New answers tagged

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 Gro├če-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 ...



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