New answers tagged

2

If I understand the question correctly, if we have any linear operator $A$ for which the exponential $\exp(A)$ is meaningful, then $A$ commutes with $-A$, and so $$ \exp(A) \exp(-A) = \exp(A + (-A)) = \exp(\mbox{the zero operator}) = I. $$ The inverse of $(I + B)^{-1}$ (again, if the latter is defined) is $(I+B)$.


0

Thanks to @user1551, please tell me if you want to post the answer yourself I delete mine. Take $S = k I $ with $k = \lambda_{max}(\frac{A + A^T}{2})$. Because $A = \frac{A + A^T}{2} + \frac{A - A^T}{2}$ But $\left< \frac{A - A^T}{2}x,x \right> = 0$


0

For any skew-symmetric matrix $K$, we have $x^TKx=(x^TKx)^T=x^TK^Tx=-x^TKx$ and therefore $x^TKx=0$. It follows that if you split $A$ into the sum of its symmetric part $\frac{A+A^T}2$ and its skew-symmetric part $\frac{A-A^T}2$, then $\langle Ax,x\rangle=\langle \frac{A+A^T}2x,x\rangle$. In other words, it suffices to find a positive definite $S$ such that ...


1

Theorem: Let $H$ be a real or complex Hilbert space, and let $\{e_{\alpha}\}_{\alpha\in\Lambda}$ be an orthonormal subset of $H$. The following are equivalent: 1. $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ is a complete orthonormal set, meaning that the only $x\in H$ that is orthogonal to every $e_{\alpha}$ is $x=0$. 2. Parseval's identity $\|x\|^2=\...


1

Hint: For any $x$ with $\|x\|_1=1$, we have $\|x\|_\infty \leq 1$.


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Alternative answer:your operator induces a continuous bijection between quotient space and image subspace. By the open mapping theorem if it were compact then the unit ball in both spaces were compact, thus finite dimensional by Riesz


1

First, note that $Z:=Im(K)$ is a closed subspace of a Banach space and thus, itself a Banach space. Thus, $K: X\to Z$ is onto. By the open mapping theorem, $K$ is open and hence, $K$ is mapping open sets to open sets. Now, assume that $K$ is compact and take the image of the open unit ball $C:=K(B_X^\circ)$ which is open in $Z$ and relatively compact in $Y$...


2

The first thing to show is that the decomposition is unique. That is, if $f$ is continuous on $\mathbb{R}$ has such a representation, then $d$ and $k$ are unique ($k$ is unique as an element of $L^1[0,\infty)$.) Equivalently, if $f=d+\int_{0}^{\infty}e^{ixt}k(t)dt$ is the $0$ function on $\mathbb{R}$, then $d=0$ and $k=0$ as an element of $L^1[0,\infty)$. ...


1

A typo slipped in; a $k$ became $j$ for no reason. Fixing that, you're almost there, re showing it's a Banach algebra: $$\begin{align}\dots=\max\limits_{0 \leq t \leq 1} \sum_{k=0}^{n}\sum_{j=0}^{k}{\dfrac{|f^{(k-j)}(t)}{(k-j)!}\dfrac{g^{(j)}(t)|}{j!}} &=\max\limits_{0 \leq t \leq 1} \sum_{j=0}^{n}\dfrac{|g^{(j)}(t)|}{j!}\sum_{k=j}^{n}\dfrac{|f^{(k-...


4

Given a linear map $T:\mathbb{C}^n\to \mathbb{C}^n$, for all $v\in\mathbb{C}^n$ we have $$\langle TT^*(v),v\rangle= \langle T^*(v),T^*(v)\rangle\geq 0,$$ because this is just the inner product of a vector with itself. Thus, $TT^*$ is positive. Furthermore, if $T$ is invertible, $T^*$ is invertible, so equality holds in the above if and only if $v=0$. ...


5

A Fredholm operator $T$ is an operator for which the solutions of the nonhomogeneous linear problem $Tx = y$ can be described using "finitely many pieces of data" just like in the finite dimensional case even though the operator acts on a possible infinite dimensional space. More explicitly, if $T$ is Fredholm then $\ker(T)$ is finite dimensional and so we ...


2

I think the intended question was something like: Given $p(x)\in\mathbb{C}[x]$ with degree $n$, such that $$\int_{0}^{1}x^k p(x)\,dx=0 $$ for every $k\in\{1,2,\ldots,n-1\}$, show that the roots of $p(x)$ lie in the circle $|x|\leq 1$. We may notice that $p(x)$ has $n+1$ coefficients and we have $n-1$ linear constraints on them, so the space of ...


3

The function $t\cdot \chi_{[-1,1]}(t)$ is orthogonal to each $t^{2k}$ in $L^2[-1,3],$ hence is orthogonal to the the linear span of $\{t^{2k} : k=0,1,\dots \}$ in $L^2[-1,3],$ hence is orthogonal to the closure of this linear span in $L^2[-1,3].$ Therfore this closure cannot be all of $L^2[-1,3].$


2

Let $A$ denote the linear span of $\{t^{2k}\}_{k \in \mathbb{N}}$. Then, if instead of $[-1,3]$ the domain was $[1,3]$ instead, we could use the Stone-Weierstrass theorem to conclude that $A$ is dense ins $C([1,3])$ and thus in $L^2([1,3])$ since $t^2$ separates the points of $[1,3]$ and $1 \in A$. However, for $[-1,3]$, we have $t^{2k}(-1) = t^{2k}(1)$ ...


3

The closure of the domain in $L^2$ is simply $L^2$: Obviously it holds $C_0^\infty(0,1)\subset D(A_0)$. The set of smooth function is dense in $L^2(0,1)$, hence its closure is $L^2(0,1)$. This implies that the closure of $D(A_0)$ is $L^2(0,1)$ as well.


2

In any Hilbert space (say $H=\ell^2(\mathbb N)$), let $T=I$ and $S=I-P$ where $P$ is a finite-rank projection. Then, as $ST=I-P$ and $\ker ST=\text{ran}\,P$, $$ 0=\ker T\subsetneq PH=\ker ST. $$ Both $S$ and $T$ are selfadjoint and Fredholm.


1

\begin{eqnarray} \langle T_z^*(\alpha),x \rangle_{H}&=&\langle \alpha , T_z(x)\rangle_\mathbb{K}\\ &=& \alpha \overline{T_z(x)}\\ &=& \alpha\langle z,x\rangle_H\\ &=& \langle \alpha z , x\rangle_H\\ \end{eqnarray} So $$T_z^*(\alpha)=\alpha z $$ It's clear now that $\|T_z\|=\|T_z^*\|=\|z\|$


0

If the kernel of $S$ contains a nonzero vector $y$ that is in the range of $T$ (let $x$ denote the pre-image of $y$, so we have ker($S$) $\ni y = Tx$), then $x$ lies in the kernel of $ST$, but not in the kernel of $T$. If there are no such vectors, then ker($S$) $\cap$ range($T$) = the zero vector space.


2

EDIT: This answer no longer applies because the OP changed the question. It is not true. Any linear map between finite dimensional vector spaces is Fredholm, so there are finite dimensional counterexamples. If you want an infinite dimensional counterexample consider $S,T:\ell^2\to\ell^2$, where $S(a_0,a_1,a_2,\ldots)=(0,a_1,a_2,\ldots)$ and $T$ is the ...


1

For the map $\Phi \colon H \to U'$, $x \mapsto \langle \cdot, x \rangle$ we have $\ker \Phi = U^\perp$. Hence $\Phi$ is injective if and only if $U^\perp = 0$. Because $H = U^\perp \oplus \overline{U}$ we have $U^\perp = 0$ if and only if $\overline{U} = H$, i.e. if $U$ is dense.


2

Both definitions are correct and common in certain contexts. Yours is common for the context of finite dimensional vector spaces (in which one may use "positive semidefinite" for your notion of positive). Your book's definition is one that is strictly applicable to infinite dimensional vector spaces. By the definition it gives, the "positive definite" ...


2

Yes, this follows from the monotone convergence theorem for nets (see Theorem IV.15 of Reed and Simon's Functional Analysis, for instance) Let $\mu$ be a regular Borel measure on a compact Hausdorff space $X$ and let $(f_{\alpha})$ be an increasing net of continuous functions converging pointwise to $f$. If $\sup \|f_{\alpha}\|_1 < \infty$, then $\...


1

let's take $\Omega = [0,1]$, $\varphi,f \in C^\infty_c((0,1))$. Then $$\langle f'',\varphi \rangle = \int_0^1 f''(x) \varphi (x) dx = f'(1)\varphi (1)-f'(0)\varphi (0) - \int_0^{1} f'(x) \varphi '(x) dx$$ $\varphi,f$ have their support strictly inside $(0,1)$ so $f'(0)\varphi (0) = f'(1)\varphi (1)= 0$ and $$\langle f'',\varphi \rangle =-\langle f',\...


1

Note that the range of $K$ is in a finite dimensional subspace $$\text{span} \{x, \cos x\},$$ thus $K$ is a compact operator and has only point spectrum ($0$ is a eigenvalue too). To look for eigenfunction, we need only restrict ourselve to the above subspace. Note that $$ K (ax+b \cos x) = b\pi x + a\frac{2\pi^3}{3} \cos x,$$ This implies that $$ ...


1

Suppose $Av=\lambda v$. Then as you noted we have $(Av)_j=\frac 1{2^{j-1}}(Av)_1=\frac \lambda{2^{j-1}}v_1$ but also $(Av)_j=\lambda v_j$. Thus we have that $\lambda v_j=\frac \lambda{2^{j-1}}v_1$ so that $v_j=\frac{v_1}{2^{j-1}}$. For $v_1$, we have that $\lambda v_1=(Av)_1=\sum_{n=1}^\infty\frac{v_1}{2^{n-1}2^n}$ which implies that $\lambda=2\sum_{n=1}^\...


2

Note that $$ \left (\sum_j a_{ij} x_j \right)^2 = \lvert \langle (a_{ij})_j, (x_j)_j \rangle \rvert^2 \leq \lVert (a_{ij})_j \rVert_1^2 \rVert x\lVert_1^2 $$ So, if $x$ is summable, you know that $\lVert x \rVert_1 < \infty$ and you can get rid of the norm of $x$ in the estimate. By $\langle \cdot,\cdot \rangle$ I mean the inner product in $\ell^1$.


0

You have to understand what $A$ looks like. It acts as an infinite matrix $(a_{ij})_{i,j=1}^{\infty}$ on your elements of $\ell^2$. Since $\ell^2$ is a Hilbert space you know that $A$ is compact if it is a limit of finite-rank operators. You can show that $A$ is the limit of $P_nAP_n$ where $P_n$ is the projection on the span of the first $n$ standard basis ...


1

Yes. This follows from the fact that Gram-Schmidt orthogonalization changes your basis by an upper-triangular matrix with $1$s on the diagonal: the first basis vector is kept the same, the second has some multiple of the first added to it, the third has some linear combination of the first two added to it, and so on. This means that $F'=FT$ for some upper-...


5

In Loewner ordering, we have $$ T_n^{-1}(T_n^{-1})^\top =\pmatrix{2&-1\\ -1&\ddots&\ddots\\ &\ddots&2&-1\\ &&-1&1} \preceq\pmatrix{2&-1\\ -1&\ddots&\ddots\\ &\ddots&2&-1\\ &&-1&2}=P. $$ Using the spectral formula for tridiagonal Toeplitz matrices, the eigenvalues of $P$ are given by $2+2\...


1

It can be easily shown, that $$\sup_{n\in\mathbb N^*} \Big\{\frac{1}{n\sqrt{n}}\|T_n\|_{\text{Tr}}\Big\} < \infty$$ In order to find a better assymptotic bound we need information about singular values of $T_n$ or some related matrices. Let $\Omega_n = T_n^TT_n + T_nT_n^T$. This is a special Toeplitz matrix and its eigenvalues are computed in Bünger, F. ...


2

Your sequence of equalities should have been $$ \|A_\lambda f\|^2=\lambda\int_0^1f(\lambda t)^2dt= \int _0^\lambda f(t)^2=\int _0^1f(t)^2\chi_{[0,\lambda]}^2\leq \|f\|\|\chi_{[0,\lambda]}\|=\lambda \|f\|. $$ Now, I don't know what inequality you are using, but I cannot make sense of it. And you get $\|f\|$ instead of $\|f\|^2$, which is a bad sign. The ...


1

This is not an answer to this interesting question, but it may shed additional light. When computing the singular values of such matrices (which is possible with R for $n \le 2000$ without trouble), I find that the $i$-th singular value $s_i$ of such a matrix satisfies very nearly $$ s_i \ge c \cdot \frac{n}{i} $$ for some constant $c$ that does not depend ...


1

You have already shown that $$ \|A_nf\|_2\leq\frac {\|K\|_1\,\log n}{\sqrt n}\, \|f\|_2, $$ which says that $$ \|A_n\|\leq\frac {\|K\|_1\,\log n}{\sqrt n}. $$


1

For Q1, by your argument, you have shown $f_n(0)\to a$ and $f_n'\to g\in C([0,1])$ in sup-norm. Then we define $f(t)=a_0+\int_0^t g(s)\,ds$ for $t\in [0,1]$. Then by fundamental theorem of calculus, it's clear that $f\in C^1([0,1])$ and $f'=g$. Then it follows that $f_n\to f$ in your norm. Some thoughts about Q2. If we consider $f''(1)$ for simplicity, ...


1

You mentioned you like Complex Analysis. So I thought I'd offer a proof using Complex Analysis applied to the resolvent. The proof comes down to evaluating the integral around all finite singularities of $(\lambda I-A)^{-1}x$ by determining the residue at $\infty$, which turns out to be $x$. This equivalence forces the completeness of spectral expansions for ...


2

Take $r > \max \sigma(A)$. Then $R_A(r)$ is self-adjoint and bounded. If $0$ is not in its spectrum, then $A = (R_A(r)^{-1}+rI$ is bounded. If $0$ is in its spectrum, it is an isolated point of the spectrum and therefore must be an eigenvalue: $R_A v = 0$ for some $v \in \mathcal H$. But that is impossible since $R_A(r) = (A-rI)^{-1}$, i.e. $R_A(r) v =...


0

Actually, this is a comment to the previous answer, but being too long, I post it as an answer. I realize now that the argument given by TrialAndError can be slightly simplified. From the fact that \begin{equation} \sigma(R_{A}(-\epsilon)) \subseteq [0, 1/ \epsilon], \end{equation} and the equality of norm and spectral radius for a bounded self-adjoint ...


0

One can define $\Delta$ in $H^1$ because $\Delta$ is in divergence form $$\Delta u = \text{div}\nabla u$$ Thus we define $\Delta : H^1_0(\Omega) \to H^1_0(\Omega)^*$ by $$\Delta f (\phi) := -\int_\Omega \nabla f \cdot \nabla \phi.$$ The right hand side is well defined for $f , \phi \in H^1_0(\Omega)$. If $f$ is in $H^2_0$, then integration by part gives ...


1

For a bounded normal operator $N$, the norm and spectral radius of $N$ are the same. That is, $\|N\|=\sup_{\lambda\in\sigma(N)}|\lambda|$. Let $\lambda \notin \sigma(A)$. Assume $A$ is unbounded. Then $(A-\lambda I)^{-1}$ is bounded and normal, with $$ \sigma((A-\lambda I)^{-1})=\frac{1}{\sigma(A)-\lambda}\cup\{0\}=\left\{ \frac{1}{\mu-\lambda} : \...


1

Suppose $A : \mathcal{D}(A)\subseteq \mathbb{H}\rightarrow\mathbb{H}$ is selfadjoint with $\sigma(A)\subseteq [0,\infty)$. Then, for every $\epsilon > 0$, $(A+\epsilon I)^{-1}=R_A(-\epsilon)$ is a bounded selfadjoint operator with $$ \sigma(R_{A}(-\epsilon)) \subseteq \mbox{closure}\left(\frac{1}{[0,\infty)+\epsilon}\right)=[0,1/\epsilon] \\ \...


2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \...


2

You can't find a positive $\gamma$ such that $\gamma \leq \frac{\langle Lu\mid u \rangle}{\langle u\mid u \rangle}$ if $\langle Lu\mid u \rangle=0$. So to find a counterexample, you might look for positive operators such that $Lu$ is sometimes $0$. The simplest example is $L=0$, the operator that sends everything to $0$.


0

If you would ask me, it is meant in the sense you mentioned. Since a Hilbert space is isometric isomorphic to its dual space due to the Riesz representation theorem you can identify the functionals in $H'$ with their riesz representers in $H$. I did that frequently in my bachelor thesis. It is not the most formal way but you can give shorter definitions and ...


3

Call this space $E$. Let $\pi : E \to B$ be the map sending an element of $E$ (a Hermitian matrix with distinct eigenvalues) to its set of eigenvalues. $B$ here is the configuration space $\text{Conf}_n(\mathbb{R})$ of $n$ distinct unordered points in $\mathbb{R}$, which is contractible; it can be described as a certain open subset of $\mathbb{R}^n$, and is ...


2

Suppose $\|A\| < 1$. You can take limits of $rA$ as $r\uparrow 1$ for a general contraction $A$, which makes the following construction generally useful for any $A\in\mathcal{L}(\mathcal{H})$ for $\|A\| \le 1$, though the limiting case involves an positive operator measure on the unit circle $\mathbb{T}=\partial D$, which is the boundary of the unit disk $...


1

Since the mapping $$ \iota\colon\mathcal{D}\to H $$ is a continuous linear operator (which is the "generalization" of bounded operators to the setting of locally convex spaces), there is a (continuous) transpose $$ \iota^t \colon H' \to \mathcal{D}' $$ satisfying $$ \langle \varphi, \iota^t(T)\rangle = \langle \iota(\varphi), T\rangle $$ for all $\varphi\in\...


1

A unitary operator is a diagonalizable operator whose eigenvalues all have unit norm. If we switch into the eigenvector basis of U, we get a matrix like: \begin{bmatrix}e^{ia}&0&0\\0&e^{ib}&0\\0&0&e^{ic}\\\end{bmatrix} which is obviously the exponential of a diagonal hermitian matrix.


2

For $f\in L^2$ and $0 \le r \le s \le 1$, \begin{align} |(Af)(r)-(Af)(s)| & \le \int_{r}^{s}|f(x)|dx \\ & \le \left(\int_{r}^{s}|f(x)|^2dx\right)^{1/2}\left(\int_{r}^{s}dx\right)^{1/2} \\ & = \|f\|\sqrt{s-r}. \end{align} Therefore the imagine of a bounded subset of $L^2$ is uniformly bounded and equicontinuous, ...


4

For Questions 1, take $\{f_n\}$ be uniformly bounded in $L^2$, consider the class $\{Af_n\}$. check 1.$\{Af_n\}$ is uniformly bounded in $L^2$. check 2.$\{DAf_n\}$ is well-define and uniformly bounded in $L^2$, where D is Differential operator. pf of check 1: This follows by Holder, $|\int_0^t f(s) ds| \leq \sqrt{t }\|f_n\|_2\leq \|f_n\|_2$, so $\|...


3

For determining the adjoint : You have that $$\langle Af, g \rangle = \int_0^1 g(t) \int_0^t f(s) ds dt $$ And this is equal (by Fubini) to $$ = \int_0^1 f(t) \int_t^1 g(s) ds dt = \langle f, Bg \rangle$$ with $Bg(t) = \int_t^1 g(s) ds$ And as this is true for every $f$ and $g$, you have that $A^* = B$



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