For questions about $L^p$ spaces, that is, given a measure space $(X,\mathcal F,\mu)$, the vector space of equivalence class of measurable functions with $p$-th power of the absolute value integrable. Question can be about properties of elements of these spaces, or when the ambient space on a ...

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45
votes
2answers
4k views

If $f_k \to f$ a.e. and the $L^p$ norms converge, then $f_k \to f$ in $L^p$

Let $1\leq p < \infty$. Suppose that $\{f_k\} \subset L^p$ (the domain here does not necessarily have to be finite), $f_k \to f$ almost everywhere, and $\|f_k\|_{L^p} \to \|f\|_{L^p}$. Why ...
41
votes
2answers
16k views

Limit of $L^p$ norm

Could someone help me prove that given a finite measure space $(X, \mathcal{M}, \sigma)$ and a measurable function $f:X\to\mathbb{R}$ in $L^\infty$ and some $L^q$, $\lim_{p\to\infty}\|f\|_p=\|f\|_\...
50
votes
2answers
17k views

$L^p$ and $L^q$ space inclusion

Let $(X, \mathcal B, m)$ be a measure space. For $1 \leq p < q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample ...
15
votes
2answers
2k views

“Scaled $L^p$ norm” and geometric mean

The $L^p$ norm in $\mathbb{R}^n$ is \begin{align} \|x\|_p = \left(\sum_{j=1}^{n} |x_j|^p\right)^{1/p}. \end{align} Playing around with WolframAlpha, I noticed that, if we define the "scaled" $L^p$ ...
14
votes
2answers
4k views

How do you show that $l_p \subset l_q$ for $p \leq q$?

I can't seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I'm assuming is the way to go about it).
11
votes
1answer
9k views

The $ l^{\infty} $-norm is equal to the limit of the $ l^{p} $-norms. [duplicate]

If we are in a sequence space, then the $ l^{p} $-norm of the sequence $ \mathbf{x} = (x_{i})_{i \in \mathbb{N}} $ is $ \displaystyle \left( \sum_{i=1}^{\infty} |x_{i}|^{p} \right)^{1/p} $. The $ l^{\...
15
votes
1answer
3k views

Strong and weak convergence in $\ell^1$

Let $\ell^1$ be the space of absolutely summable real or complex sequences. Let us say that a sequence $(x_1, x_2, \ldots)$ of vectors in $\ell^1$ converges weakly to $x \in \ell^1$ if for every ...
12
votes
4answers
1k views

Convergence of integrals in $L^p$

Stuck with this problem from Zgymund's book. Suppose that $f_{n} \rightarrow f$ almost everywhere and that $f_{n}, f \in L^{p}$ where $1<p<\infty$. Assume that $\|f_{n}\|_{p} \leq M < \...
5
votes
3answers
742 views

When $L_p = L_q$?

As we know that $L_p \subseteq L_q$ when $0 < p < q$ for probability measure, I was wondering when $L_p = L_q$ is true and why. Is it to impose some restriction on the domain space? Thanks!
13
votes
1answer
1k views

If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic

If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic unless $p=2$. Maybe I would have to use the Rademacher's functions.
8
votes
3answers
2k views

A typical $L^p$ function does not have a well-defined trace on the boundary

This question is from PDE by Evans, 1st edition, Chapter 5, Problem 14. It has been posted here previously, however, I cannot quite put all the information together from the responses there. Hopefully ...
2
votes
2answers
210 views

$f \in L^1$, but $f \not\in L^p$ for all $p > 1$

"Find an $f \in [0,1]$ such that $f \in L^1$ but $f \not\in L^p$ for any $p > 1$." I've thought about doing something like $$f(x) = \frac{1}{x}$$ where $|f|^p = \frac{1}{x^p}$ doesn't converge ...
20
votes
2answers
8k views

Why is $L^{\infty}$ not separable?

$l^p (1≤p<{\infty})$ and $L^p (1≤p<∞)$ are separable spaces. What on earth has changed when the value of $p$ turns from a finite number to ${\infty}$? Our teacher gave us some hints that ...
5
votes
1answer
3k views

$\ell_p$ is Hilbert space if and only if $p=2$

Can anybody please help me to prove this.. Let $p$ greater than or equal to $1$, show that the space of all $p$-summable sequences is an inner product space if and only if $p=2$.
2
votes
1answer
79 views

Is $L^2(0,\infty;L^2(\Omega)) = L^2((0,\infty)\times \Omega)$?

If $\Omega$ is a bounded $C^1$ domain, is $L^2(0,\infty;L^2(\Omega)) = L^2((0,\infty)\times \Omega)$? Are they the same? I know this is true when instead of $(0,\infty)$ we have a bounded interval.
13
votes
1answer
1k views

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 ...
8
votes
2answers
4k views

Smooth functions with compact support are dense in $L^1$

Here is another homework question that I did and I'd be glad if you could tell me if it's right. We now strengthen the result of Question Two for $R$ where we have the notion of differentiability. ...
5
votes
2answers
552 views

How do I prove the completeness of $\ell^p$?

Say $\{x_n\}$ is Cauchy in $\ell^p$ and $x$ is its pointwise limit. To argue that $x \in \ell^p$ would the following be correct: Let $\varepsilon > 0$ and let $N$ be s.t. $n,m > N$ $\Rightarrow$...
4
votes
2answers
763 views

Inclusion of $l^p$ space for sequences

Inclusion of $L^p$ spaces for functions has been discussed here. Does this apply to $l^p$ space of sequences similarly? I tried to show the following: For $1\leq p<q<\infty$, $l^q\subset l^p$ ...
1
vote
1answer
168 views

Integral of series with complex exponentials

Suppose that $f\in L^2(\mathbb{R}/2\pi\mathbb{Z})$ takes the form $$f(\theta)=\sum_{n=1}^\infty a_ne^{in\theta}.$$ The function $$F(z)=\sum_{n=1}^\infty a_nz^n$$ converges in $|z|<1$. How can I ...
20
votes
2answers
332 views

Why are $L^p$-spaces so ubiquitous?

It always baffled me why $L^p$-spaces are the spaces of choice in almost any area (sometimes with some added regularity (Sobolev/Besov/...)). I understand that the exponent allows for convenient ...
14
votes
1answer
849 views

Distance minimizers in $L^1$ and $L^{\infty}$

If $H$ is a Hilbert space, we have the Hilbert Projection Theorem, which tells us that given a nonempty, closed, convex subset $K \subset H$, and a point $x \in H$, there is a unique point $y \in K$ ...
14
votes
1answer
600 views

How should I prove the duality?

Rudin asked (Real Complex Analysis, First edition, Chapter 6, Problem 4): Suppose $1\le p\le \infty$, and $q$ is the exponent conjugate to $p$. Suppose $u$ is a $\sigma$-finite measure and $g$ is a ...
8
votes
1answer
1k views

Does $L^p$-convergence imply pointwise convergence for $C_0^\infty$ functions?

It is stated in my professor's notes that, given a sequence $\{f_j\}$ of $C_0^\infty(\Omega)$ functions (infinitely differentiable with compact support), and a function $g\in C_0^\infty(\Omega)$, all ...
6
votes
1answer
985 views

How do you prove that $\ell_p$ is not isomorphic to $\ell_q$?

I guess that for all $1\le p,q<\infty $, such that $p\ne q$ , the spaces $\ell_p$ and $\ell_q$ are not isomorphic, but how do you prove this?
4
votes
2answers
124 views

What is $L^p$-convergence useful for?

Why do people care about $L^p$-convergence $f_n \rightarrow f$? Are there any interesting application of $L^p$-convergence? For example, if $p=\infty$, then the limit $f$ of the sequence $f_n$ of ...
12
votes
1answer
1k views

A Hamel basis for $l^{\,p}$?

I am looking for an explicit example for a Hamel basis for $l^{\,p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one ...
5
votes
1answer
2k views

Liapunov's Inequality for $L_p$ spaces

Let $1 \leq p,q < \infty$ and $0 \leq \lambda \leq 1$. If $r = \lambda p + (1 - \lambda)q$ and $f \in L_p \cap L_q $, then $$||f||_r^r \leq ||f||_p^{\lambda p} ||f||_q^{(1 - \lambda)q} \tag{*}$$ ...
5
votes
1answer
355 views

Subsequence convergence in $L^p$

I recall a fact that for functions $f_1,f_2,\ldots\in L^1$ such that $\|f_n-f\|_1\rightarrow 0$ as $n\rightarrow\infty$, there exists a subsequence $f_{n_i}$ that converges to $f$ almost everywhere. ...
4
votes
1answer
281 views

When is a subset of $\ell^2$ compact?

I have been looking on the internet for hours now and even asking in chat without an answer. When is a set $M\subseteq\ell^2$ compact? For $L^p$, there is the Arzelà–Ascoli theorem that provides a ...
4
votes
1answer
884 views

$\ell^{\infty}(\mathbb N)$ 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. ...
1
vote
1answer
115 views

A Cauchy sequence has a rapidly Cauchy subsequence

I am trying to fill in the details of a proof related to the Riesz-Fischer Theorem. We need to show that every Cauchy sequence $\{f_n\}$ has a rapidly Cauchy subsequence. My text claims that we can ...
0
votes
3answers
307 views

show that $l^2$ is a Hilbert space

Let $l^2$ be the space of square summable sequences with the inner product $\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i$. (a) show that $l^2$ is H Hilbert space. To show that it's a ...
2
votes
2answers
629 views

Translation operator and continuity

I came across a text that proves that translation operator $T_a(f):=f(x-a)$ where $a\in\mathbb{R}^n$ and $f\in L^p(\mathbb{R}^n)$ is continuous. The proof follows: $$||f(x-a)-f(x)||_p=||f(x-a)-g(x-a)+...
11
votes
1answer
1k views

How to show convolution of an $L^p$ function and a Schwartz function is a Schwartz function

We have the Schwartz space $\mathcal{S}$ of $C^\infty(\mathbb{R^n})$ functions $h$ such that $(1+|x|^m)|\partial^\alpha h(x)|$ is bounded for all $m \in \mathbb{N_0}$ and all multi-indices $\alpha$. ...
6
votes
2answers
96 views

Do non-$\ell^2$ sequences have an $\ell^2$ functional that takes them to infinity? [duplicate]

Suppose $\{a_n\}_{n=1}^{\infty}$ is a sequence of real numbers (suppose also positive for simplicity) so that $$\sum_{n=0}^{\infty} a_n^2 = \infty$$ i.e. the sum diverges. Can you necessarily find a ...
2
votes
2answers
398 views

Compact inclusion in $L^p$

Is it true that there is a compact inclusion from $L^p$ to $L^q$ whith $q<p$? What is the counterexample if what I said is wrong? Thank you.
8
votes
1answer
2k views

Fourier transform in $L^p$

Let the $f$ be a function in $L^s$ where $s \in [1,\infty) $. For which $r$ Fourier transform $\hat{f}$ belongs to $L^r$? I'd be grateful for any kind of help including providing a literature or ...
5
votes
1answer
91 views

Examples of measures that induce certain inclusions in the Lp spaces.

I apologize for the terribly worded title, but I didn't know how else to title this questions (which comes from Rudin's Real & Complex Analysis chapter 3 questions). The question says: For ...
4
votes
1answer
61 views

Exists $C = C(\epsilon, p)$ where $\|u\|_{L^\infty(0, 1)} \le \epsilon\|u'\|_{L^p(0, 1)} + C\|u\|_{L^1(0, 1)}$ for all $u \in W^{1, p}(0, 1)$?

Let $p > 1$. For all $\epsilon > 0$, does there exist $C = C(\epsilon, p)$ such that$$\|u\|_{L^\infty(0, 1)} \le \epsilon\|u'\|_{L^p(0, 1)} + C\|u\|_{L^1(0, 1)}$$for all $u \in W^{1, p}(0, 1)$?
4
votes
1answer
115 views

If a sequence $f_n$ is bounded in $L^2$ and converges to zero a.e., then $f_n\to 0$ in $L^p$ for $0<p<2$

Let $M>0$, $\{f_n\}\subset L^2([0,1])$ such that $\int_0^1 |f_n|^2 dm\leq M$ and $f_n(x)\to 0$ as $n\to\infty$ almost everywhere, $m$ is Lebesgue measure. Show that for all $0<p<2$, $$\lim_{n\...
3
votes
1answer
283 views

Gauss–Ostrogradsky formula for Distributions

Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $u\in W^{1,p}(\Omega)$, $v\in L^{p'}(\Omega)^N$ with $p\in (1,\infty)$. Let $\operatorname{div}(v)$ be the divergence of $v$ in the sense of ...
2
votes
1answer
123 views

How can I show that if $f\in L^p(a, b)$ then $\lim_{t\to 0^{+}}\int_{a}^b |f(x+t)-f(x)|^p\ dx=0$..

can anyone help me show that if $f\in L^p(a, b)$ then $$ \lim_{t\to 0^{+}}\int_{a}^b|f(x+t)-f(x)|^p\ dx=0.$$ Thanks, any help will be useful..
1
vote
1answer
36 views

Convolution with Gaussian, without distribution theory, part 2

I only know basic $L^p$ theory (nothing about distributions) and am trying to prove the following: Let $t>0$, $f\in L^{p}(\mathbb{R}^n,m)$, $\Gamma(t,x)=(4\pi t)^{-n/2}e^{-|x|^2/4t}$ and $$ u(t,x)=...
7
votes
2answers
379 views

What does the $L^p$ norm tend to as $p\to 0$?

This is something I was thinking about, so I'm going to post it as a question and post my own answer. I hope that anyone who wants will comment, correct me if I'm wrong, and add their own knowledge ...
4
votes
3answers
826 views

Cesàro operator is bounded for $1<p<\infty$

The Cesàro operator $T\colon \ell_{p}\to\ell_{p}$ is defined by $(Tx)_{k}=\frac{1}{k}\sum_{j=1}^{k}x_{j},\: k\in\mathbb{N}$, where $x=(x_{k})_{k=1}^{\infty}$ Show that $T$ is bounded if $1<p<\...
3
votes
1answer
455 views

Pointwise a.e. convergence implies strong convergence?

Let $ 1 \leq p_1 < p_2 < \infty$, and suppose that $f_n$ is a sequence of functions in $L^{p_1}[a,b]$ such that $f_n \to f$ pointwise a.e. on $[a,b]$. Suppose in addition that $ ||f_n||_{p_2} \...
3
votes
1answer
275 views

Any positive linear functional $\phi$ on $\ell^\infty$ is a bounded linear operator and has $\|\phi \| = \phi((1,1,…)) $

This is a small exercise that I just can't seem to figure out. When I see it I'll probably go 'ahhh!', but so far I haven't made any progress. I'd like to prove that any linear functional $\phi$ on $\...
3
votes
1answer
65 views

Discontinuous functionals on $L^p$

Using the axiom of choice and a Hamel basis for a normed space, one can prove the existence of everywhere defined discontinuous linear functionals. My question: Does there exist a discontinuous ...
3
votes
3answers
147 views

How to prove that $L^p [0,1]$ isn't induced by an inner product? for $p\neq 2$

I'd like to know how could i prove that $L^p [0,1]$ isn't induced by an inner product? (For $p\neq 2$, including $p=\inf$). It is clear to me that i would need to find two functions $f$, $g$ in $L^p$ ...