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## Hot answers tagged banach-spaces

6

One useful example is the holomorphic functional calculus. It allows us to generalize Cauchy's integral formula from complex analysis in one variable to evaluate functions of operators. Let $V$ be a Banach space and let $T$ be a bounded linear operator on $V$. If $\Gamma$ is a positively oriented rectifiable Jordan curve such that the spectrum of $T$ is ...

4

First of all, in order to "calculate" something explicitly you need a reasonably nice Banach space $B$ as your target. For instance, suppose your Banach space is $B=C([a,b])$. Then a continuous function $f: [0,T]\to B$ is nothing but a continuos function of two variables $F(x,t)$, $x\in [a,b], t\in [0,T]$: $$f(t)(x)= F(x,t).$$ Now, computing the Bochner ...

4

The statement is false, as I discovered here. Since not everybody has access to the paper, let me provide a summary of the argument: Let $R=\mathbb C[t]$, $L$ the left shift operator, and view $\ell^p$ as an $R$-module by defining $t\cdot x=Lx$. Let $X=\sum \ker L^i \subset \ell^p$ be the subspace of eventually-zero sequences. Lemma: Given a PID $P$, a ...

3

This definition of quotient has metric in mind, since we ask a unit ball to be mapped to unit ball. As such, it has a natural generalization to metric spaces $X,Y$, but not necessarily to general topological spaces. (What's a generalization of "isometry" to topological spaces?) Let $B(x,r)$ be the open ball of center $x$ and radius $r$, and ...

3

Let $A:=\bigcup_{k=1}^\infty A_k$. Since $A$ is contained in a complete normed space, it suffices to show that $A$ is totally bounded. So fix $\varepsilon\gt 0$ and $k_0$ such that $\sum_{k\geqslant k_0}r_k\lt\varepsilon$. For $m\geqslant 1$, $$A_{k_0+m}\subset\left\{x=a_{k_0}+\sum_{l=1}^mx_k, x_k\in A_{k_0+k},\lVert x_k\rVert\leqslant r_k\right\}.$$ This ...

3

There is no reason to expect that the $g_n$ converge, the analogous implication does also not hold in $\mathbb{R}$. For an - admittedly boring - explicit example, take $f_n = \chi_{[0,2]}$ for all $n$ (where $\chi_A$ is the characteristic function of the measurable set $A\subset \mathbb{R}$), and $g_n = \chi_{[n,n+1]}$. Evidently the constant sequence $f_n$ ...

3

From the inclusions of sets $$\ell_p\subset c_0\subset \ell_\infty,$$ where $c_0$ denotes the set of convergent sequences and the separability of $(c_0,\lVert \cdot\rVert_\infty)$ is separable we conclude that $(\ell_p,\lVert\cdot\rVert_\infty)$ is separable. $\ell_p$ endowed with the supremum norm is not a closed subspace of $\ell_\infty$ because its ...

3

The weak topology is weaker than the norm topology. This implies that every weakly continuous map (needn't even be linear) is norm-continuous. For the other implication: The weak topology is (more or less by definition) the weakest topology making all norm-continuous linear functionals continuous. Therefore, the norm-continuous functionals are weakly ...

2

I think they are order continuous if $q < \infty$. If $0 \le f_1 \le f_2 \dots \le g$, then $\mu(g>t) \ge \mu(f - f_n > t) \to 0$ where $f = \sup f_n$. Since $$\| f - f_n\|_{p,q}^q = \int_0^\infty [(f-f_n)^*(x)]^q dx^{p/q} = \int_0^\infty [\mu(f-f_n>t)]^{p/q} dt^q,$$ we see by dominated convergence that $f-f_n \to 0$ in $L^{p,q}$. If $q = ... 2 If you indeed find extremely awkward proofs tempting, then go for it... but a proof of $$\lim_{h\to0} \frac{|\sum_{k = 1}^{+\infty}h_kh_{k+ 1}|}{\|h\|_{l_2}} = 0$$ need not be awkward. Since$2h_{k}h_{k+1}\le h_k^2+h_{k+1}^2 $, the numerator is at most$\|h\|_{l_2}^2$. Can the first solution be somehow linked with the second? In the second ... 2 The claim is that for any$a \in A$,$\sigma(a) = \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$. Hence, to prove this, you take any element$a \in A$, and show that for that fixed element$a$,$\sigma(a) \subseteq \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$and$\sigma(a) \supseteq \{\chi(a) \mid \chi \in \Omega(A)\} \cup \{0\}$. So, you must fix a ... 2 No. First of all, you need to restrict yourself to bounded$C^1$functions with bounded derivative (which we might denote by$BC^1(X,Y)$, otherwise the$C^1$"norm" may be infinite. But consider$X=Y=\mathbb{R}$. I claim$BC^1(\mathbb{R}, \mathbb{R})$is not separable, for the same reason that$BC(\mathbb{R}, \mathbb{R})$is not separable. Specifically, ... 2 Your idea I thought about constructing some$\hat{T}\colon X\times Z\to Y$such that is linear, continuous and surjective is the right one. To make things a little easier to see, we assume that$T$is injective (by considering$\tilde{T}\colon X/\ker T \to Y$, we lose no generality). Since$Y/\mathcal{R}(T)$is finite-dimensional, we can choose$Z$to ... 2 We can easily construct an inverse to$\phi^*$, it is given by$(\phi^*)^{-1}=(\phi^{-1})^*$, indeed we have: $$\left<(\phi^{-1})^*\phi^*v,x\right>=\left<\phi^*v,\phi^{-1}(x)\right>=\left<v,\phi\phi^{-1}(x)\right>=\left<v,x\right>$$ for all$x\in X$and$v\in X^*$. Thus$(\phi^{-1})^*\phi^*=1$, and similarly$\phi^*(\phi^{-1})^*=1$. 1 No. Take$A=c_0$. Then$c_0$is complemented in$\ell_\infty(c_0)$(just project down on the very first coordinate). Note that$c_0$is never complemented in a dual Banach space as this is equivalent to being complemented in$c_0^{**}\cong \ell_\infty$, which is apparently not the case. See t.b.'s answer for Complementability of von Neumann algebras ... 1 Since$x$is a Shauder basis then for all$y\in X$we have unique representaion $$y=\sum_{n=1}^\infty a_k(y) x_k$$ It is a standard fact proved in each Banach geometry course, that$a_k:Y\to\mathbb{C}:y\mapsto a_k(y)$is a bounded linear functional (see e.g. theorem 1.1.3 in Topics in Banach space theory. F. Albiac, N. Kalton). Then for$a_k^n:=a_k(y_n)$... 1 Let$T:Y\to H$be the embedding operator and$T^*:H^*\to Y^*$its adjoint. Suppose the range of$T^*$is contained in a proper closed subspace of$Y^*$. There is a bounded linear functional on$Y^*$that vanishes on that subspace. By reflexivity, this functional is evaluation on some nonzero element$y\in Y$. Let's spell out what this means: for every ... 1 I don't see a way forward with a Hamel basis (maybe it's the lack of coffee). Anyway, I think a more natural way is to identify$X$with a dense subset of$X\oplus \mathbb K$(where$\mathbb K$is your field of scalars, real or complex). The product can be equipped with any product norm, e.g.,$\|(x,t)\| =\|x\|+|t|$. Then the norm of$X\oplus \mathbb K$, ... 1 This should be a comment, but it's too long. I'm not sure I understand what you are defining. Typically a stochastic process is something like a jointly measurable function$X : \Omega \times (a,b) \to \mathbb{R}$, where$(\Omega, \mathcal{F}, P)$is a probability space and$(a,b)$is some interval (which could be replaced by a closed interval, an ... 1 At first I wanted to follow the reasoning showed in "Brezis - Functional Analysis Sobolev Spaces and Partial Differential Equations" at pages 59-60, riassumed in this identity $$B_V =\bigcap_{f \in V^*,\|f\| \leq 1} \{x \in V \mid |\langle f,x \rangle | \leq 1 \}$$ (two words about it: it relies on the characterization of open/closed set in the weak topology ... 1 First let's recall what it means to say$\displaystyle\lim_{n\to\infty} x_n = x$in a Banach space. It means$\displaystyle\lim_{n\to\infty}\|x_n-x\|=0$. The first limit is a limit of a sequence of members of the Banach space; the second is a limit of a sequence of non-negative numbers. Note that$\|A\|=\|(A-B)+B\|\le \|A-B\|+\|B\|\$, so that ...

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