+1}|f(x)|^2dx = \int_{l}^{l+1}|f(x)|^2dx,\;\;\; k,l=1,2,3,\cdots. $$It's a problem because such f could not be in L^2 ... 1 Let \mathbb T be the circle and \mathbb D the disc. The canonical unitary in C(\mathbb T) does not lift to a partial isometry in C(\mathbb D): such a lift would automatically be unitary, violating Brouwer's fixed-point theorem. The following is true, however (see Lemma 9.2.1 in Rordam-Larsen-Laustsen): Let A be unital and u\in A/I a unitary. ... 2 If A is a normal operator (and every selfadjoint operator is normal,) then the Spectral Theorem for normal operators gives you a way to define F(A) for any function F that is continuous on the spectrum \sigma(A) of A. And, this correspondence preserves algebra, meaning$$ (F+G)(A) = F(A)+G(A),\;\;\;(FG)(A)=F(A)G(A),\;\;\; 1(A)=I. $$And ... 1 Consider \psi_r(ta+x)=rt + \phi_0(x) instead. Then check that r\mapsto \|\psi_r\| is continuous on [0,\infty) and apply the intermediate value theorem to find some r>0 with \|\psi_r\|=2 and extend by Hahn-Banach. 5 The claim you are trying to prove is the following statement. Proposition. Let H be a separable Hilbert space and let (T_n)_{n=1}^\infty be a sequence of operators in B(H) which converges to some T in the weak operator topology, i.e.,$$\langle T_nx, f\rangle \to \langle Tx, f\rangle$$for all x,f\in H. Then there is a sequence ... 3 It can be anything greater than 1, take for instance$$ A=\begin{bmatrix}0&x\\1/x&0\end{bmatrix} $$4 If A is symmetric (Hermitian), then that's true (assuming you're talking about the induced Euclidean norm). However, in general, \|A\| can be very large. For example, take$$ A=\pmatrix{1&t\\0&-1} $$with t as large as you want to make it. Note that for any operator on a Hilbert space, we have \|A^*A\| = \|A\|^2 (you should have this as a ... 3 Let H=\ell^2(\mathbb{N}) and let \mathcal{A} be the set of all self-adjoint elements in the unit ball of B(H). Define S_n\colon H\to H, S_n \xi(k)=\xi(k+n). Then \|S_n\|=1, hence A_n:=\frac 1 2(S_n+S_n^\ast)\in \mathcal{A}. The adjoint of S_n is given by$$ S_n^\ast\xi(k)=\begin{cases}\xi(k-n)&\colon k\geq n\\0&\colon k<n\end{cases} ...

0

In general you can not obtain $f\in D(A)$ for each $b:\Omega\rightarrow\mathbb{R}$. Consider $f_k(x)=\sum_{l\neq i}h_k(x_l)$, where $h_k$ is a smooth sequence which converges to a non smooth function in $L^2$ in one dimension subspace. This is possible, for instance using mollifier. Then $f_k$ converges to a function in $L^2\setminus H^1$, and you still ...

1

Let us assume that $K$ is a vector space and $A$ is linear and injective (this is needed anyway in order that you have a metric). Then, $K$ equipped with your metric (or, equivalently, with the corresponding norm) is complete if and only if the image of $A$ is dense in $L^2$. In fact, $A \colon K \to \operatorname{image}(A)$ is a bijective isometry. Hence, ...

0

Answer for (1): Using the Poincare-Wirtinger inequality, there is $C>0$ so that (write $u = u(f)$) $$\tag{1} \| u\|_2 = \| u - u_D\|_2 \le C\| \nabla u\|_2,$$ where $u_D := \int_D u$ is assumed to be zero here. Then we have $$\begin{split} \|\nabla u\|_2^2 &= \int_D |\nabla u|^2 dx \\ &= \sum_{i=1}^n \int_D u_i u_i dx \\ &= ... 2 The arguments I know require that \Omega is both cyclic and separating (i.e., also cyclic for the commutant). When you only require that \Omega is also separating, I don't think that S_0 even makes sense. For instance, let \mathcal M=B(\mathcal H), \Omega=e_1. Consider the (constant) sequence E_{12}e_1=0. Then S_0E_{12}e_1=E_{21}e_1=e_2\ne0, ... -1 This is not an answer for the question but I want to comment on part (i). It seems to me that your estimate$$\lVert Tx\rVert_{\ell^2}\leqslant 2\lVert x\rVert_{\ell^2}$$is too rough. My calculation shows that$$\lVert Tx\rVert_{\ell^2}\leqslant \lVert x\rVert_{\ell^2}$$and it turns out that \|T\|=1 since \|T\| \geqslant \sqrt{1-\frac 3{4n}} for any n ... 1 Assuming N is a strict contraction you can show I+N is surjective by the usual method; linearity doesn't matter. Fix x\in\Bbb R^n. You want to show that there exists y with$$y+Ny=x.$$That's the same as$$x-Ny=y,$$which says$$Ty=y,$$where$$Ty=x-Ny.$$It's easy to show T is a strict contraction and you're done. 1 Fix x_0 \in R^d. As you have noticed, for every x \in R^d and for every k = (k_1,...,k_d) \in R^d,$$Du(x)\cdot k = \sum_{i=1}^d u_{x_i}(x)k_i.$$Now, for each i=1,...,d and h=(h_1,...,h_d) \in R^d, let us write$$u_{x_i}(x_0+h)-u_{x_i}(x_0) = \nabla u_{x_i}(x_0)\cdot h + r_i(h),$$where r_i(h) = o(\|h\|), as \|h\| \rightarrow 0. For ... 3 Its not quite correct; you have computed a directional derivative. You need that$$ |F(\phi + x) - F(\phi) - (A\phi,x)-(\phi,Ax)| = o(‖x‖_{H}) $$Which is indeed true by Cauchy-Schwarz,$$|F(\phi + x) - F(\phi) - (A\phi,x)-(\phi,Ax)| = |(x,Ax)| \leq ‖A‖_{\text{op}} ‖x‖_H^2 If $f:\Bbb R→\Bbb R$ is a differentiable function, there is indeed a chain rule, ...

1

I'm assuming that the statement in your question is that "for every $b\in B^+$ there exists a $\varphi\in Y$ such that $\|b\|=\varphi(b)$". The positivity condition is necessary, because otherwise it may be impossible to have $\varphi(b)$ positive. We can assume that $B$ is unital, since the state space of the unitization agrees with the state space of $B$. ...

4

Start with an orthonormal set $\{ f_n \}_{n=1}^{\infty}$ in $L^2[1/2,1]$, and extend the functions to be $0$ on $[0,1/2]$ in order to obtain an orthonormal set $\{ \tilde{f}_n \}_{n=1}^{\infty}$ in $L^2[0,1]$. Then, for $n\ne m$, \begin{align} \|T\tilde{f}_n-T\tilde{f_m}\|^2 & =\|x\tilde{f}_{n}-x\tilde{f}_m\|^2 \\ & ...

Top 50 recent answers are included