# Tagged Questions

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### Properties of a set in $\ell^2$ space

Let $\ell^2 = \{x= (x_1,x_2,x_3,\ldots): x_n\in \mathbb C\text{ and } \sum_{n=1}^\infty |x_n|^2 < \infty\}$ and $e_n \in \ell^2$ be the sequence whose $n$-th element is 1 and all other elements ...
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### Dense subsets of $(L^p(\Omega),\|\cdot\|_p)$

The following results hold. Theorem Let $\Omega\subset\mathbb{R}^n$ be an open set. Then $C^0_c(\Omega)$ is dense in $(L^p(\Omega),\|\cdot\|_p)$, if $1\le p<\infty$. Theorem Let ...
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### On finite measurable space $X$, the whole of $L^p(X)$ is closed in $L^1(X)$ iff there is a const $C$ st $||f||_p\leq C||f||_1, \forall f \in L^p(X)$

On finite measurable space $(X, \mathcal{M}, \mu)$, the whole of $L^p(X, \mu)(p>1)$ is closed in $L^1(X,\mu)$ iff there is a const $C$ st $||f||_p\leq C||f||_1, \forall f\in L^p(X)$, iff both ...
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### what are closed sets in $L^{1}(\mathbb R)$?

Consider, $L^{1}(\mathbb R)$= The space of Lebesgue integrable functions on $\mathbb R$; for $f\in L^{1}(\mathbb R),$ we define its norm, by $\|f\|_{L^{1}}=\int_{\mathbb R}|f(x)| dx$; It is well-known ...
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### $(\ell_p,\|.\|_{\infty})$ Banach or separable

Is $(\ell_p,\|.\|_{\infty})$ for $1\leq p<\infty$ a Banach or separable space? is there any fast way of proving it without checking separability or completeness with the usual way?
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### $\ell^{\infty}(\mathbb N)$ is not a separable space

I have to prove that $\ell^{\infty}(\mathbb N)$ is not separable. My attempt Consider a SUBSET $V$ of $\ell^{\infty}(\mathbb N)$ consisting of bounded sequences that have only $0$, $1$ entries, e.g. ...
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### Countably generated $\sigma$-algebra implies separability of $L^p$ spaces

Let $\Sigma = \sigma(\mathcal C)$ be the $\sigma$-algebra generated by the countable collection of sets $\mathcal C \subset \mathcal{P}(X)$. How can I prove that if $\mu$ is a $\sigma$-finite measure ...
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### Closed set in $l^1$ space

Let $$X := \left \{ (a_n) : \sum_{n=0}^\infty |a_n| < \infty \right\}$$ with the metric $d(a_n,b_n) := \sum_n |a_n-b_n|$. Let $\delta_j^{(n)} := 1$ if $n = j$ and $0$ otherwise. Denote ...
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### Closedness of $L^\infty_+$ in $\sigma(L^\infty,L^1)$

Let $L^\infty_+$ be the set of all $f\in L^\infty$ which are non negative. Our measure can be assumed to be finite. My goal is to prove that $L^\infty_+$ is closed in the weak-star topology ...
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### Proving that certain subspace of $\ell_1$ is non closed

I need to prove that $$L= \left\{(x_i) \in\ell_1 : \sum_{i=1}^\infty ix_i= 0\right\}$$ is non-closed in $\ell_1$. I can't really think of sequences of sequences that are in this subspace, much less ...
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### Show the subspace $\textrm{span}\{e_n:n\in\mathbb N\}$ is not closed in $\ell^2$

How to show the subspace $\textrm{span}\{e_n:n\in\mathbb N\}$ where $e_n=(\delta_{nk})_{k\in\mathbb N}$ is not closed in $\ell^2$?
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### Closure of a subspace of $l^\infty$

Let $X$ be the following subspace of $l^\infty$: $$X=\mathrm{lin}\{e_n:n\in\mathbb{Z}^+\}$$ where $e_j$ has zeroes everywhere except for one in the $j$-th entry. I want to know what the closure of ...
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### Problem 4 chapter 2: functional analysis (Rudin)

$L^1$, $L^2$: usual Lebesgue spaces on the unit interval. Show that $L^2$ is of the first category (meager) in $L^1$, in three ways: (a) Show that $F_n:=\{f:\int|f|^2 \leq n\}$ is closed in ...
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### Closed set in $\ell^1$

Show that the set $$B = \left\lbrace(x_n) \in \ell^1 : \sum_{n\geq 1} n|x_n|\leq 1\right\rbrace$$ is compact in $\ell^1$. Hint: You can use without proof the diagonalization process to ...
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### Why is $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ dense in $L^2(\mathbb{R}^n)$?

In Lieb and Loss's Analysis, I saw that they mentioned $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ dense in $L^2(\mathbb{R}^n)$ (dense wrt the $L^2$ norm, I think). But I didn't find its proof in the ...
### Is the injection $\ell^p \subset \ell^q$ continuous for $p<q$?
It is easy to show that $\ell^p \subset \ell^q$ when $1 \leq p<q \leq + \infty$, but is the injection continuous? If so is $\ell^{\infty}$ the direct limit $\lim\limits_{\rightarrow} \ \ell^p$ as ...
### what is the closure of $\mathbb{Q}^\mathbb{N}$ in $l^\infty$?
I was wondering that since $l^\infty$ is not separable, which means that there is not a countable dense set in it. However the set $\mathbb{Q}^\mathbb{N}$ is countable (am I right in this?). So what ...