Let $\Omega$ be any measurable ($\sigma$-finite, if necessary) space. let $1\leq p\leq\infty,x\in L^p(\Omega)$.  Then how to proof $\forall\varepsilon&gt;0,\exists\delta&gt;0$，so that for any $1\leq q\leq\infty$, if $|q-p|&lt;\delta$, we have $x\in L^q(\Omega)$ and $\left| ||x||_p -||x||_q\right|&lt;\varepsilon$
1759
Suppose that $\|f\|_p &lt \infty$ for all $1\leq p &lt p'$, I want to know if the the following is true and in that case how to show it     $p \mapsto \|f\|_p$ is continuous on $[1,p')$   Or maybe we need to impose some more constraints such as finite measure space. In case of finite measure space, I tried to use .
80121
Let $\Omega\subseteq \mathbb R^n$ be open, bounded and connected set with $\mathcal{L^n}(\Omega)&gt;0$. Let $\Omega=\Omega_1\cup\Omega_2$ where $\Omega_1\cap\Omega_2=\emptyset$ with $\mathcal{L^n}(\Omega_1)&gt;0$ and $\mathcal{L^n}(\Omega_2)&gt;0.$ Also $\Omega_1,$ $\Omega_2$ are $\mathcal L^n$-measurable sets.  Prove that  $\exists$ $\Phi\in C_c^{\infty}(\Omega)$ such that $\displaystyle \int_{\Omega_1}\nabla\Phi(x)dx\neq 0.$   Here $\Phi$ is real valued function and $C_c^{\infty}(\Omega)$ consists of all $C^{\infty}(\Omega)$ functions which have compact support in $\Omega.$   I tried to show by contradiction and wished to have $\mathcal{L^n}(\Omega_1)=0 $ but couldn't be able to do that.  Any help is appreciated. Thank you.
74641
This is a step in a lecture note I'm reading. It should be simple because the author considers it obvious but I can't see it. What am I missing?  Suppose $U$ and $V$ are integrable over measure space $(\Omega,\mathcal{F},\mu)$. Claim: $$ \int_\Omega\sqrt{U^2+V^2}d\mu\geq\sqrt{\left(\int_\Omega Ud\mu\right)^2+\left(\int_\Omega Vd\mu\right)^2}. $$ Alternative phrasing: with $Z=U+iV$, claim that  $$ \int_\Omega|Z|d\mu\geq\left|\int_\Omega Zd\mu\right|. $$ Thank you for your time!
120818
Let $f\in S_\infty\subset L_1(\mathbb{R},\mu)$ with $\mu$ as the Lebesgue linear measure be a Lebesgue-summable function such that $$\forall (p,q)\in\mathbb{N}^2_{\ge 0}\quad\exists C_{pq}&gt;0: \Bigg|x^p\frac{d^q}{d x^q}f(x)\Bigg|&lt; C_{pq}$$ I was wondering whether, if $\forall p\in\mathbb{N_{\ge 0}}\quad\int_{\mathbb{R}}x^pf(x)d\mu=0$, then $f$ is constantly, or almost everywhere, null. I cannot find a counterexample and therefore I think that the implication might well hold, but I cannot prove it either.  Since $x^p f(x)$ belongs to $S_\infty\subset L_1$, thanks to the fact that $f$ belongs to it, and is continuous I think that the Lebesgue integral and the Riemann improper integral $\mathscr{R}\int_{-\infty}^\infty t^pf(t)dt$ are the same. Please correct if I am wrong.  Nevertheless I cannot use calculus facts to prove the desired implication... I have also tried using the fact that the Fourier transform induces a bijection $S_\infty\to S_\infty$, but with no result. What do you think about it? Has anybody got a counterexample or proof? Thank you very much!
28355
Inclusion of $L^p$ spaces for functions has been discussed .  Does this apply to $l^p$ space of sequences similarly?   I tried to show the following: For $1\leq p&lt;q&lt;\infty$, $l^q\subset l^p$ By using Hölder inequality but it doesn't seem to work.  My question is that is this true? If yes, what's the right way to prove it and what's a good counter example for showing $l^p\subset l^q$ is not true? Thanks.
17835
Let $(X,B,\mu)$ be a complete measure space,Show that $$\lim _{q \rightarrow \infty}\|f\|_{q}=\|f\|_{\infty}, \quad \forall f \in \bigcup_{p} \bigcap_{p \leqslant q&lt;\infty} L^{q}$$ So,$\lim _{q \rightarrow \infty}\|f\|_{q}$ , $\|f\|_{\infty}$ are equal-norm with space $ L^{\infty} \cap\left(\bigcup_{p} \bigcap_{p \leqslant q} L^{q}\right)$. Case 1: $m(X)&lt;\infty $.It's easy to prove that. Case 2: $m(X)=\infty $. I have no idea about it,And I started to doubt the correctness of this conclusion. Can somebody give me a hint for this problem or just give an example to prove that this is a wrong conclusion when $m(X)=\infty $. Thanks in advance.
58752
Let $X$ be in $\mathfrak{L}^1(\Omega,\mathfrak{F},P)$ and $\mathfrak{G}\subset \mathfrak{F}$.  Prove that if $X$ and $E(X|\mathfrak{G})$ have same distribution, then they are equal almost surely.  I know what I have to show, that $X$ is $\mathfrak{G}$ measurable, but I don't know how...
117432
Preliminaries  Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a complete probability space.  Let $D$ be a complete, separable, metrizable topological space with Borel $\sigma$-algebra $\mathcal{B}(D)$ (such as $D = \mathbb{R}^q$ with $\sigma$-algebra $\mathcal{B}(D) = \mathcal{B}(\mathbb{R}^d)$).  Let $\mathbb{R}$ be equipped with its canonical Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$.  Let $g: \Omega \times D \rightarrow \mathbb{R}$ be a bounded $(\mathcal{G} \otimes \mathcal{B}(D) ) / \mathcal{B}(\mathbb{R})$-measurable function.  Let $\Pi: \Omega \rightarrow D$ be a $\mathcal{G}/\mathcal{B}(D)$-measurable random variable.  Let $H : \Omega \rightarrow \mathbb{R}$ be a $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable random variable, defined by $$ H(\omega) := g(\omega, \Pi(\omega)).$$ Note, that, since $g$ is bounded, we have $H \in \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P})$.  Let $j: D \rightarrow \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P}) $ be defined by $$ j(\pi)(\omega) := g(\omega, \pi)  $$  For all $\pi \in D$, let $j(\pi)$ be independent of $\Pi$.  Question  I am interested in the conditional expectation $$\mathbb{E}[H \mid \Pi] :\Omega \rightarrow \mathbb{R}$$ of $H$ with respect to $\Pi$. More specifically, I suspect that (a $\mathbb{P}$-unique version of) this condititional expectation is given by  $$ \mathbb{E}[H \mid \Pi] (\omega) = \mathbb{E}[j(\Pi(\omega))], \quad (\dagger)  $$ whereby $\mathbb{E}[j(\Pi(\omega))]$ can of course also be written as $$\mathbb{E}[j(\Pi(\omega))] = \int_{\Omega} j(\Pi(\omega))(\tilde{\omega}) d\mathbb{P}(\tilde{\omega}) . $$  How can I prove, that $(\dagger)$ is the case? I have tried, tracking the definition of conditional expectation and using Fubini, but with little success so far.  Thanks for any advice!
29266
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 book. So I wonder why it is?    Does this conclusion hold for $L^1(\Omega, \mathcal{F}, \mu) \cap L^2(\Omega, \mathcal{F}, \mu)$ for any measure space $(\Omega, \mathcal{F}, \mu)$?  Are there similar statements if replace $L^1$ with $L^p$, and  $ L^2$ with $L^q$ for $p \leq q \in (0, \infty ]$ or $\in [1, \infty]$?  Thanks and regards!
18402
Let $f$ integrable and let $E_\alpha=\{x\mid |f(x)|&gt;\alpha\}$ for $\alpha\geq 0$. Show that $$\int\left|f(x)\right|dx=\int_0^\infty m(E_\alpha)d\alpha$$ where $m$ is the Lebesgue measure.   By hypothesis, $f$ is measurable, therefore there exists a sequence of simple function such that $\varphi_n\nearrow |f|.$ Moreover, $E_\alpha^n\nearrow E^\alpha$. By the continuity of the measure, $m(E_\alpha^n)\to m(E_\alpha)$ for all $\alpha\geq 0$. By the convergence monotone theorem, $$\int_0^\infty  m(E_\alpha)d\alpha=\lim_{n\to\infty }\int_0^\infty  m(E_\alpha^n)\underset{(*)}{=}\lim_{n\to\infty }\int\varphi_n(x)dx=\int\left|f(x)\right|dx.$$   I don't understand the equality $(*)$, why $m(E^n_\alpha)=\varphi_n(x)$ ?
122536
Suppose $(\Omega,\mathcal F,P)$ is a probability space and that $\mathcal G$ is a sub-sigma-algebra of $\mathcal F$. If $X$ is an integrable, non-negative random variable with the same distribution as $E[X\mid \mathcal G]$, how does one show that $X=E[X\mid \mathcal G]$ a.s.?
76827
currently I am working on the following two exercises as a revision for my exam.  Let $\mu$ be the counting measure on $\mathbb N$.   Show that if $1 \le p &lt; s &lt; \infty$ then $f \in L^p$ implies $f \in L^s$ and $\|f\|_s \le \|f\|_p$. Show that if $f \in L^p$ for some $p \ge 1$ then $\lim_{s \to \infty} \|f\|_s = \|f\|_\infty$.   My attempts so far:   It is $f \in L^s$ iff $\sum_{k = 1}^\infty |f(k)|^s$ converges. Since $f \in L^p$ we know that $\sum_{k = 1}^\infty |f(k)|^p$ converges. Thus for any $\varepsilon &gt; 0$ there is $N \in \mathbb N$ s.t. \begin{align*} k \ge N \Longrightarrow |f(k)|^p &lt; \varepsilon \Longrightarrow |f(k)| &lt; \varepsilon^{1/p} \end{align*} We have $\forall k \ge N$ that $|f(k)|^s &lt; \varepsilon^{s/p}$. Originally I wanted to use the comparison test now, showing that \begin{align*} \sum_{k = N}^\infty \varepsilon^{s/p} &lt; \infty, \end{align*} but that does not work since \begin{align*} \sum_{k = N}^\infty \varepsilon^{s/p} = \varepsilon^{s/p} \underbrace{\sum_{k \ge N} 1}_{= \infty},  \end{align*} so I don't know how to go on here. Let $K := \|f\|_\infty$. We need to show that $\forall \varepsilon &gt; 0$ $\exists M \in \mathbb N$ s.t. \begin{align*} s \ge M \Longrightarrow \left|\left(\sum_{k = 1}^\infty |f(k)|^s\right)^{1/s} - K\right| &lt; \varepsilon. \end{align*} Choose $M$ s.t. $M &gt; p$. Then $s &gt; p$ and from part (i) we know that \begin{align*} \left(\sum_{k = 1}^\infty |f(k)|^s\right)^{1/s} \le \left(\sum_{k = 1}^\infty |f(k)|^p\right)^{1/p}. \end{align*} But how can I use this?
43906
I have a question that I need help with getting started (possibly I would be back for more help).     I have a measure space $(X,A,\mu)$ that is finite, and $f \in L^{\infty}(\mu)$. Also, defined is $a_n = \int_X\,|f|^n\,d\mu$. I need to show that the limit is: $$\lim_{n\to\infty}\,\frac{a_{n+1}}{a_n} = \|f\|_{\infty} .$$   I am stuck on getting started, anybody have any suggestions?  thanks much
102050
Let $(X,S,\mu)$ be a measure space. Suppose $\mu(X)&lt;\infty$. Put $E_n=\{$ $x\in X:$ $n-1\leq |f(x)|&lt;n \}$ Then, $f\in \mathcal{L}^p$ if and only if $\sum_{n=1}^{\infty}n^p\mu(E_n)&lt;\infty$. My attempt: Notice, $X=\bigcup_{n=1}^{\infty}E_n$. Since $(E_n)_n$ are pairwise disjoint, we deduce by a standard property of measures that $\mu(E_n)&lt;\infty$ for each $n$. As a consequence, $\int |f|^p d\mu=\sum_{n=1}^{\infty}\int_{E_n}|f|^p d\mu$ Now, observe that for each $n$, $n^p\chi_{E_n} \geq |f|^p\chi_{E_n}\geq (n-1)^p\chi_{E_n}$ Upon taking integrals, we obtain: $ \sum_{n=1}^{\infty}(n-1)^p \mu(E_n)\leq\int |f|^p d\mu \leq \sum_{n=1}^{\infty} n^p\mu(E_n)$ So, if $\sum_{n=1}^{\infty}n^p\mu(E_n)&lt;\infty$ then $f\in \mathcal{L}^p$. Now for the reverse, not that if $\mu(E_n)=0$ for each $n$ then we are done. If there is some $n$ such that $\mu(E_n)=0$  if $f\in \mathcal{L}^p$, then $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)&lt;\infty$. Now the question is how do we determine whether a series coneverges given another one converges? The answer is the $\textbf{limit comparison test}$: Theorem: Let $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ be two series such that $a_n\geq 0$ and $b_n&gt;0 0$ for all $n$.  Put $c= \lim_{n\rightarrow \infty}\frac{a_n}{b_n}$.  If $0&lt;c&lt;\infty$ then either both series converge or both series diverge.  To invoke this theorem notice that if for all $n$, $\mu(E_n)=0$ then we are immediately done. So, we may assume that for some n , $\mu(E_n)&gt;0$. If for all $n$, this is the case, then we are immediately done. If only finitely many satisfy this property, then we may disregard the ones with 0 measure. Thus, we may assume for each $n$, $\mu(E_n)&gt;0$. In which case, the limit comparison test above allows us to deduce that $c=1$. Since $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)$ converges, so must $\sum_{n=1}^{\infty}n^p \mu(E_n)$.
56456
Let's consider the space $\mathcal{L}^p(X, \Sigma, \mu)$ of all functions $f\colon X \to \mathbb{R}$ (or $\mathbb{C}$) for which: $$ \int\limits_X|f|^p \mu(dx) &lt; \infty. $$ Here $X$ is a metric space, $\Sigma\subset 2^X$ — sigma-algebra on $X$ and $\mu$ — is the measure on $\Sigma$. As usual, functions that are equal almost everywhere considered to be equivalent, so technically we are dealing with classes $$[f]=\{g\in \mathcal{L}^\infty(X,\Sigma, \mu) \colon \mu \{x\in X\colon f(x)\neq g(x)\}=0\}.$$ This space is endowed with integral metric $$\rho_p(f,g)=\left(\int\limits_X|f-g|^p \mu(dx)\right)^{1/p}$$  The statement is that space $\mathcal{L}^p(X, \Sigma, \mu)$ is separable iff $(\Sigma, \rho_\Delta)$ is separable, where $\forall A,B\in\Sigma$ $$\rho_\Delta(A,B)=\mu(A\Delta B)$$ Here, again, we consider $A\in \Sigma$ as a class $[A]=\{A'\in\Sigma\colon \mu(A\Delta A')=0\}$ to have the identity of indiscernibles.  How do one prove the statement?
73546
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 defined in an open set $\Omega\in\mathbb{R}^n$:  $$\|f_j-g\|_{L_p(\Omega)}\to0\implies|f_j(x)-g(x)|\to0\,\forall x\in\Omega$$  I was not able to prove it, however. Searching on the Internet, I found many counterexamples to the statement that $L^p$-convergence implies pointwise convergence, but they do not deal with $C_0^\infty(\Omega)$ functions, so it may also be true. Can anyone point to a proof or a counterexample?
64805
I've been learning about $L^p$-spaces and came across this problem in Rudin's Real and Complex Analysis:     Suppose $\mu(\Omega)=1$ and $h:\Omega \to [0,\infty]$ is measurable. If $A= \int_\Omega {h d\mu}$, prove that:   $$ \sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}d\mu \leq 1+A $$   I've already shown that $\sqrt{1+A^2} \leq \int_\Omega \sqrt{1+h^2}d\mu$ using Jensen's inequality applied to the convex function $\varphi(x)=\sqrt{1+x^2}$. I can't seem to show the latter inequality. Any help or hints would be appreciated. Thank you in advance!