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3 votes
2 answers
92 views

Limit depending on parameter and $L^1$ function

What is the $\lim_{n\to\infty} n^a\int_0^1 \frac{f(x)dx}{1+n^2x^2}$ depending on $a\in\mathbb{R}$, if $f\in L^1(0,1)$? By Banach-Steinhaus theorem I deduced that the limit is zero for $a\leq 0$, but I ...
alans's user avatar
  • 6,515
2 votes
1 answer
62 views

For $1\le p < +\infty$ $L^p$ is a Banach space: Real and abstract analysis, Hewitt - Stromberg

I have some doubts about the proof of this theorem. From time to time I will put my justification. For $1\le p < +\infty$, $L^p$ is a Banach space Let $(f_n)_n$ be a Cauchy sequence in $L^p$, i.e., ...
MathMister's user avatar
2 votes
0 answers
71 views

Can one construct an isometric embedding $\phi_p:\ell^p\to\ell^\infty$ without Hahn-Banach?

It is well-known (for example, see this question) that, as a consequence of the Hahn-Banach theorem, every separable Banach space can be isometrically embedded in $\ell^\infty$. In particular, for $1\...
csch2's user avatar
  • 4,817
2 votes
0 answers
39 views

Can angles be defined by norms that are not induced by inner products?

Can angles be (well-)defined in a normed vector space where the parallelogram law does not hold? In other words, if the norm is not induced by an inner product in a normed space (say $L^1$ space), can ...
chaohuang's user avatar
  • 6,399
1 vote
2 answers
52 views

Computing the spectrum of the operator $A(f)(t)=tf(t), A:L^1([0,1])\to L^1([0,1])$

Let $A$ be the bounded linear operator from the Banach space $L^1([0,1])$ to $L^1([0,1])$ defined as $A(f)(t) = tf(t)$. I've come to learn that then $A$'s spectrum should be $[0,1]$, but I'm a bit ...
Cartesian Bear's user avatar
2 votes
2 answers
82 views

Reference request: $\ell^1$, $\ell^2$, $\ell^p$

I would like to recive from you some references (books and or good notes) about the spaces $\ell^p$. I have searched over here already, but I really didn't find any good match to what I am asking for. ...
Heidegger's user avatar
  • 3,482
1 vote
1 answer
38 views

Minimizing vector for a closed set in $L^3$

Let $\Omega$ be a measurable set in $R$. $X =L^{3}(\Omega)$ and $Z$ be a closed subspace of $X$. Let $x \in X$. Show that $\exists y \in Z$ such that $\lVert x-y\rVert=\inf_{z \in Z} ||x-z||=\delta$. ...
lone_wolf's user avatar
  • 307
2 votes
1 answer
90 views

Calculating operator norm of operator $(Ax)(k) = \int_{0}^{1} x(t)\operatorname{arcctg}\left((-1)^kt\right) \,\mathrm{d}t$

I have no idea how to solve this: $$A: CL_1[0,1] \rightarrow \ell_\infty$$ $$(Ax)(k) = \int_{0}^{1} x(t)\operatorname{arcctg}\left((-1)^kt\right) \,\mathrm{d}t,\quad\forall x \in CL_1[0,1],\quad\...
worst functional analytic's user avatar
0 votes
0 answers
73 views

Is it possible for a sequence in $L^p$ to converge to an element of $L^q$ ($q>p$) w.r.t. the $q$-norm?

Suppose that $1 \leq p < q \leq \infty$ and that $f_n$ is a sequence in $L^p(\mathbb{R})$. If $f_n$ is a Cauchy sequence with respect to the metric induced by the $L^p$ norm, then (I believe that) ...
tparker's user avatar
  • 6,280
0 votes
1 answer
53 views

Weak star convergence in $L^{\infty}$ and strong convergence in $L^2$ imply strong convergence in $L^p$

I can't see why the following would hold. Can someone please point out why? If $f_n$ convergence weak-* in $L^{\infty}(\Omega)$, and converges strongly in $L^2(\Omega)$, then it must also converge ...
Morcus's user avatar
  • 585
2 votes
1 answer
115 views

$\|f \otimes g\|_{L^p} \leq \|f\|_{L^p} \|g\|_{L^p}$ for $1 \leq p \leq \infty$

I am looking to prove the inequality $$\|f \otimes g\|_{L^p} \leq \|f\|_{L^p} \|g\|_{L^p}$$ for $1 \leq p \leq \infty$. Here $f$ takes values in a Banach space $X_1$ and $g$ takes values in a Banach ...
CBBAM's user avatar
  • 6,275
3 votes
0 answers
28 views

If for a.e. $x \in X$ the sequence $(f_n(x, \cdot))_n$ is Cauchy in $L^p_{\text{loc}} (Y)$, then $(f_n)$ is Cauchy in $(L^0 (Z), \rho_Z)$

Let $T>0$ and $p \in [1, \infty)$, $X :=[0, T]$ and $Y:= \mathbb R^d$, $\cal A$ the Lebesgue $\sigma$-algebra of $X$, $\cal B$ the Lebesgue $\sigma$-algebra of $Y$, $\mu, \nu$ complete finite ...
Analyst's user avatar
  • 5,817
0 votes
0 answers
28 views

Convergence in marginal measure implies that in product measure

Let $T>0$ and $p \in [1, \infty)$, $X :=[0, T]$ and $Y:= \mathbb R^d$, $\cal A$ the Lebesgue $\sigma$-algebra of $X$, $\cal B$ the Lebesgue $\sigma$-algebra of $Y$, $\mu, \nu$ complete finite ...
Analyst's user avatar
  • 5,817
1 vote
0 answers
51 views

Let $\|f_n-f\|_{L^p_{\text{loc}}} \to 0$. There is a subsequence $(n_k)$ such that $f_{n_k} \xrightarrow{k \to \infty} f$ a.e.

Let $p \in [1, \infty)$ and $Y:= \mathbb R^d$. Let $L^p_{\text{loc}} (Y)$ be the space of measurable functions $f:Y \to \mathbb R$ such that $$ \|f\|_{L^p_{\text{loc}}} := \sup_{y \in Y} \|1_{B(y, 1)} ...
Akira's user avatar
  • 17.6k
2 votes
0 answers
34 views

Assume $f(x, \cdot) \in L^p_{\text{loc}} (Y)$ for a.e. $x \in X$. Then the map $x \mapsto \|f(x, \cdot)\|_{L^p_{\text{loc}}}$ is measurable

Below we use Bochner measurability and Bochner integral. Let $T>0$ and $p \in [1, \infty)$. Let $X :=[0, T]$ and $Y:= \mathbb R^d$. Let $L^p_{\text{loc}} (Y)$ be the space of measurable functions $...
Akira's user avatar
  • 17.6k

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