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3

The support is contained in a interval of the form $[-m,m]$ for $m$ large enough. The function is uniformly continuous on $A=[-m-1,m+1]$ since $A$ is compact. Take $\epsilon >0$. From uniform continuity on $A$ for there exists $\delta>0$ such that for $|x-y|< \delta$ it implies that $ |f(x) - f(y)| < \epsilon$. We prove continuity on ...


0

Use the $\varepsilon$/$\delta$-definitions of continuity and uniform continuity. (The $\delta$ that works in the support works outside as well.)


0

Let $$ f_n(x)=\sqrt{x^2+\frac1{n^2}},\quad-1\le x\le1. $$ $f_n\in C^1([-1,1])$ and converges uniformly to $f(x)=|x|$, but $f\not\in C^1([-1,1])$. This shows that $C^1([-1,1])$ is not complete with respect to the uniform norm. $C^1([-1,1])$ is a normed space with the uniform norm, but bot a Banach space.


1

Recall that a subspace of a Banach space is closed if and only if it is complete if and only if it is a Banach subspace. Thus, and proper, dense subspace cannot be a Banach subspace. Observe then, by the Stone-Weierstrass Theorem, that the polynomials are dense in $C[a,b]$. In particular, $C^1[a,b]$ is dense in $C[a,b]$.


0

The question is answered in the paper: S. Matsumoto: Orbit equivalence types of circle diffeomorphisms with a Liouville rotation number, Nonlinearity 26 (2013), 1401-1414, at least for Liouville numbers. The general case is covered by the Theorem in: W. Krieger: On Borel Automorphismsand Their Quasi-Invariant Measures, Math. Z. 151 (1976), 19-24.


0

A cyclic representation is nondegenerate. A direct sum of nondegenerate representations is nondegenerate. Hence, the theorem could only hold for nondegenerate representations. You are right that in Conway's proof it can be seen that the algebra must be nondegenerate, as we have $\mathscr{H}_0$ defined to be a subset of $\mathrm{cl}[\pi(A)\mathscr{H}]$, ...


2

If $\sum_{n=1}^{\infty}d(a_n, b_n)$ is absolutely convergent (which is equivalent to convergent because $d(a_n, b_n) \geq 0$), then $\lim\limits_{n\to\infty} d(a_n, b_n) = 0$. Suppose $a_n$ converges to $a$, then I claim that $b_n$ converges to $a$ as well. Note that we have $$d(a_n, b_n) \geq |d(a_n, a) - d(b_n, a)|$$ by the reverse triangle inequality. ...


1

If the series converges, then $d(a_n,b_n)\to 0$, and so if $a_n\to L$, then since $d(b_n,L)\le d(a_n,b_n)+d(a_n,L)$, it follows that $b_n\to L$. In other words, the sequences converge and diverge together, and necessarily to the same limit. However, this only uses the convergence $d(a_n,b_n)\to 0$ and not the convergence of the series, which is much ...


3

Take $(\Bbb R,|.|)$ the ususal metric on $\Bbb R$ and $a_n=n,b_n=n+\frac1{n^2}$ so what we can conclude?


1

Nothing, because $a_n$ could be any sequence and $b_n$ could equal $a_n$.


1

It appears to be a typo. It should say the C*-algebra generated by $h$ and the closed $*$-algebra generated by $k$, or equivalent. There is no assumptions that $k$ sits in a C*-algebra. For instance, later it is noted that $\|k\|\geq \|k\|_{\mathrm{sp}}$, but this would be equality if it were in a C*-algebra.


1

No, it isn't correct, because you are trying to use "$|h(x)|$" without it having been defined.


1

It appears you saying that you want to assume that $1=1'$. Also, presumably $A$ and $A'$ sit together within a larger algebra (perhaps you want to think of them as subalgebras of the same $B(H)$?). Let $B$ be the $C^*$-algebra generated by $A$ and $A'$. Then the identity $1$ for $A$ and $A'$ is also the identity for $B$. Thus you know that the spectra ...


1

Let $C$ be the C*-algebra generated by $\pi(A)$. Then $\pi(1)$ is a multiplicative identity for $C$, because it is one for $\pi(A)$ and multiplication is continuous. Now you can apply the proof you know to $\pi:A\to C$.


1

For a subsets of topological vector space $X$ the following theorem hols true. Theorem: If $A,B,C\subset X$ and $B$ is bounded $C$ is closed and convex and $A+B\subset C+B,$ then $A\subset C$ e have $B_Y \subset T(MB_X )+ \delta B_Y ,$ hence $$\delta B_Y +(1-\delta ) B_Y \subset \overline{T(MB_X )}+ \delta B_Y ,$$ and from the theorem we obtain that ...


2

The equation must be considered in the context of the Hilbert space $L^{2}(-1,1)$ because the filter of requiring eigenfunctions to be in $L^{2}$ is what eliminates the non-regular solutions, and it is what determines the eigenvalues. For $m = 1,2,3,\cdots$, the operators $$ L_{m}f = -\frac{d}{dx}(1-x^{2})\frac{d}{dx}f + \frac{m^{2}}{1-x^{2}}f ...


1

The process is good. Maybe some wording could be added. For example, the first displayed inequality only holds for $m$ and $n$ large enough. The existence of the limit of the sequence $(x_n)$ has to be justified (by completeness of a Hilbert space).


0

Hint: Show that the kernel of $T^*$ is the annihilator of $\operatorname{Im} T$. Show that the only closed subspace with $0$ annihilator is the whole space.


2

Assuming that $\Omega$ is a bounded $C^1$ domain, we can argue as follows. I - Note that if $u\in C^\infty(\overline{\Omega})$ then, $u\in H^1(\Omega)$ and $\Delta u\in L^2(\Omega)$, therefore $$C^\infty(\overline{\Omega})\subset \{u\in H^1(\Omega):\ \Delta u\in L^2(\Omega)\}\subset H^1(\Omega).$$ II - Remember that $C^\infty(\overline{\Omega})$ is dense ...


2

Hints: Let $A - B = \{a-b \;|\; a \in A, b \in B\}$. Show that $A-B$ is closed, convex, and $0 \notin A - B$. Since $A-B$ is closed, there is a small $\varepsilon > 0$ so that $T := \{x\in X \;|\; \|x\| < \varepsilon\}$ and $A-B$ are disjoint. Separate them with a functional. $T$ is open, so the functional is continuous. And because of the ...


1

Since $\lVert L\rVert = 1$, it follows that $L-\lambda I$ is invertible for all $\lambda$ with $\lvert\lambda\rvert > 1$, we can see that using the Neumann series. For $\lvert \lambda\rvert > \lVert L\rVert$, the series $$\sum_{n=0}^\infty \lambda^{-n}L^n$$ is absolutely convergent, and since $B(H)$ is a Banach space, it is convergent. One computes ...


1

Since for each non-negative $a$ and $b$, $(a+b)^2\leqslant 2a^2+2b^2$, we have $$\|u\|_{\mathcal{H}_0^1}^2 = \left(\left\|\frac{\partial u}{\partial x}\right\|_2+\left\|\frac{\partial u}{\partial y}\right\|_2 \right)^2\leqslant 2\left\|\frac{\partial u}{\partial x}\right\|_2^2+2\left\|\frac{\partial u}{\partial y}\right\|_2^2=2\cdot a(u,u),$$ hence ...


1

Yes it can happen! Given the real line $\mathbb{R}$. Consider the metrics $d(x,y):=|x-y|$ and $d(x,y):=|\arctan x-\arctan y|$. So both give rise to the same topology. But they cannot be equivalent as: $x_n:=n:\quad d(x_m,x_n)'\to0$ (Note how sublteties arise on the uniqueness of finite dimensional TVS.) (Caution also that it reveals incompatibility ...


1

First of all, see this question. If $Tx=xT=0$ for every $x\in I$ then for every $h\in H$ we have $Txh=0$. Now the set $\{xh,\ x\in I, h\in H\}$ is dense in $H$, since the representation is non-degenerate by the question above. Since $\ker(T)$ is closed in $H$ and contains $\{xh,\ x\in I, h\in H\}$, which is dense in $H$ then $\ker(T)=H$ and $T=0$.


1

The equality implies that if $\lvert v\rvert\xi_1 = \lvert v\rvert\xi_2$, then also $v\xi_1 = v\xi_2$, so it is immaterial which element $\xi\in \lvert v\rvert^{-1}(\eta)$ is chosen to define $u(\eta) = v(\xi)$. All choices yield the same result. Thus $u = v\circ \lvert v\rvert^{-1}$ is well-defined although in general $\lvert v\rvert^{-1}$ is not a map ...


2

Notice that $\|u(1-u^*u)\xi'\|^2=\|u^*u(1-u^*u)\xi'\|^2=\|(u^*u-u^*u)\xi'\|^2=0$, for every $\xi'$. Thus, $u(1-u^*u)=0$.


5

Frobenius Norm Case We first introduce a lemma. Lemma 1. If $A, B$ are respectively $m \times n$ and $n \times p$ matrices, then $\|AB\| \le \|A\|\|B\|$. Proof. Note that $AB$ has columns $A\vec{\beta}_1, \dots, A\vec{\beta}_p$ where $\vec{\beta}_j$ is the $j$th column of $\beta$. So$$\|AB\|^2 = \sum_{j=1}^p \|A\vec{\beta}_j\|^2 \le \sum_{j=1}^p ...


1

Note that in this setting boundedness and continuity are equivalent. Let $X,Y$ be Banach spaces and $A:X \rightarrow Y $ a linear, bijective and bounded operator. Since $A$ is bounded it is also continuous. Now you could use the open mapping theorem to prove that also the inverse $A^{-1}$ is continuous, and hence bounded. Edit: The result is valid only ...


1

WARNING. This is not a counterexample as it works only in real Hilbert spaces. See comments. The statement is false if $u$ and $v$ are not assumed to be symmetric. Consider the Hilbert space $\mathbb{R}^2$. The operators $$ u\mathbf{x}=(-x_2,x_1)$$ and $$ v\mathbf{x}=(-2x_2,2x_1)$$ are such that $$ (u\mathbf{x}, \mathbf{x})=(v\mathbf{x}, ...


2

This is a corollary of the previous statement in the book: the polarization identity says that for a sesquilinear form $\sigma$, you can write $$4\sigma(x,y) = \sigma(x+y,x+y)+i\sigma(x+iy,x+iy) - \sigma(x-y,x-y) -i\sigma(x-iy,x-iy).$$ In particular, this shows that given two sesquilinear forms $\sigma, \sigma'$, if $\sigma(v,v) = \sigma'(v,v)$ for all $v$, ...


2

Outlining the construction by Mariano Suárez-Alvare... Problem Does every plain vector space admit a norm? Construction Given a plain vector space $V$. Choose a Hamel basis $\mathcal{B}$. Denote functions with finite support by $\mathbb{R}^\mathcal{B}_0$. Decide for a norm there $\|(\lambda_b)_{b\in\mathcal{B}}\|$. Now, regard the isomorphism: ...


2

For the $s=0$ case, let $s_{n}=\sum_{k=1}^{n}c_{k}$ for $n \ge 1$ and let $s_{0}=0$. Then $s_{n}-s_{n-1}=c_{n}$ for all $n \ge 1$ and $\lim_{n} s_{n}=0$ by assumption. So the following are absolutely convergent for $0 \le r < 1$: $$ \begin{align} \sum_{n=1}^{\infty}r^{n}c_{n} & = \sum_{n=1}^{\infty}r^{n}(s_{n}-s_{n-1}) \\ & = ...


1

The contraction mapping theorem seems to do the job, without needing closedness nor convexity.


1

$(x,u^*\alpha y)=(ux,\alpha y)=\bar\alpha(x,u^*y)=(x,\alpha u^*y)$.


13

Vector spaces are, by default, unnormed. A norm is extra structure we add to a vector space, to define a normed vector space.


1

Of course there are. :-) First of all, there are linear spaces which are not endowed by a topology, and there are topological vector spaces which are not endowed by a norm. So a wise question is: when a topology of a topological vector space $X$ is generated by a norm? It seems it is iff $X$ is a locally convex Hausdorff space containing such a neighborhood ...


7

Every (real or complex) vector space admits a norm. Indeed, every vector space has a basis you can consider the corresponding «$\ell^1$» norm.


8

While your question could have multiple answers, perhaps the closest to what you are looking for is the notion of a non-metrizable vector space. In the general setting of topological vector spaces, we consider (as one might guess from the name) vector spaces endowed with a topology so that we can discuss ideas like the continuity of linear operators. Normed ...


5

If $V$ is finite dimensional, it is normable, in the sense that you can use an isomorphism into $\mathbb R^n$ to pull back the $\mathbb R^n$-norm . And then this norm is equivalent to any other norm topology-wise, i.e., in a finite- dimensional space, all norms are equivalent in this sense.


1

No. Let $V$ be the orthogonal complement of $L^2(X,\mu)$ in $H$ and let $f$ be a function in $V$. Then, $\int fg d \mu =0$ for all $g$ in $L^2(X,\mu)$. This condition easily implies that $f$ is identically zero.


2

Let $V$ be a finite-dimensional linear space over a field $\mathbb{F}$. Let $\mathcal{L}(V)$ denote the linear operators from $V$ into itself. Then I would expect that the determinant to be the unique multiplicative function $\tau : \mathcal{L}(V)\rightarrow\mathbb{F}$ for which $\tau(I)=1$. That is, $\tau(AB)=\tau(A)\tau(B)$. Is it?


2

The polynomials are dense in $C^{0}[0,1]$. You need to show that $l$ is bounded; then you can extend by continuity. The norm of $l$ is $M=l(1)$, which is non-negative by assumption. Indeed, if $p$ is a real polynomial, then $\|p\|\pm p \ge 0$ which gives you $$ 0 \le l(\|p\|1)\pm l(p),\\ \mp l(p) \le \|p\|l(1),\\ ...


2

First you should extend $l$ from $P$, to $\bar l:U\to \mathbb R$ where $U$ is the space of all polynomials (hint: note that every monomial is a non-negative polynomial in $[0,1]$, and extend linearly). $U$ is clearly a subspace of $C^0([0,1])$, and the extended $\bar l:U\to \mathbb R$ can be checked to be sub-linear (actually it is linear). Apply ...


3

We need only to bound the sum $$T(M)=\sum_{k_1\geq k_2\geq 1, k_1\geq M}\frac{1}{(k_1^2+k_2^2)^2}$$ We have: $$T(M)=\sum_{k_1\geq M}\frac{1}{k_1^3}A(k_1)$$ with $$A(k_1)=\frac{1}{k_1}\sum_{k_2=1}^{k_1}\frac{1}{(1+(k_2/k_1)^2)^2}$$ $A(k_1)$ is a Riemann sum, and has as limit $\displaystyle \int_0^1\frac{dx}{(1+x^2)^2}$. Hence $A(k_1)$ is bounded, say by $c$. ...


0

This is not true, even for this particular $F$. To see it, let $Q:H^1_0(\Omega)\to\mathbb{R}$ be the linear functional, defined by $$Q\phi=\int_\Omega \nabla w\nabla \phi.$$ If $w\neq 0$ then, the image of $Q$ is $\mathbb{R}$. Therefore, there is $v\in H_0^1(\Omega)$ such that $Qv\neq 0$. Note that for this particular $v$, we have that $Tv=0$, because ...


3

This is an attempted simplification of the answer by Michael. The determinant of degree (?) $n$ is the unique multilinear alternating form $f:V^n\to k$ such that $f(\rm id)=1$. Here multilinear alternating means $f$ is linear in each coordinate and if we let $S_n$ act on $V^n$ by permuting the order of the vectors, $f((ij)x)=-f(x)$ for any $x\in V^n$. By ...


3

You need to prove that $uv\in H^1(\mathbb{R})$, or equivalently, that $uv\in L^2(\mathbb{R})$ and there is $g\in L^2(\mathbb{R})$ such that $$\int_\mathbb{R}(uv)\varphi'=-\int_\mathbb{R}g\varphi,\ \forall\ \varphi\in C_0^\infty(\mathbb{R}).\tag{1}$$ The first question is: does $uv\in L^2(\mathbb{R})$? The answer is yes, because $H^1(\mathbb{R})$ is a subset ...


1

Do you like the following? If we take the definition proposed in the linked question, we can do the following: Take $v_1, \dots, v_n \in V$ with $v_1 \wedge \dots \wedge v_n \neq 0$ (with $n = \dim V$). Such $v_1, \dots, v_n$ exist, because otherwise $\Lambda^n V = \{0\}$. By the definition in the other post, we have $$ (Tv_1) \wedge \dots \wedge (T v_n) ...


0

Following on from the comments we do not needed to use the bounded inverse theorem as I had originally thought. As $T$ is continuous and surjective by the open mapping theorem $T$ is an open map. This means that for some $r >0$ we have $r T(B_{l_\infty}) \subset T(B_{\mathbb{H}^{\infty}})$ where the $B$ are the open unit balls in the respective spaces. ...


9

1. We first talk a bit about the underlying method of the proof. We want look for $(I + \epsilon T)^{-1}$ as an infinite series, similar to the way we would expand $1/(1+x)$ into a power series for $|x| < 1$ over the real or complex field. We then show that our infinite series of operators converges. Along the way, we need to use the fact that $\|TS\| ...



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