New answers tagged

0

Consider the vector $v=(b_1,b_2)\in \mathbb R^2.$ Then there exists a nonzero vector $u=(a_1,a_2)\in \mathbb R^2$ such that $u\perp v.$ That is the same as saying $a_1b_1+a_2b_2=0.$ The same will be true of $cu$ for all $c\in \mathbb R.$ Thus the set $A$ contains $\{ c(a_1,a_2,0,0,\dots), c\in \mathbb R\}.$ That is an unbounded set in $l^1.$


0

Yes, $Y\cap M$ is dense in $M$. Here is why. Write $Y=(Y\cap M) \oplus Y'$ (algebraic direct sum) and notice that the projection $\pi :X\to X/M$ is injective on $Y'$ because $Y'\cap M=\{0\}$. Since $\text{dim}(X/M)=1$ we deduce that $\text{dim}(Y')\leq 1$. If $\text{dim}(Y')=0$ then $Y\subseteq M$, and hence $Y$ is obviously dense in $M$, so let us ...


0

I'm sure you've already proved $\int_{-L/2}^{L/2}e^{i2\pi(n-m)x/L}dx=L\delta_{mn}$ for $m,\,n\in\Bbb Z$, so the $u_n(x):=\tfrac{1}{\sqrt{L}}e^{i2\pi nx/L}$ form an orthonormal basis of the vector space they span. The hard part is how "diverse" a space of functions that is. The way you've tried to write an arbitrary $f$ as a linear combination of ...


2

Fix an eigenvalue $\lambda$ of $T_1$ and let $E_1$ be the eigenspace of $T_1$ corresponding to $\lambda_1.$ Observe that $E_1$ is invariant under $T_2$ because if $v\in E_1$ then $T_1(T_2(v))=T_2(\lambda v)=\lambda T_2(v)$ that is $T_2(v)\in E_1.$ Now restrict $T_2$ to $E_1$ and let $f\in E_1$ be an eigenvector of the restriction of $T_2.$ Note that $f$ is ...


0

This is not really on the topic of the question, but the situation on the unbounded domain does not work the way you said it works. It is much more subtle than that. The situation in the bounded domain does work like you said. And yes, you have the counterintuitive situation that, given a $L^2$ function $f$ on $[-L/2,L/2]$, you can find a sequence of ...


0

Using $\lbrace e_1,e_2,\dots \rbrace \subseteq \text{ran}(B)$ strictly only shows $\ell^1 = \overline{\lbrace e_1,e_2,\dots \rbrace} \subseteq \overline{\text{ran}(B)}$. To show the opposite inclusion: Take any $x \in \overline{\text{ran}(B)}$. Then there exists a sequence $(x^{(n)})_{n\in \mathbb{N}} \subseteq \text{ran}(B)$ such that $$ x^{(n)} \to x $$ as ...


0

The answer is affirmative for all Borel functions $f$ and $g$. Here is the reason. First of all let me say that the most concrete form of the Spectral Theorem I know asserts that, given any normal operator $T$ on a separable Hilbert space $H$, there exists a $\sigma $-finite measure space $(X, \mathscr A, \mu )$, and a unitary operator $U:H\to L^2(X)$, ...


0

If you mean 2 dim. Fourier transform as mentioned in comment by reuns and in the format of $F(s,w)=\frac 1{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)\exp(\!-\!isx\!-\!iwy)\:dx\:dy\;$ so $\;f(x,y)=2\pi\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} F(s,w)\exp(isx\!+\!iwy)\:ds\:dw\;$ and you really meant $f(x,y)=e^{(-x^2-y^2\!-xy)}\;$ then the ...


1

I know at least one variation that is true: Consider the following theorem from Murphy's excellent text "$C^*$-algebras and operator theory": Theorem 2.5.7 (p73): Let $u$ be a normal operator on the Hilbert space $H$ and let $g: \mathbb{C}\to \mathbb{C}$ be a continuous function. Then $(g\circ f)(u) = g(f(u))$ for all $f \in B_\infty(\sigma(u)).$


1

First notice that any constant function is in $A^\perp$. Take any $g \in A^\perp$. Then $(f,g) = 0$ for all $f \in A$. Set $\overline g = \frac 12 \int^1_{-1} g(t) dt$, and take $$f(x) = g(x) - \overline g, \,\,\,\,x\in [-1,1].$$ Then $f \in A$ (you should check this), and so $$0 = (f,g) = (f,g) - (f,\overline g) = (f,g-\overline g) = (g-\overline g, g-\...


2

Claim: $A\psi=\lambda\psi$ for some $\psi\ne 0$ iff $$ P(\{\lambda\})\psi = \psi. $$ Proof: First assume that $A\psi=\lambda\psi$ for some $\psi\ne 0$. Then $$ 0=\|(A-\lambda I)\psi\|^2=\int_{-\infty}^{\infty}|\mu-\lambda|^2d\rho(\mu), $$ where $\rho(S)=\langle P(S)\psi,\psi\rangle= \|P(S)\psi\|^2$ is the measure associated with $\psi$. It ...


1

In any neighborhood of a point of the form $(0,y_0)$ you may find distinct points $(a,y_0)$ and $(-a,y_0)$ by choosing $a$ sufficiently small. Since $$ f(a,y_0) = (a^2, a^2 + y_0) = f(-a,y_0), $$ we see that $f$ is not injective on any neighborhood of $(y_0,0)$. Because of this $f$ is not locally injective at $(0, y_0)$. On the other hand, regarding ...


1

You can say that your function is injective on U if and only if : $$ f(x,y)=f(x',y') \Rightarrow (x,y)=(x',y')$$ So we are trying to find a subset U where F is injective. So first you can take $(x,y,x',y')\in\mathbb{R}^4, f(x,y)=f(x',y')$ Then you have : $$x^2=x'^2, x^2+y=x'^2+y' \\x^2=x'^2, y=y' $$ In this case, you can easily see that f will be injective ...


1

\begin{align*}I=\int_{\mathbb{D}}|f'(z)|^2(1-|z|^2)dA(z)&=\int_0^1\int_0^{2\pi}|f'(re^{i\theta})|^2(1-r^2)rd\theta dr\\ &=2\pi\sum_{k \ge 1} k^2|a_k|^2\int_0^1r^{2k-2}r(1-r^2)dr \end{align*} (by applying orthogonality in $\theta$ and switching sum and integral by absolute convergence) Integrating in $r$ one gets that: $I=2\pi\sum_{k \ge 1} k^2|a_k|^2 ...


2

I'm not sure if there's a way to see this directly from the definition of $P_A$, but here is a proof using the resolvent, which is quite natural in view of the proof of the spectral theorem. With $R_A(z)=(A-zI)^{-1}$, we know that $$\langle\psi,R_A(z)\psi\rangle=\langle \psi,\frac{1}{z_0-z}\psi\rangle=\frac{1}{z_0-z}||\psi||^2$$ This is in turn, by the ...


1

The fact that $u \in W^{2,1}_p(\Omega \times (\sigma,T))$ does not follow from $L^2$-theory using energy estimates; you need to use the fact that $f$ lies in $L^p.$ Your idea is correct to show this and deduce the result using Sobolev embedding, but you would need to use a suitable $L^p$ estimates for the form $$ \lVert u \rVert_{W^{2,1}_p(\Omega \times (\...


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The proof is in Reed/Simon “Methods of Modern Mathematicsl Physics”, Vol. 2, Section X: Theorem X.1


1

We obviously have $L^2_a(G)\subset L^2_c(G)$, where $L^2_c(G)$ is the space of continuous square integrable functions. Then $L^2_c(G)\subset L^2(\mu)$ in the following sense: There is an injection $j:L^2_c(G)\to L^2(\mu)$ such that for any $u\in L^2_c(G)$ the only continuous representative of $ju\in L^2(\mu)$ is $u$.


0

Say $f\sim g$ if $f=g$ ae. Say $[f]$ is the corresponding equivalence class. It's not literally true that $L^2_a\subset L^2$, but it's clear that $$L^2_a/\sim\,\,:=\{[f]:f\in L^2_a\}\subset L^2.$$ And there's no real difference between $L^2_a$ and $L^2_a/\sim$. The Point: It happens a lot that people say something about $X$ when they really mean $X/\sim$. ...


1

I'm not aware of any such reference that isn't either a text on functional analysis or a text on commutative algebra. It would help if you were more specific about what you wanted to know about infinite-dimensional vector spaces. Most of what is worth knowing about them is a list of things that don't generalize from finite-dimensional vector spaces. If $V$ ...


0

Here is a proof which works more generally for $x,y\ge 0$ in a $C^*$-algebra, where the spectral radius satisfies $r(z)\le \|z\|=\sqrt{\|z^*z\|}$ for every element $z$, and $r(t)=\|t\|$ for every normal element $t$. The main difference with @user1551's argument is that we will use the invertibility of $y$. Other than that, the idea is essentially the same. ...


1

Let $\{x_n\}_n$ be any sequence in $A$. For every $k$ one has that the sequence of scalars $$ \{x_n(k)\}_{n\in {\mathbb N}} $$ is bounded by $|y(k)|$, so it admits a converging sub-sequence. By doing this for each $k$, and using Cantor's diagonal argument, we may find a sequence $n_1<n_2<\cdots \quad$ of natural numbers, such that $$ z(k) := ...


1

It is very easy to exhibit a non-converging Cauchy sequence in $H\odot K$, but actually proving that it does not converge is a bit more difficult. The approach suggested by @Max, although somewhat sophisticated, is perhaps the easiest way to pin down all of the details. Here is a pedestrian way to describe this method. For each pair of vectors $(x, y)\in ...


2

Because the exponential functions $e^{cx}$ with $c\in\mathbb{R}$ exist in $C^\infty([0,1],\mathbb{R})$: $$\|D(e^{cx})\|=\|ce^{cx}\|=|c|\cdot\|e^{cx}\|,$$ and there is no constant $M\geqslant0$ such that this norm is $\leqslant M\cdot\|e^{cx}\|$ for all $c \in \mathbb{R}$.


0

Suppose $(c_n)$ is bounded. $\|\sum c_n\langle x, e_n \rangle e_n\|^{2}=\sum |c_n\langle x, e_n \rangle |^{2} \leq M \sum |\langle x, e_n \rangle |^{2}=M^{2}\|x\|^{2}$ where $M =\sup_m |c_n|$. Hence $T$ is bounded with$\|T\| \leq \sup_m |c_n|$. Also $\|Te_n\| =|c_n|$ and this implies (by definition of norm of an operator) that $\|T\| \geq |c_n|$. This is ...


1

AH, found it. There was no way the inequality you wanted was true : I had to see the context and get the answer. There's a hanging variable. It is this : let us write $dx = dx_1dx_2...dx_n$ for the integral over $Q$. Then the line before the question mark is actually (by definition of $Q$): $$ d^{p-1}\int_Q \int_{a_1}^{a_1+d} |D_1u(s,x_2,...,x_n)|^p ...


2

The adjoint $T^*$ is defined as the set of $g\in L^2(\mathbb{R})$ for which there exists a constant $C_{g}$ such that $$ |\langle Tf,g\rangle_{L^2}| \le C_g\|f\|_{L^2},\;\;\; \forall f\in \mathcal{D}(T). $$ This inequality holds iff there is a unique $T^*g\in L^2$ such that $$ \langle Tf,g\rangle = \langle f,T^*g\rangle,\;...


0

Lemma: The algebra of operators on $\ell^2$ generated by $\{I, L, R\}$ contains all operators whose matrix has finitely many nonzero entries. Proof: Denoting the canonical basis of $\ell^2$ by $\{e_n\}_{n=0}^\infty $, notice that $$ p:= I-RL $$ is the projection onto the space spanned by $e_0$. Moreover $R^mpL^n$ is the operator which sends all basis ...


1

This is simply the standard way of writing a finite sum as an integral relative to a corresponding atomic measure. In particular: if the operator in question can be written as a countable sum $$ A = \sum_i \lambda_i P_i $$ (with the $P_i$ equal to spectral projectors and hence mutually orthogonal and "complete"), then the measure associated with $A$...


1

The answer is no. I'll explicitly construct a triple $(p,q_1,q_2)$ on four points that violates the relation. This can be extended to however rich a Polish space you like (as long as it has at least $4$ points) by convolving with appropriately thin densities. Let $$p = \frac{(1,1,1,1)}{4}\\ q_1^x = \frac{(2-x, 2-x, x,x)}{4}\\ q_2 = \frac{(3, 1, 1, 1)}{6},$...


2

Since $$\|Tx\|^2 = 0 + 16x_1^2 + x_2^2 + 16 x_3^2 + \cdots \le 16(x_1^2 + x_2^2 + x_3^2 + \cdots) = 16 \|x\|^2$$ you have $\|T\| \le 4$. Equality is attained because in the special case $x = (1,0,0,\ldots)$ you have $$4 = \|Tx\| \le \|T\| \|x\| = \|T\|.$$


1

Note that $\|\tau_yf-f\|_p\leq2\|f\|_p$, and for $m$ sufficiently large, $\int_{|y|>\rho}\delta_m(y)dy$ is going to be small.


4

Okay, so it's not obvious (or even true) that $G_n$ is closed because the interval on which your bound holds might shrink. Indeed, let $f_m=n 1_{(0,2^{-m})}+2n 1_{[2^{-m},1)}$. Then, clearly, $f_m\in G_n$, but $f_m\to 2n$ in $L^1$. Instead, define $$ G_{n,m,k}=\{ f\in L^1((0,1))|\; \|f|_{(k2^{-m},(k+1)2^{-m})}\|_{\infty}\leq n\} $$ where $k$ ranges over $\...


0

Let $H = \mathbb{R}^2$ and $A = \begin{pmatrix}0& 1\\ -1 & 0 \end{pmatrix}$. Then $\langle Ax,x \rangle = 0$ for all $x$ but $A$ is non zero. For an infinite dimensional Hilbert space, just choose an Hilbert basis $(e_k)$, and define $A$ diagonally to be such as in the above example on pairs $(e_{2k},e_{2k+1})$.


1

It just implies that $x$ is orthogonal to $Ax$. Can you think of such a linear transformation? Hint: think about $\mathbb{R}^2$.


1

This is an exercise on Lebesgue decomposition. We can write $\lambda$ as $\lambda_1+\lambda_2$ where $\lambda_1 << \mu$ and $\lambda_2 \perp \mu$. There exists $B$ such that $\mu (B)=0$ and $\lambda_2 (B^{c})=0$. Also there exists $f \in L^{1}(\mu)$ such that $\lambda_1 (E) =\int _E fd\mu$ for all $E$. Now $\lambda (A\setminus B)=\int_{A\setminus B} ...


0

The following is a full solution of (1). If I have time, I will solve other parts. (1) Firstly, we prove the lamma: Given an open neighborhood $U$ of $e\in G$, there exits an open neighborhood $V$ of $e$ such that $V=V^{-1}$ and $VV\subseteq U$. Proof of the lemma: Let $\theta:G\times G\rightarrow G$, $\theta(x,y)=xy$. Note that $\theta$ is continuous with ...


2

The statement is false. Consider the sequence $ \mathbf{a}_n \in \ell^1 $ given by $$ \mathbf{a}_n = \langle a^n_i \rangle_{i=1}^\infty \quad a^n_i = \begin{cases} \frac{1}{i^{1/(p-1)}} &\mbox{if } i \leq n \\ 0 & \mbox{if } i > n \end{cases} $$ Define $$ \mathbf{a}^* = \left\langle \frac{1}{i^{1/(p-1)}} \right\rangle_{i=1}^{\infty} $$ Then we ...


1

Main idea: For simplicity, let the space be the reals. Let $f\in L^p(\mathbb{R})$. Let $K$ be an arbitrary compact subset. Then $m(K)<\infty$. Hence, $\int_K |f|\leq \|f\|_p\|1\|_q$. When $q<\infty$, we have $\|1\|_q<\infty$ since $m(K)<\infty$.


1

By the spectral Theorem there exists an orthonormal set $\{x_n\}_{n\in \mathbb N}$, and a sequence $\{\lambda_n\}_{n\in \mathbb N}$ of real numbers converging to zero, such that $$ T(x) = \sum_{n\in \mathbb N} \lambda _n\langle x,x_n\rangle x_n, \quad \forall x\in \mathcal H $$ It follows that $$ J(x) = \langle T(x),x\rangle = \sum_{n\in \mathbb N} \...


0

Lemma. Let $X$ be a Banach space equipped with a separately continuous bilinear map $$ \mu :X\times X\to X. $$ Then there exists a constant $C$ such that $$ \|\mu (x,y)\|\leq C\|x\|\|y\|, \quad \forall x,y\in X. $$ Proof. For each $a$ in $X$, let $$ L_a : x\in X\mapsto \mu (a,x)\in X, $$ and $$ R_a : x\in X\mapsto \mu (x, a)\in X. $$ ...


1

A simple counterexample: A very easy way to get a counterexample is to consider any finite dimensional space $X \not= \{0\}$. Then every $C_0$-semigroup is given by a matrix exponential function, and thus, every operator that occurs in a $C_0$-semigroup is necessarily invertible. So just take $B$ to be any non-invertible operator on a finite dimensional ...


3

Defining a new measure $\mu $ on $[\alpha , \beta ]$ by $$ \mu (E) = \lambda (f^{-1}(E)), $$ for every Borel measurable set $E\subseteq [\alpha , \beta ]$, one has that $$ F(g) = \int_a^b g(f(x))\, dx = \int_\alpha ^\beta g(y)\, d\mu (y). $$ Regarding boundedness of $F$, the first important point is whether or not $\mu $ is absolutely continuous ...


0

Hint 1: You made a mistake, the following equality is not true: $$ \sup_{x\in X}\frac{||\sum_{j=1}^n c_j x(t_j)||}{||x||}\neq \sup_{x\in X} \sum_{j=1}^n \gamma_j \frac{||x(t_j)||}{||x||}$$ Hint 2: For all $c_j \neq 0$ denote by $\omega_i:= \frac{c_j}{||c_j||}$. Then $$\|\sum_{j=1}^n c_j x(t_j)\| \leq \sum_{j=1}^n| c_j| | x(t_j)| \leq \sum_{j=1}^n| c_j| \| x\...


2

Let $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ be the unbounded operator defined on the dense set $C_0^{\infty}(\mathbb{R})$ by $T=-\frac{d^2}{dx^2}$ then for $f,g\in C_0^{\infty}(\mathbb{R})$ we have by integration by parts: $$\langle Tf,g\rangle=\int_{-\infty}^{\infty}-\frac{d^2}{dx^2}f(x)\overline{g(x)}dx=\int_{-\infty}^{\infty}\frac{d}{dx}f(x)\...


1

The question: $p_0 \in I$ or not ... $X = [0,1/2]$ with Lebesgue measure. $$ f(x) = \frac{1}{x},\qquad I = (0,1), \\ f(x) = \frac{1}{x\log^2(1/x)},\qquad I = (0,1]. $$ You can get similar examples where $I=(1,+\infty)$ or $I=[1,+\infty)$. Add two examples (on disjoint subsets of the measure space), one example with $I = (0,1]$ and one with $I = [1,+\infty)$, ...


0

This indeed seems to be easier than I (did not) thought of as supinf intimates. Sketch of proof: a Cauchy net in the topology of uniform convergence on compacta is pointwise Cauchy so it has a pointwise limit. It is easy to see that the pointwise limit is a linear map, and by taking pointwise limits in the Cauchy condition, convergence to this limit is ...


1

This is not the case. Take $X=\mathbb{R}^2$, $Y=\{(0,t):t\in\mathbb{R}\}$, $f(x):=x\cdot j$, $g(x):=2x\cdot i$ (so $gY=\{0\}$). Let $x:=i=(1,0)$, then whether one chooses $y=(0,1)$ or $(0,-1)$ (of unit norm), $$|f(x)-g(x)|=|0-2|=2>1=|f(y)|$$


1

The proof is faulty because $T$ is not linear ($T(-y)=T(y)$) and even so, it is not true that $T(y)$ is contained in $B(y,\epsilon)$. (Take $n=2$, $y=(1,0)$, then $T(y)=(1,1)$ is not in a small ball around $y$.) In fact, $B(y,\epsilon)\cap T(\mathbb{R}^n)=\emptyset$ for $\epsilon$ small enough. The mapping $T : \ell^2\to\ell^2$, defined by $T (a_n) := (a_0, ...


1

Let X be a non reflexive Banach space and $Y$ be a reflexive Banach space. Suppose that $A \colon X \to Y$ is an injective bounded linear operator. I claim that the range of $A$, denoted by $R(A)$, can't be closed in $Y$. Arguing by contradiction, suppose that $R(A)$ is closed and thus reflexive. Then, the operator $ A \colon X \to R(A) $ is a continuous ...


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