A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.

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Compactness of a bounded operator $T\colon c_0 \to \ell^1$

Pitt Theorem says that any bounded linear operator $T\colon \ell^r \to \ell^p$, $1 \leq p < r < \infty$, or $T\colon c_0 \to \ell^p$ is compact. I know how to prove this in case $\ell^r \to ...
8
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2answers
360 views

Is a completion of an algebraically closed field with respect to a norm also algebraically closed?

Assume we have an algebraically closed field $F$ with a norm (where $F$ is considered as a vector space over itself), so that $F$ is not complete as a normed space. Let $\overline F$ be its completion ...
6
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3answers
633 views

Bounded Linear Mappings of Banach Spaces

This problem has been giving me some troubles. Does anyone have any ideas on how to go about proving this? Let $X$ and $Y$ be Banach spaces. If $T: X \to Y$ is a linear map such that $f \circ T \in ...
5
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3answers
290 views

Is a sequence of disjointly supported functions in $L^\infty$ complemented?

Let $(f_n)_{n \geq 1}$ be disjointly supported sequence of functions in $L^\infty(0,1)$. Is the space $\overline{\mathrm{span}(f_n)}$ (the closure of linear span) complemented in $L^\infty(0,1)$? By ...
8
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1answer
272 views

What is the ''right" norm for the Banach space tensor product in this situation?

Let $X,Y$ denote (real) vector spaces. The vector space of $n$-linear maps $X^n \to Y$ will be denoted by $L^n(X,Y)$. Unless I'm much mistaken $$L(X,L(X,Y)) \ \ \ L^2(X,Y) \ \ \ L(X \otimes X,Y)$$ ...
5
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2answers
643 views

construction of a linear functional in $\mathcal{C}([0,1])$

Can someone help me to construct a linear functional in $\mathcal{C}([0,1])$ that does not attain its norm? Actually, I want to prove that $\mathcal{C}([0,1])$ is not reflexive Banach space. Is it ...
8
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2answers
991 views

On the density of $C[0,1]$ in the space $L^{\infty}[0,1]$

It's easy to show $C[0,1]$ is not dense in $L^{\infty}[0,1]$ in the norm topology, but $C[0,1]$ is dense in $L^{\infty}[0,1]$ in the weak*-topology when take $L^{\infty}$ as the dual of $L^{1}$. how ...
2
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1answer
209 views

Preservation of Lipschitz Constant by Convolutions

The following is a step in a proof: $f$ is a Lipschitz function from $E$ to $F$ where $E$ is a finite-dimensional Banach space and $F$ an arbitrary Banach space. $\phi\geq 0$ is a $C^\infty$ function ...
12
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1answer
765 views

Nested sequences of balls in a Banach space

This seems to be a fairly easy question but I'm looking for new points of view on it and was wondering if anyone might be able to help (by the way- this question does come from home-work, but I've ...
1
vote
0answers
172 views

Rotund and smooth space

Suppose $X$ is normed space. Prove that $X^\ast$ is rotund iff $X/M$ smooth whenever $M$ is a closed subspace of $X$ such that $X/M$ is two-dimensional. I know that "a normed space is smooth iff each ...
4
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Uniform Boundedness/Hahn-Banach Question

Let $X=C(S)$ where $S$ is compact. Suppose $T\subset S$ is a closed subset such that for every $g\in C(T),$ there is an $f\in C(S)$ such that: $f\mid_{T}=g$. Show that there exists a constant $C>0$ ...
1
vote
1answer
2k views

Space of all continuously differentiable functions

Let $C^1[0,1]$ be space of all real valued continuous function which are continuously differentiable on $(0,1)$ and whose derivative can be continuously extended to $[0,1]$. For $f$ in that set define ...
5
votes
1answer
355 views

Completeness of BMO without duality to $H^1$

for $f \in L^1_{loc}$, let the sharp-function be defined as $f^\sharp(x) := \sup_{B \in x} |B|^{-1} \int_B |f(y) - |B|^{-1} \int_B f(z) dz | dy$ We define the space $BMO \subset L^1_{loc}$ via $BMO ...
6
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3answers
283 views

$L^1$ space with values in a Banach Space

I have been reading a bit about the Bochner integral and now I'm wondering the following: For the theory to be "nice", one would expect that $$L^1([0, \tau], L^1([0, \tau])) \cong L^1([0,\tau] ...
7
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2answers
879 views

Why is the space of surjective operators open?

Suppose $E$ and $F$ are given Banach spaces. Let $A$ be a continuous surjective map. Why is there a small ball around $A$ in the operator topology, such that all elements in this ball are surjective?
7
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2answers
436 views

Finding all linear operators $L: C([0,1]) \to C([0,1])$ which satisfy 2 conditions

As above, I'm trying to find all linear operators $L: C([0,1]) \to C([0,1])$ which satisfy the following 2 conditions: I) $Lf \, \geq \, 0$ for all non-negative $f\in C([0,1])$. II) $Lf = f$ for ...
0
votes
1answer
357 views

A formal series in the Banach space $c_0$

Let $(e_n)$ be the standard unit vector basis for $c_0$, let $x_1=e_1$, $x_n=e_n-e_{n-1}$ when $n\gt1$. Prove that the formal series $\sum x_n$ is not weakly subseries convergent.
25
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3answers
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Norms on C[0, 1] inducing the same topology as the sup norm

This is an old homework problem of mine that I was never able to solve. The solution may or may not involve the Baire category theorem, which I am terrible at applying. Let $C[0, 1]$ denote the ...
2
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1answer
176 views

Lipschitz flows on Banach spaces

I have a simple question on a Lipschitz flow. Recall that a flow on a non-empty set $X$ is a map $\Phi: \mathbb R \times X \to X$ satisfying $$\Phi(0,x) = x \text{ and } \Phi(s, \Phi(t, x)) = \Phi(s ...
3
votes
1answer
364 views

Limit point of basic sequence in Banach space

This question is about Schauder bases in Banach spaces. Suppose $(x_n)$ is a basic sequence in a Banach space $E$. Prove that $0$ is the only possible weak limit point of $\{x_n \in E \;\vert\; n \in ...
3
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2answers
636 views

Converging series in Banach space

Does someone know if the following is true: Let $\mathbb{X}$ be some arbitrary Banach space. $\{x_k \}_{k=1}^{\infty} \in \mathbb{X}$ is a sequence chosen from $\mathbb{X}$. Now, if the series ...
5
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1answer
380 views

Is there a notion of basis for Banach spaces?

Consider the Banach space $\ell^1(\mathbb N)$. The sequence $(e_n)_{n\in\mathbb N}$ feels like a kind of basis because every element $a\in\ell^1(\mathbb N)$ can be written as an absolutely convergent ...