New answers tagged

1

I recently wrote up a solution to this very problem, which I've copied below. Note that in this context, $\mathfrak A$ is the space of bounded operators on $B$. I hope you find this helpful. $ \newcommand{\f}{\mathfrak} \DeclareMathOperator{\rad}{r} \newcommand{\eps}{\varepsilon} \DeclareMathOperator{\spec}{spec} \newcommand{\lp}{\left(} \newcommand{\rp}{\...


0

Since $E^*$ is reflexive you got that $E^{**}$ is reflexive (you showed that). Since $E$ is isometric isomorphic to a closed, complete subset of $E^{**}$ you know that $E$ is also reflexive because every closed subspace of a reflexive space is also reflexive. Hope that it helps you :)


0

This is well known exercise many times solved on MSE. Here is one example. $\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}$


1

Consider $u\in B(H_1\oplus H_2, H_1\oplus H_2)$ and since $u^*u$ is projection, then $u$ is partial isometry, which means that $uu^*u=u.$


0

Pick any two-dimensional subspace of $X$, thinking of it as a plane $\mathbb{R}^2$ with another norm $\|\cdot \|$. Let $B=\{(x,y)\in \mathbb{R}^2 : \|(x,y)\|\le 1\}$; this is a bounded closed convex set. Let $P$ be the point of $B$ that is the furthest from $(0,0)$ in the Euclidean norm. We can choose coordinates so that this point is $(1,0)$. In particular,...


2

By this answer, you are looking at the set $$ B_{w^{\ast}}(X) = \{T^{\ast} : T \in B(X_{\ast})\} $$ which is convex. Now if $\sigma$ denotes the weak-$\ast$-topology on $B(X)$, then $(B(X),\sigma)$ is a topological vector space whose topology is generated by the family of semi-norms given by $$ p_{f,y}(T) := |T(f)(y)| $$ for $f\in X, y\in X_{\ast}$. Hence, ...


0

Consider the map $$\phi: l_r \to l_s : f \mapsto f.$$ This map is well defined, that is $$f \in l_r \Rightarrow f \in l_s;$$ to see this just observe the following inequalities for $f \in l_r$: $$||f||_s < || f ||_r < \infty.$$ This shows that $f \in l_s$. Clearly this map is injective. Suppose $\phi(f) = \phi(g)$, then $f = \phi(f) = \phi(g) = g$, ...


2

Ben has answered your question, but here is an even easier route to a counter-example (modulo the hard part of the existence of a space without the AP). Take your favourite separable space without the AP. You may embed it into $C[0,1]$, which actually has a basis (the Schauder system, a prototypical example of a Schauder basis), so in particular it has the ...


2

One can even have $T|_{c_0}=0$. Indeed, first note that if $X\subseteq \ell_\infty$ is isomorphic to $\ell_2$, then $X\cap c_0$ is finite-dimensional so there is a subspace $X\subseteq \ell_\infty$ isometric to $\ell_2$ with $X\cap c_0=\{0\}$. (Actually, by simple repetition of the coordinates in the standard way of embedding separable spaces into $\ell_\...


4

There are two things that will help: First, whenever $c_0$ is isomorphic to a subspace of a separable Banach space $X$ then that subspace is complemented in $X$. Second, whenever $Y$ is a subspace of a normed space $X$ and $T:Y\to\ell_\infty$ is a continuous linear operator then there exists a continuous linear extension $\widetilde{T}:X\to\ell_\infty$ ...


3

A more general case is when $B$ is a real normed space. You vcan apply Mazur–Ulam theorem : If ${\displaystyle V}$ and ${\displaystyle W}$ are normed spaces over $\mathbb{R}$ and the mapping ${\displaystyle f\colon V\to W} $ is a surjective isometry, then ${\displaystyle f}$ is affine. So $f$ is affine with $f(0)=0$, so $f$ is linear. So if $B$ ...


2

Every $\ell_p$, $p\neq 2$, and $c_0$ have closed subspaces without the ap. So the answer to your question is no.


1

The key point is to recognize that when you write $$x= \sum^\infty_{i=1}{\alpha_i}{e_i},$$ you mean $$ x=\lim_{n\to\infty} \sum^n_{i=1}{\alpha_i}{e_i} $$ (in other words, a series is not a sum, it is a limits of sums). This also requires that you recognize which topology you are using for the limit, which in this case is the norm topology. So your ...


3

So my real question is the following: suppose $Y = \{(x_i) :$ those sequences for which $ \lim_n \frac{1}{n} \sum_{i=1}^n \|x_i\| $ converges $\}$. Does $\|x\|_{\mathcal{U}} = \lim_n \frac{1}{n} \sum_{i=1}^n \|x_i\|$ on $Y$? This is in fact a question about real sequences. You are asking whether $$\newcommand{\limti}[1]{\lim\limits_{#1\to\infty}}\limti ...


1

I think your proof is fine. From your last inequations, $$|x(t)-x(s)|-|x_n(t)-x_n(s)|\leq|(x_n−x)(t)−(x_n−x)(s)|\leq \epsilon|t−s|,$$ for some sufficient large fixed $n$. Adding $|x_n(t)-x_n(s)|$ on the both sides gives $$|x(t)-x(s)|\leq\epsilon|t-s|+|x_n(t)-x_n(s)|\leq \epsilon|t-s|+\alpha|t-s|\leq(\epsilon+\alpha)|t-s|,$$ for some $\alpha>0$. ...


2

A cauchy sequence is bounded (in norm), so $$\|x_n\| \le C$$ for some $C$, which means $$|x_n(s) - x_n(t)|\le C|t-s|$$ for all $s, t, n$. Take $n\to \infty$ gives $$|x(s) - x(t)|\le C|t-s|$$ and so $x\in X$. Note that this paragraph should be mentioned before you wrote "So it follows $\|x_n-x\| \le\epsilon$" Otherwise your proof is good.


1

Suppose some rearrangement of this series converged to a vector $v\neq e_1$, and choose an index $k$ such that these vectors differ in the $k$-th component. That is, $p_k(v)\neq p_k(e_1)$, where $p_k$ is the $k$-th projection. Then, since $p_k$ is linear and continuous, applying $p_k$ to every term of the rearranged series, we get a series of numbers ...


1

There's nothing to see, actually. You can find a sequence $(x_n) \subset Y$ such that $x_n \to x \in X \setminus Y$. Then $(I_0(x_n))$ must converge to $I_0(x) \in Y$ (because we want $I_0$ continuous). But $I_0(x_n) = I(x_n) = x_n \to x \not \in Y$.


0

I don't follow your solution: you write "let $(x_n, y_n)$...", and then "let $(x_n)$...", which already makes it unclear what is going on. One can't "let" a thing be something twice. I would prove that $\operatorname{dom}(A)$ equipped with the graph norm is isomorphic to the graph of $A$, using the isomorphism $x\mapsto (x,Ax)$. So one of these is complete ...


2

Notice $$ \int_0^1 e^{\tau + t - 3} f(\tau) \, d\tau = e^t \int_0^1 e^{\tau - 3} f(\tau) \, d\tau .$$ So the equation can be rewritten as $$ f(t) = 1 - C e^t ,\tag 1$$ where $$ C = \int_0^1 e^{\tau - 3} f(\tau) \, d\tau .\tag 2$$ Now substitute equation (1) into equation (2), and solve for $C$.


4

Neither conjecture is true. Conjecture 1 fails trivially if $f=0$. For a less trivial counterexample that also violates Conjecture 2, let $X=\mathbb{R}^2$ with the sup norm, let $E=\{(x,0):x\in\mathbb{R}\}\subset X$, and let $f(x,y)=x$. Then $\|f\|=1$. Now take $x_0=(-2,1)$. If $g\in X^*$ agrees with $f$ on $E$ and is such that $g(x_0)>0$, then $$g(0,...


2

$C(X)$ is reflexive if and only if $X$ is finite (in which case $C(X)$ is finite-dimensional). Otherwise, you will find a seqeunce of norm-one disjointly supported functions. These functions span an (isometric) copy of $c_0$, which is certainly not reflexive. $C(X)$ is never isomorphic to $C(X)^{**}$ unless $X$ is finite. This is because $$C(X)^{**}\cong (\...


0

You should know that for each $f \in L^{p}_{loc}(\Omega)$ we can define a Distribution by: $T_{f}: \mathcal{D}(\Omega) \rightarrow\mathbb{R}$, where $T_{f}(\phi) = \displaystyle\int_{\Omega}f\phi$. In the case, $T_{f}$ is called regular distribution. The Du Bouis Reymond Lemma states that the map $f\mapsto T_{f}$ is one-to-one. However, there are ...


2

$W_C = \{u + iv \in X_C \mid u, v \in W\}$ is invariant under $T_C$, so we may as well forget about the real Banach spaces and consider an operator $T$ on a complex Banach space $X$ with a finite-codimensional invariant subspace $W$, such that $\sigma(T) \subseteq \mathbb R$. The claim is that $\sigma(T|_W) \subseteq \mathbb R$. If $\lambda \notin \sigma(T)...


1

You need the result that the topological closure of a set $A$ is the set of limits points of nets in $A$. To show it let $x$ be an element of $\overline{A}$, the set $\mathcal V_x$ of neighbourhoods of $x$ is a directed set and each neighbourhood of $x$ intersects $A$. So take as a net the map $\mathcal V_x \to A$ where each neighbourhood $V$ is sent to an ...


2

If $(f_n)$ is Cauchy in that norm then you can show that the two sequences $(\int_0^1 f_n)$ and $(\int_0^1 tf(t)\,dt)$ are Cauchy...


2

Looking at the definition of a norm on wikipedia (which matches the definitions I encountered in textbooks) (link here) if $V$ is a vector space, then a norm is a function $\rho: V\to \mathbb R$ that satisfies certain properties so it is quite explicitly stated that the norm of every vector is a real number, thus finite.


1

The following result proves (a). Proposition. Suppose that $X$ is a Banach space, and that $f \colon X \to X$ has the form $f(x)=x+r(x)$ where $\operatorname{Lip}(r)<1$. Then $f$ is a lipeomorphism and $\operatorname{Lip}(f^{-1}) \leq (1-\operatorname{Lip}(r))^{-1}$. Proof. Pick $u \in X$ and define $\xi_u$ by $\xi_u(x) = u-r(x)$. Thus $f(x)=u$ if and ...


2

The reverse triangle inequality yields $|G(x) -G(y)|\ge (1-L)|x-y|$. This gives the Lipschitz continuity of the inverse map, as soon as $G$ is shown to be surjective. For surjectivity, you are right to use the Banach fixed point theorem: the equation $x+F(x)=y$ has a solution for $x$ because the map $x\mapsto y-F(x)$ has a fixed point. The supremum norm of ...


3

Hint regarding the usefulness of the completeness of $C[0,1]$: Say $f_n$ is Cauchy in $X$. You can show that in general $||f||_\infty\le ||f||_X$, hence $f_n$ is also Cauchy in $C[0,1]$. So you have $f$ with $||f_n-f||_\infty\to0$. That does not show by itself that $||f_n-f||_X\to0$, but it does give you a handle on things - at least now you have the limit,...


1

No, this is not true. For example, let $X=\ell_1$, $X^*=\ell_\infty$, and $Y=c_0\subset \ell_\infty$. Then $Y$ is a weak* dense subspace, but not norm-dense. Fix a nonzero vector $v\in \ell_\infty$ and define $T:\ell_\infty\to\ell_\infty$ as $T(x) = \phi(x)v$ where $\phi$ is a Banach limit. Then $T$ is identically zero on $Y$, but not on $X^*$.


1

The space $c_0$ (sequences converging to $0$, with supremum norm) is separable but is not even isomorphic to any $\ell_p$ (never mind isometrically isomorphic). Indeed, its dual is $\ell_1$ — a separable nonreflexive space. The dual of $\ell_p$ is either nonseparable ($p=1,\infty$) or reflexive ($1<p<\infty$). Finite dimensions Every $2$-...


1

Since $\varphi(A)$ is an $\mathbb{R}$-interval, it is of the form $(a,b)$, $[a,b]$, $(a,b]$, $[a,b)$, with $-\infty\le a,b\le\infty$. Supose $\varphi(A)$ is not open. Then, WLG we can assume $\varphi(A)=[a,b)$. Then, there is $x\in A$ such that $\varphi(x)=a$. Since $A$ is open, there is $r>0$ such that if $\|x-y\|<r$ we have $y\in A$. Now, if $\|z\|&...


2

Let $\mathcal D$ be the subspace of $X$ consisting of functions of the form $f_1 \chi_{[0,1]} + f_2 \chi_{[2,3]}$ where $f_1$ and $f_2$ are polynomials and $\chi_I$ is the indicator function of $I$. On this domain consider the operator $T$ defined by $T(f_1 \chi_{[0,1]} + f_2 \chi_{[2,3]}) = f_2 \chi_{[0,1]} + f_1 \chi_{[2,3]}$. To see that this has dense ...


3

No. Say $X=L^1(\Bbb T)$, and let $$Y=U=H^1(\Bbb T)=\{f\in L^1(\Bbb T):\hat f(n)=0,n<0\}.$$So $H^1$ is just the classical Hardy space, the class of functions that extend to holomorphic functions in the unit disk. Define $T:U\to Y$ by $Tf=f$. If $T$ extended to an operator on $X$ that would be a projection from $L^1$ onto $H^1$, and there is no such ...


1

I cannot contribute references. But, regarding question 1, note that for $\lambda\in\mathbb R$, you have that $T-\lambda I$ is invertible if and only if $T_{\mathbb C}-\lambda I$ is invertible. Then $$ \sigma_\mathbb{R}(T)=\mathbb{R}\cap\sigma(T). $$ To show the above invertibility, let us write $T$ and $T_{\mathbb C}$. If $ST=TS=I$, then $$ S_{\mathbb C}...


1

Your question is unclear to me but I will comment on it anyway. What generating relations do you impose on your generators? Note that it is an open problem whether the reduced group algebra of the free group on two generators has a Schauder basis. Before Enflo showed that there exist Banach spaces without the AP, people had looked at this object as a natural ...


1

I suggest the book by Pazy on Semigroups of Linear Operators. It's one of the most elegant and readable books in Functional Analysis I've encountered. On page 14, there is a theorem: Theorem: A Linear operator is dissipative if and only if $$ \|(\lambda I-A)x\| \ge \lambda \|x\|,\;\; x\in\mathcal{D}(A),\; \lambda >0. $$ His setting for ...


3

You will find one proof here. Here's a possible different route. Show that for every compact Hausdorff space $K$, the Banach space $C(K)$ has Pełczyński's property (V): A. Pełczyński, Banach spaces in which every unconditionally converging operator is weakly compact, Bull. Acad. Polon. Sci. 10 (1962), 641-648. Show that dual spaces with property (V) ...


0

Hint for the $C(X)$ question: Suppose $X$ is a compact metric space with more than one point. (Note that if $X$ is not compact, it's not clear that $\| \,\|_\infty$ is a norm on $C(X).$) Consider $f(x) = 1/(1+d(x,a)),g(x) = 1/(1+d(x,a))^2.$


1

B is a sufficient condition to invoke the proof of the open mapping theorem (see Wiki). I guess that $T$ being open is sufficient to imply A.


1

Suppose $0<\mu (A)<\infty$ and $0<\mu (B)<\infty,$ and $A\cap B=\phi.$ (1).Let $f(x)=1/\mu (A)$ when $x\in A$ and $f(x)=0$ when $x\not \in A.$ Let $g(x)=1/\mu (B)$ when $x \in B$ and $g(x)=0$ when $x\not \in B.$ Then $f, g$ are linearly independent, and $\|(f+g)/2\|_1=\|f\|_1=\|g\|_1=1.$ (2). Let $f(x)=1$ when $x\in A$ and $f(x)=0$ when $x \...


1

I'll prove it showing both inequality's, first since $y_n = (x_1,x_2,...,x_n,0,0,...)\in c_0$ for all $n\in\mathbb{N}$ one has $$ \| x + c_0 (\mathbb{N}) \|_{\ell^{\infty} / c_0} \leq \inf_n \| x-y_n\|_\infty = \inf_n \sup_{m\geq n} |x_n| = \limsup_n |x_n|. $$ On the other hand by definition, for all $y=(y_1,y_2,y_3,...)\in c_0$ there is an $n\geq 1$ such ...


3

Suppose each $f_n$ is bounded, i.e. for each $n$, we have $\|f_n\|=\sup_{\|x\|=1}|f_n(x)|<\infty$. Define $T_n:X\to \ell^1(\mathbb N)$ by $x\mapsto \sum_{i=1}^n f_i(x)e_i$, where $\{e_i\}$ is the canonical basis for $\ell^1(\mathbb N)$. If $\|x\|=1$ then $$\sup_n\|T_n x\|=\sum_{i=1}^\infty|f_i(x)|<\infty, $$ so by Banach-Steinhaus we have $$\sup_n\...


0

It is well known in narrow circles that $$ d(Z,\ell_2^n)\leq\sqrt{n}\tag{1} $$ Furthermore it is easy to see straight from the definition that $$ d(U, V)\leq d(U,W)d(W,V)\tag{2} $$ Now set $U=X$, $V=\ell_\infty^n$, $W=\ell_2^n$ and apply $(1)$.



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