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1

Let $P$ be an $L$-projection. From $$\lVert x\rVert = \lVert Px\rVert + \lVert x-Px\rVert$$ for all $x\in X$ it follows that $\lVert P\rVert \leqslant 1$ and $\lVert I-P\rVert \leqslant 1$, and hence $$\lVert P^\ast f\rVert = \lVert f\circ P\rVert \leqslant \lVert f\rVert\cdot \lVert P\rVert \leqslant \lVert f\rVert,$$ and analogously $\lVert (I-P)^\ast ...


1

If I'm reading Wojtaszczyk's book "Banach spaces for analysts" correctly, any absolutely summing operator maps weakly convergent sequences to norm convergent sequences. The Paley projection does not: if you denote by $(f_k)$ the trigonometric system and by $(e_n)$ the canonical basis of $\ell_2$, then $Pf_{2^n}=e_n$ for all $n\geq 0$, but $f_{2^n}\to 0$ ...


-1

Ok to bad, the answer is again: No! For $s_n:=n\chi_{[0,\frac{1}{n}]}$ it follows $s_n\to 0$ but $I(s_n)\to 1$.


2

This is basically continuation of the solution suggested in Daniel Fischer's comment. I will use notation $(w^{(n)})$ for the sequence of elements of $\ell_1$. They are sequences, I will use $w^{(n)}_k$ for the $k$-th term of the $n$-th sequence. We want to show $$\limsup_{n\to\infty} \|w^{(n)}-w\|_1 = \limsup_{n\to\infty} \|w^{(n)}\|_1 + \|w\|_1.$$ ...


-1

1.(Riesz-Dunford Functional Calculus) Consider the function $f(z):=|z|^2$ defined on the real and imaginary axis only. Then around every point it has an extension to a continuously differentiable function within some neighborhood. But that extension is confined to the Cauchy Riemann equations and therefore it must be $f(z)=+z^2$ and $f(z)=-z^2$ ...


0

You don't want to prove this by writing down the formula for all the derivatives. The formula would be an extension of the Campbell-Baker-Hausdorff formula to infinite dimensions and more non-commutative terms. It would have $p+1$ non-commuting terms for the $p$th derivative where C-B-H only has $2$ for all derivatives. Even the differentiability of a ...


0

This was resolved in comments: $\|T^*\|=\|T\|$ for every bounded linear operator between normed spaces. In fact, we only need the easier part here: $\|T^*\|\le \|T\|$, which comes directly from the fact that $T^*$ acts on linear functionals by composing them with $T$.


1

My suggestion is to consider the formal differential $$ \sum_{n=1}^\infty \frac{1}{(n-1)!}A^{n-1}h, $$ prove that it is (totally) convergent, and conclude that it must coincide with the derivative of $\exp (A)$ evaluated at $h$. Then iterate.


0

The claim is proven if $$\lim_{h\rightarrow 0} \frac{||\exp(A+h) - \exp(A) - \exp(A) h||}{||h||} = 0.$$ Now $$\exp(A+h) = \sum_{n=0}^{\infty} \frac{(A+h)^n}{n!} = I + \sum_{n=1}^{\infty} \frac{A^n}{n!} + \sum_{n=1}^{\infty} \sum_{m=1}^{n} \frac{A^{m} h A^{n-m-1}}{n!} + o(||h||)$$ so $$ \exp(A + h) = \exp(A) + \sum_{n=1}^{\infty} \sum_{m=1}^{n} ...


1

The duality $\ell^p(X)^*=\ell^q(X^*)$ for $1<p<\infty$ holds for every Banach space. Indeed, $c_{00}(X)$, the space of finitely supported sequences, is dense in $l^p(X)$. Therefore, every linear functional on $l^p(X)$ is determined by its values on sequences with one nonzero element. This identifies such a functional with an $X^*$-valued sequence ...


1

You want to show that there is a distortable subspace if and only if there is an equivalent norm that is not oscillation stable. You can make the connection between the two concepts the following: if $Y$ is a distortable subspace, then for an equivalent norm $|\cdot |$, $$ |y_1| / |y_2| \geq 1 + \delta$$ for some $\delta > 0$, and for $y_1, y_2$ on the ...


1

We can show actually more that $\ell_1$ and $\ell_\infty^*$ are not Banach-space isomorphic. (There are non-reflexive Banach spaces isometrically isomorphic to their second duals.) If you accept the fact that $\ell_\infty \cong C(\beta \mathbb{N})$ (which follows from the very definition of the Stone–Čech compactification applied to the discrete space of ...


2

This is the answer I've unravelled: Let $z \in (\ell^p)^{\ast\ast}$. I want to prove that exists $x \in \ell^p$ such that $\langle z,f\rangle=\langle f,x \rangle$ for every $f \in (\ell^p)^\ast$. I know that there are the isomorphisms: $j_p: \ell^q \rightarrow (\ell^p)^\ast$ and $j_q: \ell^p \rightarrow (\ell^q)^\ast$ Now, fix $z \in ...


0

It is not true that $$x=\sum_{n=1}^{\infty} x_ne_n$$ for $x\in\ell^{\infty}$, because convergence does not necessarily occur in the $\ell^{\infty}$-norm. Indeed, if $x=(1,1,1,\ldots)$, then, for any $N\in\mathbb N$, $$\left\|x-\sum_{n=1}^{N}x_ne_n\right\|_{\infty}=\|(\underbrace{0,\ldots,0}_{N\text{ times}},1,1,1,\ldots)\|_{\infty}=1,$$ so $x$ cannot be the ...


1

Note that "$x = \sum_n x_n e_n$" does not hold in norm if $x \in \ell^\infty \setminus c_0$, neither does it weakly, as for $x \not\in c_0$ there is some $x^*\in (\ell^\infty)^*$ with $x^*|_{c_0} = 0$ and $x^*(x) = 1$. Then $$ x^*\left(\sum_{n=1}^N x_ne_n\right) = 0 \not\to 1 = x^*(x) $$ It holds only weakly$^*$, as given $y \in \ell^1$, we have $xy\in ...


0

The sum you give is not norm convergent, unless $x\in c_0$. (It is weakly$\!^*$ convergent, but that is beside the point.) You have here a Schauder basis for $c_0$, a (comparatively) tiny subspace of $l^\infty$.


1

So you have the canonical maps: $$\begin{cases} f_q: \ell^p\to (\ell^q)^* \\ f_p: (\ell^q)^*\to (\ell^p)^{**}\end{cases}$$ which are the isomorphisms between $\ell^p$ and $(\ell^q)^*$ and $\ell^q$ and $(\ell^p)^*$ respectively. Then just write $f_p\circ f_q=j_p:\ell^p\to (\ell^p)^{**}$.


3

Being identified with is a loose term, it does not have a precise definition. It is used when we have a canonical isomorphism, i.e., an isomorphism that is naturally defined in terms of the objects themselves, without any choices made on our part. In our case, there is one canonical map from $X$ to $X^{**}$, namely $J$. If this map is an isomorphism, we can ...


2

The point is, when the authors of your book write "hence $X$ can be identified with $X^{**}$" they mean that $J$ is onto (and hence an isomorphism, as $J$ is isometric by Hahn-Banach), but want to say that it is more or less straightforward to see it. Let us show this explicitly for Hilbert spaces. Let $H$ be a Hilbert space with inner product ...


3

The point is the following: There are bounded functionals on $\ell^\infty$, which are not of the form $$ f(y) = \sum_k x_k y_k $$ for some $x$. I do not know if such a functional can be given explicitly, but they do exist. Let $f \colon c \to \mathbb R$ (where $c \subseteq \ell^\infty$ denotes the set of convergent sequences) be given by $f(x) = \lim_n x_n$. ...


1

This is less smart solution than Christian's. For the begining, every measure space $(\Omega,\mu)$ have continuous $(\Omega_c,\mu_c)$ and atomic part $(\Omega_a,\mu_a)$. Note, for $\sigma$-finite spaces, all atoms are of finite measure. See this question. Case 1. If atomic part contains infinite amount of atoms we may choose a countalble collection of of ...


1

Call $A\in\Omega$ an atom if $\mu(A)>0$, but if $B\subset A$ is measurable, then $\mu(B)=0$ or $=\mu(A)$. Since $\mu$ is $\sigma$-finite, we can cover all atoms by at most countably many atoms $A_n$. Let $Y=X\setminus \bigcup A_n$. Then $\mu(Y)=0$ if $L^1(X,\mu)$ is a dual space because the unit ball of $L^1(Y,\mu)$ has no extreme points: if $\|f\|=1$, we ...


4

Let us try to estimate $|B|$ from below. Every space $B_{n-1}^*$ ($n\geqslant 1)$ is of the form $C(K_n)$ for some compact Hausdorff space. For example, $K_1 = \beta\mathbb{N}$ and $|K_1| = \beth_1$. In particular, by the Riesz–Markov–Kakutani representation theorem each space $B_n$ is isometric to the space $M(K_n)$ of Radon measures on $K_n$. Moreover, we ...


2

You can use the criterion that reflexivity is equivalent to the unit ball being compact in the weak topology. Then the intersection of the unit ball with the closed subspace (which is still closed in the weak topology) is also compact, hence $E\subseteq X$ is reflexive. Edit: If you want to do it directly, you can use the Hahn Banach theorem to do it ...


0

The definition of a Besov space depends on a number of parameters. Some Besov spaces can boil down to certain Sobolev spaces and Sobolev spaces on nice enough domains tend to be isomorphic to $L_p$. In general, however, Besov spaces need not embed to $L_p$. A standard reference for completeness of Besov spaces is Chapter 3 of: J. Peetre, New thoughts on ...


0

NOTE: There is no such thing as a boundary condition $f(0)=0$ or $f(\pi)=0$ on $L^{2}[0,\pi]$. Functions in $L^{2}[0,\pi]$ are equivalence classes of functions which are equal to each other a.e.. That part should be tossed out of your question. What is true is that $$\left\{ \sqrt{\frac{2}{\pi}}\sin(nx)\right\}_{n=1}^{\infty}$$ is a complete orthonormal ...


1

Try the following, if you want to avoid any Fourier series methods: first throw in the function $1$ into your set $\{\sin{kx}\}$. The resulting linear span is a subalgebra of $C(T,\mathbb{R})$, where $T$ is the one-dimensional torus. Then apply Stone-Weierstrass to obtain density in the $C$-norm. Now use the fact that $C(T,\mathbb{R})$ is dense in $L^2(T, ...


1

Hints: Let's use the max norm: $|| x||=\max |x_i|$. The $(i,j)$ entry of the Jacobian: $Jf(x)_{ij} =-\dfrac{d_{ij}}{(1+D_{i*}x)^2}$. Now $\forall y$ such that $||y||=1$, we have $|| Jf(x) \cdot y ||\le \max_{i,j} d_{ij}<1$. Why? The are gaps here for you to fill. You need to use the fact $x\in [0,1]^n$. So the norm of the Jacobian ...


1

Your interpretation of $(x^\ast \otimes y)(x) = x^\ast(x)y$ is the correct one. Ryan explains several different interpretations of tensors in section 1.3 of his book, among which the one intended here. With this interpretation, the algebraic tensor product $X^\ast \otimes Y$ is simply the space of finite rank operators from $X$ to $Y$. In particular, ...


1

You need to show that $B=i(B(0,1))$ is relatively compact. Suppose $y_n \in B$. Then since $\|y_n\|_0 + \|y_n'\|_0 < 1$, we see that $\|y_n\|_0 < 1$ for all $n$, hence $y_n$ are uniformly bounded. Furthermore, since $\|y_n'\|_0 <1$, we see that the $y_n$ are Lipschitz with rank at most one, hence equicontinuous. Applying the Arzelà–Ascoli theorem, ...


1

Suggestion: Apply the Arzelà–Ascoli theorem.


1

Here's a couple more examples. If $f : \Omega\subseteq\mathbb{C} \rightarrow \mathcal{L}(X)$ is a function from an open subset $\Omega$ of the complex plane into the bounded linear operators $\mathcal{L}(X)$ on a complex Banach space $X$, then $f$ is holomorphic iff $\lambda\mapsto x^{\star}(f(\lambda)x)$ is holomorphic on $\Omega$ for all $x \in X$, ...


1

Let $f\in L^{p}(\mathbb{T})$, for some $1\leq p<\infty$, where $\mathbb{T}$ denotes the one-dimensional torus. Let $(a_{m})_{m\in\mathbb{Z}}$ be a bounded complex sequence. For $R\geq 0$, let $(a_{m}(R))_{m=1}^{\infty}$ be a compactly supported sequence such that $a_{m}(R)=a_{m}$ for all $\left|m\right|\leq R$. Define ...


5

The statement is equivalent to the Continuum hypothesis. Indeed, taking a quotient cannot increase the density character, and the density character of $l_\infty$ is $c$. This gives one implication. If $\aleph_1=c$, then pick a dense subset of cardinality $\aleph_1$ in the unit ball of $l_\infty$. Map the standard basis vectors of $l_1(\aleph_1)$ ...


2

Use the following: Let $X$ be a Banach spacce. 1) $(y_n)$ is a basis of $X$ if and only if each $x\in X$ has a unique representation $\sum\limits_{i=1}^\infty \alpha_i y_i$. 2) If $\sum\limits_{i=1}^\infty z_i$ converges unconditionally to $z$, then any rearangement of $\sum\limits_{i=1}^\infty z_i$ converges to $z$. 3) If $(x_n)$ is a basis of $X$ with ...


2

Yes, the chain rule applies here, assuming $T$ is at least Gateaux differentiable. The derivative you consider is the direct analog of directional derivatives of multivariable calculus, $D_v f = \frac{d}{dt}f(x+tv)$ where $x,v$ are elements of vector space and $t$ is a scalar. Chain rule applies: $D_v f = \nabla f \cdot v$. Yes, the inner derivative can be ...


1

Note that every separable space can be given an equivalent strictly convex norm. Indeed, if $X$ is separable then you can easily construct an injective operator $T\colon X\to \ell_2$. Then define a new norm on $X$ by $\|x\|^\prime = \|x\|+\|Tx\|_{\ell_2}$, which is strictly convex. (I guess this trick is attributed to Victor Klee.) Now, to answer your ...


1

$P$ isn't even well-defined. For the corresponding map $\tilde P \colon X \times Y \to X$, $\tilde P(x,y) = x$ is not bilinear (if $X \ne 0$). If $\tilde P$ is bilinear and hence $P$ well-defined, we must have $$ x = P\bigl(x \otimes (y_1+y_2)\bigr) = P(x \otimes y_1 + x \otimes y_2) = P(x\otimes y_1) +P(x \otimes y_2) = 2x $$ for all $x$, hence $X = 0$. ...


3

It's false. For a strictly convex norm, given $f$, $x$ is unique (if it exists); the dual notion, where, given $x$, $f$ is unique, is called "smoothness" in convex geometry. For a concrete example, take $\mathbb{R}^2$ with the norm whose unit ball is the intersection of two disks of radius $2$, centred at $(0,\pm 1)$. The boundary of the unit ball ...


1

Yes, I believe so (if by invertible, you mean boundedly invertible, and by $BL$ you mean bounded linear operators). First, recall the sequential characterization of closure: $R$ is a closed linear operator $A \to C$ if and only if, whenever $(\phi_n) \in \text{dom} (R)$ satisfies $$\phi_n \to \phi \, \text{(in $A$)}, \quad R\phi_n \to \psi \, \text{(in ...


1

The map $(a,b) \mapsto (a,Sb)$ is a homeomorphism $A \times B \to A \times C$ and it maps the graph of $T$ to the graph of $ST$. Therefore $T$ is closed if and only if $ST$ is closed.



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