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0

The claim follows from the triangle inequality and positive homogeneity of the norm: Take $t_1,t_2\in\mathbb R$, $\lambda\in (0,1)$. Then $$ \|x-(\lambda t_1 + (1-\lambda)t_2)y \| = \| \lambda(x-t_1y) +(1-\lambda) (x-t_2 y)\| \le \lambda \|x-t_1y\| + (1-\lambda) \|x-t_2 y\| $$


2

The converse implication holds by James' theorem, which states that a Banach space $E$ is reflexive if and only if every continuous linear functional on $E$ attains its maximum on the closed unit ball of $E$. If a normed space $E$ has the property that every nested sequence of non-empty bounded closed convex sets has non-empty intersection, then $E$ is a ...


1

It might be useful to point out that there is a standard technique at work here. Often one wants to prove or disprove an inequality with a multiplicative constant $C>0$: $$\tag{1} A\le C B. $$ The contrapositive of the statement I) "there exists $C>0$ such that for all values of $A$ and $B$ one has $A\le CB$" is II) "for any constant $C>0$, ...


0

As you say, you need to prove that $$ \overline{span(\psi_n(x))} = L^2(\mathbb{R}). $$ For that, is sufficient to prove that, if $f\in L^2(\mathbb{R})$ satisfies $$ (f, \psi_n) = 0, $$ for all $n$, then $f = 0$, because the orthogonal complement of a dense subspace in a Hilbert space is $0$. Note that the linear span of Hermite polynomials is equal to the ...


2

In addition to the other answers and comments here, I think it might be useful to quickly drop the proof here that the linear operator $T$ is continuous if there exists $C\in \mathbb{R}$ such that $\Vert Tx\Vert \leq C\Vert x\Vert$ for all $x\in X$. Here it is: Assume we have $C \in \mathbb{R}$ so that $$ \forall{x\in X\colon\ } \Vert Tx\Vert \leq C\Vert ...


2

By definition, to be continuous means there is a fixed $C>0$ s.t. $$\Vert Tx\Vert\leq C\Vert x\Vert,\forall x\in X$$ So if $T$ is not continuous, then $\forall C>0$, $\exists x_C\in X$ (here $x_C$ depends on the choice of $C$) s.t. $$\Vert Tx_C\Vert > C\Vert x_C\Vert$$ In particular, you can choose $C=n$ for any $n\in\mathbb{N}$, and in which case ...


2

For an operator $$B: \mathcal{D}(B) \subseteq L \to \mathcal{R}(B) \subseteq L$$ the inverse $B^{-1}$ is a mapping $$B^{-1}: \mathcal{R}(B) \to \mathcal{D}(B);$$ in particular, $B^{-1} g \in \mathcal{D}(B)$ for any $g \in \mathcal{R}(B)$. Since the operator $B := (\lambda_n-A)$ satisfies $\mathcal{D}(B) = \mathcal{D}(A)$, we have $(\lambda_n-A)^{-1}: L \to ...


2

It is true: $C^*$-algebras are a special class of Banach algebras. But they have a very rich structure as they appear as selfadjoint algebras of operators on a complex Hilbert space. The theory of $C^*$-algebras is very developt. On the other hand, general Banach algebras are a much wider class of algebras and working on them one has to assume some ...


2

By exercise 4.8 in Tensor Norms and Operator Ideals. A. Defant, K. Floret $\ell_\infty \check\otimes E$ is isometrically isomorphic to a subspace of $\ell_\infty(E)$ consisting of relatively compact sequences. For the case of $c_0$ we have a criterion of relative compactness. Applying it we see that the sequence $(x_n)_{n\in\mathbb{N}}$ with ...


1

The example $f:c_0\to \mathbb R$, $(x_n)_{n\in\mathbb N} \mapsto \sum\limits_{n=1}^\infty x_n/2^n$ shows that the statement for the closed balls does not hold in general.


2

This answer is community wiki. The meet and join can be rewritten as: $$f\wedge g=\frac12\{(f+g)-|f-g|\}\quad f\vee g=\frac12\{(f+g)+|f-g|\}$$ (Note that addition, multiplication and modulus are continuous.)


1

Consider any separable metric space $X$ and a complete metric space $Y$. Then endow $C(X,Y)$ with the compact convergence topology: a sequence $(f_n)$ converges to $f\in C(X,Y)$ if and only if for every compact subset $K$ of $X$, $f_n\mid K$ converges uniformly to $f\mid K$. You can show this is actually metrizable, so that compact and and sequentially ...


0

The separability argument does not lose any generality. The reason is that a compact metric space is totally bounded (which you are using), and a totally bounded metric space is separable. So if you assume that the domain is compact then you get that it is separable for free. Here's a proof of the totally bounded implies separable argument: Suppose $X$ is ...


1

Writing $X = M \oplus N$ implies that the projection $\pi_1: X \to M$ is continuous. Then $N = \pi^{-1}(\{0\})$ must be closed.


2

You're probably looking for material related to the Stone Weierstrass Theorem. This is taken from Royden's Real Analysis: Theorem [Stone-Weierstrass]: Let $X$ be a compact space and $A$ an algebra of continuous real-valued functions on $X$ which separates the points of $X$ and which contains the constant funtions. Then given any continuous real-valued ...


0

Of course they are measurable since they are continuous. I found a counterexample to differentiability though, but forgot the source, it was $$f:[0,1]\rightarrow L^1([0,1]):t\mapsto{\bf 1}_{[0,t]}$$


1

Let me try to outline the solution. I. First, we describe the set of characters (continuous nonzero linear multiplicative functionals) $\varphi$ on $A,$ which is the same as $\hat A=\{\ker\varphi\mid\varphi:A\to\mathbb C\ \mbox{is a character}\}.$ As you have mentioned, there is a set of characters $\varphi_z\in\hat A$ which vanish on $C_\infty$ and ...


2

Yes, it is. It is obvious that $C_\infty (\Omega , E)$ is a normed linear space with norm $$||F||_\infty := \sup_{x\in \Omega} ||F(x)||_E . $$ To show that it is complete, let $F_n$ be a sequence in $C_\infty(\Omega , E)$ so that $\{F_n\}$ is Cauchy with respect to the norm. The completeness of $E$ imply that $F_n(x)$ converges to some $F(x) \in E$, for ...


3

How about this for a proof that $T$ is bounded? Consider the map $U:Y^*\to X^*$ given by $Uf = f \circ T$. We first show that $U$ has a closed graph: if $f_n \to f$ in $Y^*$ and $Uf_n = f_n\circ T \to g$ in $X^*$, then for all $x \in X$, it follows that $f_n\circ Tx \to f\circ Tx$ and $f_n\circ Tx \to g(x)$. Hence $g = f \circ T$. Since $X^*$ and $Y^*$ ...


0

Yes, choose $x^*\in E^*$ with $\|x^*\|=1$ and $|x^*(I_f)|=\|I_f\|$. Then $$\|I_f\|=|x^*(I_f)|=|\int x^*\circ f|\le\int|x^*\circ f|\le\int\|f\|.$$


0

I don't know if this is exactly what David had in mind, but here is one construction: Given a Hamel basis $\{e_\alpha\}$ you can make any vector space $X$ a normed space (norm properties can be checked) by setting $$ x = \max{\{|c_1|,\ldots,|c_k|\}} $$ where $x = c_1e_1 + \cdots + c_ke_k$ is the unique expression of $x$ as a finite linear combination of ...


1

Example Consider a nilpotent operator: $$N\in\mathcal{B}(E):\quad N^K\neq0\quad N^{K+1}=0$$ Then its resolvent has a singularity at zero only: $$\sigma(N)=(0):\quad(\lambda-N)^{-1}=\lambda\sum_{k=0}^K\lambda^{-k}N^k\quad(\lambda\neq0)$$ So the inequality necessarily becomes strict! Explanation Nilpotent operators have singularities of higher order at ...


0

Given a continuous morphism, the preimage of the unit ball in the target contains a ball in the source. If we scale the given morphism by some scalar of sufficiently small absolute value (which we can assume lies in $\mathbb Q$, if $K$ has char. zero but positive char. residue field, which seems to be implicit in the question), it will take the unit ball in ...


0

Hint: $(f_n)$ has a subsequence $(f_{n_k})$ that coverges to $f$ a.e.. Note $f^2_{n_k}\buildrel{k\rightarrow\infty}\over\longrightarrow f^2$ a.e., and apply Fatou's Lemma to $(f^2_{n_k})$.


1

Normally this would mean describe the points of $\hat{A}$ as functionals, and also describe their topology (so, up to homeomorphism if you like). Note though that you have only described those multiplicative functionals that vanish on $C_{\infty}$; there are lots of other multiplicative functionals that don't vanish on $C_{\infty}$.


0

This is a modified version from: Cohn: Measure Theory Enumerate a countable dense set: $$\#S\leq\mathfrak{n}:\quad S=\{s_1,\ldots\}\quad(\overline{S}=F\Omega)$$ Regard the finite subsets: $$S_K:=\{s_1,\ldots,s_K\}$$ Construct the domains by: $$A_k:=A_n(s_k):=\{\omega:\|s_k\|\leq\|F(\omega)\|\}\cap\{\omega:\|F(\omega)-s_k\|<\tfrac{1}{n}\}$$ And sum up ...


2

The answer is ‘no’. In fact, we can derive the statement from the following theorem: Theorem. Let $ X $ be a non-empty Baire space where every point is a limit point. Then there is no function $ f: X \to \mathbb{R}_{\geq 0} $ such that $$ \lim_{\substack{x \in X \setminus \{ a \} \\ x \to a}} f(x) = \infty $$ for every $ a \in X $. Proof Assume ...


0

This is taken from Bochner's original paper. By strong measurability there are simple functions: $$S_n\in\mathcal{S}(\mathbb{R}^d,E):\quad S_n\to F$$ So one can decompose the above into: $$\|D_r(F;z)-F(z)\|\leq D_r(\|F-S_N\|;z)+\|D_r(S_n;z)-S_N(z)\|+\|F(z)-S_N(z)\|$$ By linearity the second term reduces to: ...


1

Not if $E \ne F$. There's no multiplication.


2

Yes. It's ok. You better use $limsup$ and $liminf$. But the limit exists,so it's ok. With this you can see the mechanics of the solution. $|Tx|=|T(x-x_j)+Tx_j|\leq \|T\|\cdot|x-x_j|+|Tx_j|\leq \epsilon+k$ for $j$ big enough and arbitrary $\epsilon$


0

This is not true. Let $Y=C([0,1])$ itself. Define $$f_n(x):=\begin{cases} 0&\text{if}\ 0\le x\le\tfrac{n-2}{2n},\\ nx-\tfrac{n-2}2&\text{if}\ \tfrac{n-2}{2n}<x\le\frac12,\\ 1&\text{if}\ \tfrac12<x\le1. \end{cases}$$ Define $f=\mathbf1_{[\frac12,1]}$. Clearly $f_n\to f$ pointwise, and $|f_n-f|^2\le2\in L^1(0,1)$, so by the dominated ...


0

The answer is no. Let $D$ be the double arrow space and consider $D_1\sqcup D_2$, the disjoint union of two copies of $D$. This space is hereditraily separable. Yet the ideal generated by the indicator function of $D_1$ is non-separable.


0

I never heard of result you stated. I take it for granted. Thanks to the parallelogram law it is enough to show that any two-dimensional subspace is a Hilbert space. Now take any two-dimensional subspace $E$ of $X$. From assumption any $1$-dimensional subspace of $E$ is one-complemented in $X$ and a fortiori in $E$. Therefore, from result you stated in the ...


0

See cross-thread on MO: Approximation Property: Decomposition Astonishingly, almost finite rank operators always have a decomposition as series.


1

The space of all bounded functions on $[0,1]$ is nothing more than $\ell_\infty(\mathfrak{c})$. The latter space is injective. Assume $C[0,1]$ is complemented in $B[0,1]$, then it is also injective. Note that any injective Banach space contains a copy of $\ell_\infty$, and therefore non-separable. So $C[0,1]$ must be non-separable too. Contradiction.


4

As a general principle, if we have a unique possible limit, together with some form of sequential compactness, we get convergence. This is a corollary of the following: $x_n \to x$ if and only if for every subsequence $(x_{n_k})$, there is a further subsequence $x_{n_{k_j}} \to x$ (To prove the less obvious implication, suppose it doesn't converge and ...


0

This is not always true! Banach spaces which admit a compact approximation are said to possess the CAP. (Enflo actually gave a counterexample for the CAP.)


0

This is a nontrivial result by Grothendieck! (See Lindenstrauss & Tzafriri, Theorem 1.e.4, Volume I.)


2

You can use Minkowski's inequality to repair the proof. First of all, note that for every $k$ the numeric series $\sum_n x_n(k)$ converges absolutely; let $x(k)$ denote its sum. The goal is to prove that $x\in \ell^p$. By Minkowski's inequality, the partial sums $S_N=\sum_{n=1}^Nx_n$ are uniformly bounded in $\ell^p$: $$ \|S_N\|_p=\left\| \sum_{n=1}^Nx_n ...


-2

Edited. Clearly the reverse inequality is always true with $C = 1$. Here is why: $$\inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \} \le \|b - 0\|_B + t \|0\|_A = \| b\|_B $$ and hence $$ \sup_{t > 0} \inf_{\tilde{b} \in B} \{ \|b - \tilde{b}\|_B + t \|\tilde{b}\|_A \} \le \|b\|_B $$ for all $b \in B$. Therefore the inequality in your ...


1

Pick $\beta \in \mathbb R$ then $$\beta \langle f, x\rangle=\langle f, \beta x\rangle < \alpha $$ Since this happens for all $\beta \in \mathbb R$, it follows that $<f,x>=0$.


0

Let us abbreviate $j:=D(g)\circ f$, $h:=D(f)$. We can then write $j\cdot h$ as $b\circ(j,h)$ with $(j,h):E\rightarrow L(F,G)\times L(E,F)$ where we understand $b$ as the bilinear map $(g,f)\mapsto g\circ f$, then the rest consists of applying the chain rule in combination with the well-known derivative of bilinear functions.


1

There are several ways to prove the existence of such an $x$. All ways that I can think of make essential use of the fact that the closed unit ball of a reflexive space is weakly compact. That should not be surprising, since in non-reflexive spaces, such an $x$ need not exist. I will give two arguments, both use (the second argument only marginally) the ...


2

Hamilton's 1982 paper The inverse function theorem of Nash and Moser available here and here is also a very good reference. The space of Riemannian metrics on a compact manifold is a Frechet manifold.


1

This is a slightly overkill answer. Consider the following theorem: Every one-to-one bounded linear operator from a Banach space onto a Banach space is an isomorphism. Let $\ell^*$ denote the collection of $\ell^1$ sequences and endow it with the norm $\|\cdot\|$, for which it is a normed vector space. Suppose it is indeed complete, ie a Banach space (and ...


1

Hint: take some sequence which is just outside $\ell^1$ and play around with signs a little.


2

Observe that your norm is identical to the usual $\ell^1$ norm on sequences with all entries positive. So you'll have to think about alternating signs. The sequence with infinitely many terms of each sign which is closest to $\ell^1$ without being in it (this is an imprecise statement-just indicating why it's my guess) is $(1,-1/2,1/3,-1/4,...)$. The ...


2

A normed space is Banach if and only if whenever a series converges absolutely, it converges. Try to find a sequence $(x_n)$ with $$\sum \lVert x_n\rVert<\infty$$ but that cannot possibly converge.


0

For the inverse, consider $X+Y$ is closed, but (1) is not sure. This means $\forall c\geqslant 0$ we can find a $x\in X$ and $y\in Y$ such that, $\Vert x\Vert>c\Vert x+y\Vert $ divided by $\Vert x\Vert$ we can assume that $x$ has norm 1. then for any $\frac{1}{c}>0$ we can find $\Vert x+y\Vert<\frac{1}{c}$, now $\vert\Vert x\Vert-\Vert ...


1

A suggestion for the converse: since $X \cap Y = 0$, then you have a direct sum. This means that any vector $v \in X \oplus Y$ has a unique decomposition as $x+y$, with $x \in X$ and $y \in Y$. Define the operator $U : X \oplus Y \rightarrow X$, $U(v)=x$ (the projection onto the first summand). Choose $c= \Vert U\Vert$ (prove that it is finite).



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