# Tag Info

4

Notice that for each vector $x$, one has $$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$ Thus $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$

1

Diagonalization will help here. (When in doubt and working with normal matrices, try utilizing diagonalization!) Write $T = UDU^*$, then $T^* = UD^* U^*$, giving that $T^*T^2 = UD^*D^2U^*$. However $T^3 = UD^3 U^*$. Since unitary conjugation does not change the operator norm, this boils down to considering $D^*D^2$ and $D^3$. Here $Dg(x) = f(x)g(x)$ for ...

1

Yes. First $f$ is surjective. Indeed, let $y \in Y$. $\frac{y}{||y||} \in B_Y = f(B_X)$ hence there is $x \in B_X$ s.t $f(x)=\frac{y}{||y||}$ and so $f(||y||x)=y$ by linearity. We conclude that $f$ is surjective and hence bijective (we already know it is injective by assumption) Because of the facts that $B_Y = f(B_X)$ and that $f$ is bijective, we have ...

0

This is also exercise 1) page $111$ points a)-b) in Functional Analysis by Rudin. By a discussion in another network the proof it's correct. For second part, it is follows essentially from ii) of convex separation theorem, taking $\lbrace w_0 \rbrace$ as a compact set, and $\overline{B}_E$ as closed set, equivalently considering corollary in A).

1

The notion of equivalent norms that I have is the following: Two norms $||\cdot||_1,||\cdot||_2$ over a vector space $V$ are equivalent if there are constants $C_1,C_2>0$ such that for every $x\in V$, $$C_1||x||_1\leq ||x||_2\leq C_2||x||_1.$$ Is it the same definition that you have? If so, under this definition, we can show the norms $||\cdot ... 1 It seems that the uniqueness part has already been settled. In your argument regarding the existence part, you have proved that if the sequence$(x_n)_{n=1}^\infty$you have constructed converges, then its limit belongs to$C$. But so far you have only that$\|x_n\| \rightarrow s$. It remains to be shown that$(x_n)_{n=1}^\infty$does converge. To this end, ... 3 There are two cases:$0\in C$and$0\notin C$. In the first case the minimum norm is$=0$, and$x_0=0$is the unique element with this property. If$x_0$with minimum norm has already been found (and your question indicates that this is already known to you) assume$x_1\neq x_0$has the same norm and it is an element of$C$. It is easy to see that$x_1$and ... 0 Suppose for example$\displaystyle \mathbb R ^\infty = \prod_{i=1}^\infty \mathbb R$is normable. Then there is an open neighborhood (the unit ball)$B$of$\overline 0$such that$\displaystyle \bigcap_{n=1}^\infty\frac{1}{n} B = \{\overline 0\}$. You should prove this if it is not obvious to you. But every neighborhood of the origin in$\mathbb R ...

0

I think you can prove it by showing if the index is infinite, then any non empty open set is not bounded (for definition of "bounded" in this case see here

2

It is sufficient to prove that if $\mathrm{dim}(E) < \infty$ then $\sigma(E,E') = \mathcal{T}_E$ where $\sigma(E,E')$ is weak topology and $\mathcal{T}_E$ strong topology, or topology induced by norm on $E$. Equivalently every open $\mathcal{T}_E$-neighborhood of origin $B_E(\epsilon)$ it's also an open $\sigma(E,E')$-neighborhood. Let $x=(x_1,...,x_n)$, ...

1

Notice that we can write $$c_1^2 + c_2^2 = |\nabla v|^2 = (\nabla v \cdot \tau)^2 + (\nabla v \cdot \nu)^2,$$ where $\tau$ is the direction of the tangent line. By assumption, we know that the second term on the RHS is $0$ on $\Gamma_0$. What is left to show is that also the tangential part of the gradient is $0$. Do you see why this is true? EDIT: take ...

3

Note, that $\|\cdot\|$ is somehow the composition of the $2$-norm and the $3$-norm on $\mathbb R^2$. We have $$\|x\| = \|(\|(x_1,x_2)\|_2,x_3)\|_3$$ and with this at hand the proof is easy (but a bit ugly). We have by the triangle inequality for the $2$-norm $$a:= \|(x_1+y_1,x_2+y_2)\|_2 \leq \|(x_1,x_2)\|_2 + \|(y_1,y_2)\|_2 =:b$$ and since for all ...

2

Yes, the inequalities hold. Let $B$ be the closed unit ball for the norm $\|\cdot \|$. The assumptions imply that $\pm e_j$, the standard basis vectors, are in $B$. Hence, their convex hull is contained in $B$. This convex hull is the unit ball for the $\ell^1$-norm, which implies $\|\cdot \|\le \|\cdot \|_1$. Similarly, we need to prove that $B$ is ...

3

The answer for infinitely many functionals is no; there's a counterexample here. For finitely many functionals it must be yes... Right. First, there is a norm on any real vector space $X$, for example if $B$ is a (Hamel) basis define $$\left\vert\left\vert\sum_{b\in B}c_b b\right\vert\right\vert=\sum_{b\in B}|c_b|.$$Now if $||\cdot||$ is a norm on $X$ ...

1

Brian's comments are spot on. Let me add a little (too long for a comment): Properties like "Cauchy" or "convergent" depend on the metric space you're living in. It doesn't make sense to say that the sequence $(3, 3.1, 3.14, . . . )$ converges or doesn't converge - you have to say what metric space you're in. In $\mathbb{R}$ with the usual metric, this does ...

1

Using the Axiom of Choice, one can construct in a given infinite dimensional normed linear space $X$, a discontinuous linear functional (see this, e.g.). The kernel of such a functional is a proper and dense subset of $X$ (see this). This would provide a subspace $Y$ of $X$ that is proper and such that for every $x\in X$, $\text{dist}\,(x,Y)=0$. From ...

5

As a complement to the earlier (good) answer and comments: the space of all sequences (whether real or complex) arises in at least one fairly natural way, namely, as the continuous dual to the LF-space (strict inductive limit of Frechet spaces) $\mathbb R^\infty=\bigcup_n \mathbb R^n$, where $\mathbb R^n$ has its usual topology and is included in $\mathbb ... 13 The answer to the question exactly as you asked it is yes; your space is isomorphic as a vector space, with no topology, to various Banach spaces. (See various comments for details.) Edit: The assertion that the answer is yes has met with vigorous disbelief. Also there's a technical point that I realized after some thought I simply didn't know how to do. ... 0 Hint: If$R$is open,$R(B_X(0,1))$contains$B_Y(0,r), r>0$, this implies that$R(B_X(0,n))$contains$B_Y(0,nr)$. Then use the fact that$\cup_nB_Y(0,nr)=Y$. 1 No, not even if$X=Y=H$, a Hilbert space. Say$H$is an infinite-dimensional Hilbert space and let$B$be a Hamel basis for$H$. Say$m:B\to(0,\infty)$is unbounded and define$T:H\to H$by $$T\left(\sum_{b\in B}c_b b\right)=\sum_{b\in B}m(b)c_b b$$(where$c_b=0$except for finitely many$b\in B$.) Then the kernel of$T$is$\{0\}$, but$T$is unbounded ... 1 This is perhaps best understood from a topological perspective. A base for the weak topology is formed by finite intersections of sets of the form, $$\{u:a < \phi(u) < b\},$$ for any continuous linear functionals$\phi$. Geoometrically, one of these sets looks like the infinite slab between two parallel hyperplanes. In finite dimensions, you can ... -1 Edit: Here is a complete modification of my first answer. I had at first leaned (without having any proof in mind) to a finite dimensional answer. I had indicated Riesz's lemma https://en.wikipedia.org/wiki/Riesz's_lemma and the classical theorem (mentionned in this article): The unit ball in a Normed Linear Space is compact iff this space is finite ... 0 No. What you want holds for every closed proper subspace of a Hilbert space. Simply take any$x \in Y^\perp $with$\|x\|=1$. 0 No, there are no hypercyclic operators on finite dimensional spaces. See e.g the article of GroÃŸe-Erdmann in the Transactions of the AMS. 0 Your set is countable. Thus the answer is no, at least when$X$is not separable, e.g. the space of bounded real sequences w.r.t. the infinity norm. 0 Assume that$r>s$. Then you can find points$A,B$in$B(x,r)$of distance$2r>2s$. Then$d(A,B)\leq d(A,y)+d(y,B)\leq 2s$, a contradiction. 3 This is just an elaboration on what others have already said in the comments. Let$V$be a finite-dimensional normed linear space (over a subfield$\mathbb F$of$\mathbb C$). As you say, all norms on a finite-dimensional vector space are equivalent, so we may assume that we have the usual 2-norm on$\mathbb C^n$. The definition of weak convergence is ... 1 Suppose$F$is such an extension. Then$F(x) = x$for$x \in Y$. But since$Y$is dense, that implies$F(x) = x$for all$x \in X$. So we must have$Y = X$. -1 Yes, let$P: Y\to T(W) $be a continuous projection on$T(W),$and let$A: X\to Y $be extension of$T$then define$\tilde{T} : X\to Y,\tilde{T} = P\circ A .$3$y=(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...) \notin c_{00}$,$y_1=(1,0,0,...)$,$y_2=(1,\frac{1}{2},0,0,...)$and so on.$y_n \in c_{00}$and$\|y_n -y \|_{\infty}=\frac{1}{n+1}$which tends to$0$when$n$tends to$\infty$. 0 Your estimate holds only in a neighborhood of$x$. Hence, you need can not control whether you may choose such a$C$independent of$x$. You need to rely on the linearity of your operator in the following way: By continuity of$f$in$0$, there exists$\delta>0$such that for all$x\in X$with$\Vert x \Vert_X <\delta$holds $$\vert f(x) - f(0) \vert ... 1 Since f is continuous (at 0), there is a neighbourhood U of 0 such that f(U)\subset(-1,1). Choose \delta>0 such that \{x\in X|\|x\|\leq\delta\}\subseteq U. Then, if x\in X is such that \|x\|\leq \delta, we have x\in U, and hence, |f(x)|\leq 1. Since \|\frac{\delta x}{\|x\|}\|=\delta, it follows that for all x\in X we have$$ ... 0 I was looking for a proof of the separability of$\mathcal{L}_p(\mathbb{R})$for$1 \leq p < \infty$when I came across you question. Although it's been asked a year ago, I will attempt an answer for the future generations. As you mentioned in a comment, you should choose$C_i \in \Big[ \inf f_1 \restriction R_i, \sup f_1 \restriction R_i \Big]$. This ... 0 If$T$is onto, this is the open mapping theorem: You first choose$x$with$Tx=y$. Then you look at the images of small open balls around$x$. These images are open and contain$y$and thus contain all but finitely many$y_n$. Hence, you can choose$x_n$in the balls as needed. 3 The$\ell^p$spaces are a special case of the$L^p$spaces obtained by using the counting measure on the set of natural numbers. If you squint closely at the integral it looks like a sum or indeed as Forever Mozart points out: summation is just integration with the trivial measure on$\mathbb{N}$. 1 Suppose$w$is in the image of$T$then$w=(b_1/1,b_2/2,\dots$) for some bounded sequence$(b_n)$, but then $$\frac{b_n}{n}=\frac{1}{\sqrt n}$$ hence$b_n=\sqrt n$which is not bounded. An accumulation point is the same as a limit point it just means that for any neighbourhood$U$of$w$there is a point$x\in \mathrm{Im}(T)$such that$x\in U$. 0 I apologize, in setting of problem i have arisen typographical errors. I will repeat text. Problem. Let$X$,$Y$real normed vector spaces and$f$isometry of space$X$in the space$Y$. Show that there is isomorphism A of spaces$X$on the space$Y$and vector$c\in Y$such that$f(x)=Ax+c$for each$x\in X$. I know that is true If$X$and$Y$normed ... 1 I think the answer is yes. A sign- and permutation- invariant norm defined on$\mathbb C^n$is called a symmetric gauge function. It is known that every unitarily invariant norm on$M_n(\mathbb C)$is induced by a symmetric gauge function. See, e.g. theorem 7.4.24 on pp.438-440 of Horn and Johnson (1985), Matrix Analysis, 1/e, Cambridge University Press. To ... 0 Consider the$\ell^2$space of maps$\mathbb{N} \to \mathbb{R}$. Then the$\ell^2$space of maps$\mathbb{N} \to \mathbb{R}$which vanish on odd numbers is a proper subspace. The isomorphism comes from the bijection$2\mathbb{N} \to \mathbb{N}$. 1 The first question is straight forward if you apply the definition of the norm, and I encourage you to do it on your own in order to assimilate the concept of norm. For the second question, you need to produce a sequence of functions$(f_n\in C[a,b]^{\mathbb N})$such that$f_n$converges (with respect with the norm defined in the exercise) towards a ... 1 An infinite-dimensional vector$x = (x_1, x_2, x_3, \ldots)$(or I guess we can just say sequence) is in$\ell^2$iff$\displaystyle \sum_{k=1}^{+\infty} |x_k|^2 < +\infty$. Well, no matter what the value of$n$is in$x = (1, 0, 0, \dots, 0, -n^2, 0, 0, 0, \dots)$, we'll always have$\displaystyle \sum_{k=1}^{+\infty} |x_k|^2 = 1 + n^4$, and this is ... 0 The claim is not true in general. And I think looks true because your intuitive picture of$x$and$y$being in the "same direction" is not what the conditions$||x|| \leq ||x + y||$and$||y|| \leq ||x + y||$imply. I'll give a counterexample in$\mathbb{R}^2$itself: Take the vectors, in$(r,\theta)$coordinates, to be$x = (1,0)$and$y = (1, ...

3

Your answer is incorrect. $\langle f,g \rangle$ is allowed to take any value, but $\langle f,f \rangle$ must be non-negative. Try to come up with an example of two vectors whose dot-product is negative, noting that the dot-product is the prototypical inner product. The property of inner products that fails here is that $$\|f\|^2 = \langle f,f \rangle = 0 ... 2 Yes, it is. In fact, we can say more. If K has the trivial valuation, X is a finite-dimensional K-vector space, and p is any norm on X, then there exist positive constants c_1 and c_2 such that c_1\leq p(x)\leq c_2 for all nonzero x\in X. To get c_2, note that if \{e_1,\dots,e_d\} is a basis for X, then writing x=\sum a_i e_i, we ... 1 Let V be a vector space and \|\cdot\|_n, n\in\Bbb N be a sequence of norms and assume that \|\cdot\|_n\to\|\cdot\| pointwise. Then \|\cdot\| is a norm iff \|x\|\ne 0 for all x\ne 0. Indeed, all properties of norm clearly transfer to the limit except the condition that non-zero vectors have non-zero norm. 1 Consider x_n= {\rm sign}(a_n)|a_n|^{q-1}. 5 By definition of bilinearity, \langle x,y\rangle - \langle x,z\rangle = \langle x,y-z\rangle. On the other hand in general there is no formula for \langle a,b\rangle \pm \langle c,d\rangle. 1 If a=\lim a_n and b=\lim b_n , then you have to show that a+b=\lim (a_n+b_n) . 1 (i) is correct. (ii) Show that \{ f \in X : f(0)=0\} and \{ f \in X : f(1)=0\} are closed subspaces of X: from this, you will have that V is an intersection of closed sets, hence closed. This can be done using sequences, and (i). Pick a sequence f_n \subset \{ f \in X : f(0)=0\} converging to some f in X. But (i) shows that$$f(0) = \lim_n ...

0

As @TrialAndError says, $C$ need not meet $S$ at all. As such your derivation is false. But the statement is true. First note that in metric spaces the concept of separability and second-countability are equivalent: If a space is second-countable with a countable open base $\{U_n\}_{n\in \mathbb N}$ then a sequence $\{x_n\}_{n \in \mathbb N}$ where \$x_n ...

Top 50 recent answers are included