New answers tagged

0

Define the operator $T$ on an admissable class of functions (I'll use $V = \{ f: [0, \infty) \to \mathbb R : f \text{ is continuous and bounded} \}$ which is a Banach space with the supremum norm, $\|\cdot\|_\infty$) by $$(Tf)(x) = e^{-2x} + \int^\infty_0 e^{-2x-2y} \sin(x-y) f(y) dy, \,\,\,\,\, x \ge 0.$$ Then for $f,g \in V$, $x \in [0\infty)$, $$\lvert ...


0

The unit sphere of $c_0$ is certainly not compact as the set $\{e_n\colon n\in \mathbb{N}\}\subset S_{c_0}$ is closed and discrete. (The unit sphere of a normed space is compact if and only if the space is finite-dimensional.) However, you may adapt this proof to see that that the unit ball as well as the unit spehre of a normed space has the property that ...


1

You need to use that $f$ is uniform continuous. By definition, it means that: For all $\epsilon >0$, there is $\delta >0$ so that $|f(y) - f(x)| <\epsilon$ whenever $\| x- y\| <\delta$. Now pick $\epsilon =1$ and $\delta_1$ be the corresponding $\delta$ in the statement. Let $x\in S_{c_0}$. Let $$ e_1 = (1, 0, 0, \cdots ) \in S_{c_0}.$$ We ...


2

The point spectrum of your operator is empty. Let first $0\neq \lambda \in \sigma_p(T).$ Then there exists an eigenvector $0\neq x=(x_i)_{i\in \mathbb{N}}\in c_0$ such that $T(x) = \lambda x$. This leads to the equations $$ \lambda x_1 =0 \quad \text{and} \quad \lambda x_{i+1} = w_i x_i \quad \text{for } i\in \mathbb{N}.$$ As $\lambda\neq 0$, we get by ...


0

You solution looks good to me. I separately solved it the following way: We can notice that $l_1$ is a direct sum of the closed subspaces $G = \{(x_n) \in l_1: x_{2n+1} = 0, \forall n \in \mathbb{N} \}$ and $H = \{(x_n) \in l_1: x_{2n} = 0, \forall n \in \mathbb{N} \}$. Also, $||x+y|| = ||x|| + ||y||$ $ \forall x \in G, y \in H$. Thus, given bounded ...


0

If your goal is to ultimately show that this is a norm on $ C^\infty([0,1])$ there are many obstacles which you won't overcome: Take a Cauchy sequence in your subspace. Even if you prove that it converges in you $ \ell^p $-norm, how do you prove that this limit is actually $ (\|f^{(n)}\|_\infty)_{n\in\mathbb N} $ for some $ f \in {C}^{\infty} $ Your ...


1

First note that $M=\left\{(3y,y,x_3,x_4,x_5,\ldots)\in l_1:y,x_i\in\mathbb{R}\right\}$, so $l_1=M\oplus\mathbb{R}(1,0,0,0,\ldots)$. Now note that $|f(3,1,0,0,\ldots)|=3$, so $\Vert f\Vert\geq 3/4$. Show that in fact for every $x\in M$, $|f(x)|\leq\frac{3}{4}\Vert x\Vert$, and then conclude that $\Vert f\Vert=3/4$. This seems weird, but it happens because we ...


0

$S_n$ converge weakly to $0$: Let $y\in E'$, $\epsilon > 0$ and $n_0$ the rank where $\forall k > n_0, |\langle x_k-x, y \rangle |\leq \epsilon$ Then $$\langle S_n, y \rangle = \langle \frac{n_0}{n} S_{n_0} + \frac{1}{n} \sum_{k=n_0+1}^n (-1)^k x_k , y \rangle$$ $$ = \frac{1}{n} \langle n_0s_{n_0}, y \rangle + \frac{1}{n}\sum_{k=n_0+1}^n (-1)^k ...


0

For any ultrafilter $\mathscr U$ the function $$\newcommand{\Ulim}{\operatorname{{\mathscr U}-lim}}f \colon x = (x_n) \mapsto \Ulim x_n$$ is a bounded linear function from $\ell_\infty$ to $\mathbb R$. Since $f(e^{i})=0$, this function is not from $\ell_1$. The limit of a sequence $(x_n)$ along an ultrafilter $\mathscr U$ or ultralimit is defined as: ...


2

Let's show that $\mathbb{C}^2$ with $\ell^1$ norm cannot be embedded in any space $\mathbb{C}^k$ with $\ell^\infty$ norm. Suppose otherwise, and let $T:(\mathbb{C}^2,\ell^1)\to(\mathbb{C}^k,\ell^\infty)$ be an embedding. We have $\Vert T(1,0)\Vert_\infty=\Vert(1,0)\Vert_1=1$, so by multiplying the coordinates of $T$ by suitable complex numbers of absolute ...


2

No reference, but here is a proof that $(\mathbb{C}^2, \ell^1)$ does not admit a linear isometric embedding into $(\mathbb{C}^k,\ell^\infty)$ for any $k$. Indeed, such an embedding can be described as a map $$(z,w) \mapsto (\alpha_j z+\beta_j w)_{1\le j\le k}$$ where $|z|+|w| = \max_j|\alpha_j z+\beta_j w|$. Plugging $(1,0)$ and $(0,1)$ for $(z,w)$ shows ...


1

You have already basically proved what you want when you wrote $|T_n(x)| \le \left(\sum_{k=1}^{n}|a_k|^2\right)^{1/2}\|x\|$. All you have to note is that $\{ a_k \} \in \ell^2$, which gives $|T_n(x)| \le \|a\|_2\|x\|_2$ for all $n$ and fixed $x$. That's enough to apply the uniform boundedness principle. The uniform boundedness principle is the simplest way ...


0

You have the following typos: the display after "yields" should read $$ \|f_n(x)-f(x)\|<\epsilon $$ because you've taken the limit for $m\to\infty$ with $n$ fixed and you've used the continuity of the norm in $X$. You're missing a $\|$ bracket before $f(x)\|$ elsewhere. Now using the inequality in 1. and the triangle inequality (in the form ...


1

Seminormalised, weakly-null sequences in $J_p$ have subsequences equivalent to the canonical basis of $\ell_p$. To see this modify Theorem 2.1 here. Suppose that $T\colon \ell_2\to J_p$ is a non-compact operator ($p<2$). Let $(x_n)$ be a bounded sequence in $\ell_2$ so that $(Tx_n)$ does not have a convergent subsequence. By reflexivity and weak-to-weak ...


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

Let $(X. \vert \cdot \vert)$ be some Banach space. Let $(x^{(n)})_{n\in \mathbb{N}} \subseteq c$ be a Cauchy sequence, where $x^{(n)}=(x_k^{(n)})_{k\in \mathbb{N}} \subseteq X$ is a convergent sequence. We define $$ x_{\infty}^{(n)}:= \lim_{k\rightarrow \infty} x_k^{(n)}.$$ We need to construct $x=(x_k)_{k\in \mathbb{N}}\in c$ such that $x^{(n)} ...


1

Let $M = \sup \|x_n\| + \|x\|$. Select any $f \in X'$ and let $g \in E'$. Then \begin{align*}|f(x_n) - f(x)| & \le |f(x_n) - g(x_n)| + |g(x_n) - g(x)| + |g(x) - f(x)|\\ &\le 2M \|f-g\|_{X'} + |g(x_n) - g(x)|.\end{align*} Let $n \to \infty$ to conclude $$\limsup_{n \to \infty} |f(x_n) - f(x)| \le 2M \|f-g\|_{X'}.$$ Since $E'$ is dense in $X'$ the ...


1

The first term contains a typo, it should read $B(x,1/2)$ instead of $B(x + 1/2)$. Moreover, $F + B(0,1/2)$ is the Minkowski sum, which is defined via $$F + B(0,1/2) = \{x + y \colon x \in F, y \in B(0,1/2)\}.$$ Hence, this equality is essentially the definition of the Minkowski sum.


2

Presumably the author assumed that all readers are familiar with diagonal sequence constructions and therefore just gave a broad outline, skipping some important details of the procedure. You don't choose the subsequences independently. One uses the diagonal sequence construction to make sure that the selected subsequence (the diagonal sequence) is a ...


1

Another exmple to show that $B[0,1]$ is not compact: For $n\in N $ let $0<a_n<b_n<a_{n+1}<1.$ For $x\in [0,a_n]\cup [a_{n+1},1]$ let $f_n(x)=0.$ For $x\in [a_n,b_n]$ let $f_n(x)$ be linear with $f_n(b_n)=1.$ For $x\in [b_n,a_{n+1}]$ let $f_n(x)$ be linear. Then $1=\|f_n\|=\|f_n-f_m\|$ for $m,n\in N$ with $m\ne n$. So $\{f_n:n\in N\}$ is an ...


1

Hint: The proof sketch is correct If a subsequence of $(f_n)$ converges, it converges pointwise towards $f$ which is not continue at 1, so $B(0,1)\subset C([0,1])$ is not compact.


0

There are versions of the Hahn-Banach theorem from which this follows more or less directly. We can get it from the Hahn-Banach that everybody knows by a little trick: Say $D$ (for "difference") is the space of all $x\in\ell^\infty$ such that $$x=(y_1-y_2,y_2-y_3,\dots)$$for some $y\in\ell^\infty$. Let $X$ be the span of $D$ and $c$: $$X=D+c.$$We want to ...


2

Let $|z| < 1$ and choose $|z| < |w| < 1$. By assumption, the series $\sum_n w^n x_n$ is weakly convergent. In particular, $w^n x_n \to 0$ weakly, so that $(w^n x_n)_n$ is a bounded sequence (essentially, this is the uniform boundedness principle). But then $$ \|z^n x_n\| = \left|\frac{z}{w}\right|^n\|w^n x_n\| \leq C \cdot ...


1

Well, $T_zx$ is just the convolution of $x$ and $z$, so the inequality you want is just a special case of Young's inequality. In a discrete case like this Young's inequality should follow just from the triangle inequality in $\ell^2$ if you look at it right. Let's see: $$\begin{aligned}T_zx&=(0,x_1z_1,x_1z_2+x_2z_1,\dots) \\&=x_1(0,z_1,z_2,\dots) ...


0

I figured out what is wrong in my example. Of course $D$ as I defined it is not dense. I mixed up two spaces and didn't saw it later on. Thanks for any comment - they helped alot to rethink my example!


3

The result holds, and $X$ need not even be Banach. To see this, note that if $X$ is an infinite dimensional normed space, then there exists a discontinuous linear functional $\varphi$ defined on $X$. Since a linear functional is continuous if and only if its kernel is closed, it follows readily that the kernel of this functional $\varphi$ is a subspace of ...


0

Here is a counterexample. Let $X=Y=H$ be an infinite-dimensional Hilbert space with orthonormal basis $\{e_j\}$. Let $T_k$ be the projection onto the span of $\{e_1,\ldots,e_k\}$. Let $D$ be the span of $\{e_1,e_2,\ldots\}$. For each $x\in D$, for $k$ big enough $T_kx=x$, so the sequence $\{T_kx\}$ is Cauchy. As we have $T_kx\to x$, we have $T_k\to I$ ...


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 $x=(x_1,...,x_n)$, ...


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 ...


0

Hint: For every $a \in \mathbb{Z}$ try to combine $f$'s in such an order that as a sum you will get constant function over interval $(0;1)$ equal $a$. After that try to think of a rearrangement of other terms which converges to $0$.


1

A) Have a look at https://en.wikipedia.org/wiki/Complete_metric_space ex: $\mathbb{R}$ is the completion of $\mathbb{Q}$ for the natural distance. B) Here, your scalar product defines a norm, so a distance. For this distance, the completion of $\mathcal{C}^{\infty}(S)$ is $L^2(S)$. To be more precise, any Cauchy sequence of $\mathcal{C}^{\infty}(S)$ cannot ...


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

In general no. Take $Mv = \frac 12 v$. Fix $c \in V$ (nonzero, since otherwise the inequality holds trivially) and let $v = -c$. Then $$\|Mv + c\| = \frac 12 \|c\|$$ but $$\|v+c\| = 0.$$


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 ...


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.


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

As discussed in the comments, this boils down to showing that $C[0,1]\oplus\mathbb C\cong C[0,1]$. This works because $C[0,1]$ has a Schauder basis $g_n$; in fact, I want to work with specifically the Faber-Schauder basis. With respect to this basis, partial sums (up to $N=2^n$) of an expansion $f=\sum a_n g_n$ are piecewise linear interpolations of $f$ at ...


2

Let's define for $a\in A$ the left multiplication operator and the right multiplication operator $$ L_a : A \rightarrow A, L_a(x)=ax \quad \text{and} \quad R_a: A \rightarrow A, R_a(x)=xa.$$ As multiplication is assumed to be continuous we get that $L_a$ and $R_a$ are both continuous for all $a\in A$. Let $B=B_1(0,A)$ denote the unit ball in $A$. Then ...


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

Since $M+N(T)$ is closed, its complementary subspace $U$ is open, since $T$ is surjective, the open map theorem implies that $T(U)$ is open, but $T(U)$ is the complementary of $T(M)$, thus $T(M)$ is closed.


2

To show completeness, take a Cauchy sequence $(f_n)_{n\in \mathbb{N}}\subseteq F_b(\Omega, X)$ and show, there exists $f\in F_b(\Omega, X)$ such that $$ \lim_{n\rightarrow \infty} \Vert f - f_n \Vert_{b}=0.$$ As you already noted, one can construct a candidate for the limit of our sequence. Namely, we note that for fixed $x\in \Omega$ we have $$ \Vert ...


3

Because $\lambda\in\partial\sigma(A)$, then there exists $\{ \lambda_n \} \subset\rho(A)$ that converges to $\lambda$. Suppose for the moment that the following holds for all $x$: $$ l(x)=\sup_{n}\|(\lambda_n I-A)^{-1}x\| < \infty. $$ Then, by the uniform boundedness principle, $$ M=\sup_{n} \|(\lambda_n I -A)^{-1} \| < ...


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.


1

I think this is easier than it initially looks. First of all, $\mathcal G(\mathcal A)$ is an open subset in the Banach space $\mathcal A$, so there is no need to worry about coordinates. On Banach spaces, differentiation basically works as you are used to in open subsets of $\mathbb R^n$. Moreover, the maps you consider all are induces by linear (or ...


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

Yes. Fix $v \in A$, and fix $\varepsilon > 0$. Clearly if $\|v\| = 0$ then $\|L(v)\| = 0 = \|L\|\|v\|$, so suppose that $v \not=0$. Since $$ \|L\| = \inf\{M \ge 0 : \|L(w)\| \le M\|w\|\text{ for all } w \in A\}, $$ there exists $M \ge 0$ such that $\|L(v)\| \le M\|v\|$ for all $v \in A$ and such that $M \le \|L\| + \frac{\varepsilon}{\|v\|}$. In ...



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