# Tag Info

7

The problem is that $L^p$, for $p \in (0,1)$ is not locally convex. Your professor showed that its dual is zero, hence it wouldn't make sense to talk about weak topology as it is defined using elements in the dual space. An easy way to convince yourself such spaces are not locally convex is to define the $L^p$ norm in $\mathbb R^2$ and draw the unit ...

6

"$\Rightarrow$": Use $$|\mathbb{E}(X_n 1_A)-\mathbb{E}(X 1_A)| \leq \mathbb{E}(|X_n-X|).$$ "$\Leftarrow$": Show $$\mathbb{E}(|X_n-X|) = \mathbb{E}[(X_n-X) 1_{\{X_n-X \geq 0\}})]+ \mathbb{E}[(X-X_n) 1_{\{X_n-X<0\}}],$$ and conclude that $$\mathbb{E}(|X_n-X|) \leq 2 \sup_{A \in \mathcal{F}} |\mathbb{E}[(X_n-X) 1_A]|.$$

3

I was asked to look at this. Looking at it, I think it's possible to simplify Urgje's answer. Give $\bf{f}$ the same meaning as in Urgje's answer, i.e., $(1,1/2,1/3,\ldots,1/n,\ldots)$. We have ${\bf f}\in l^2$ because $||{\bf f}||^2 = \sum_{n=1}^\infty 1/n^2$ by definition of the norm and $\sum_{n=1}^\infty 1/n^2$ is a well-known convergent series, so ...

3

Your conclusions about the function $f$ are correct. The inclusions are generally false (they are true for finite measure spaces).

3

We have to assume $\mathbb{E}(|X|^r)<\infty$; otherwise the expession $\|X_n-X\|_{L^r}$ might not even be finite. Note that $$\|X_n-X\|_{L^r}^r = \int_{|X_n-X| \leq \epsilon} |X_n-X|^r \, d\mathbb{P} + \int_{|X_n-X| > \epsilon} |X_n-X|^r \, d\mathbb{P} \tag{1}$$ for any $\epsilon>0$ and $n \in \mathbb{N}$. Obviously, $$\int_{|X_n-X| \leq ... 2 Remember that f \in W^{1,1} (\gamma_1) means that f, f' \in L^1 (\gamma_1) where f' is understood in the weak sense. Remember also that f' in the Sobolev sense in the same as f' in the distributional sense, therefore let us compute the latter one. Finally, we shall use that \gamma_1 = \dfrac 1 {\sqrt {2 \pi}} {\rm e}^{-\frac {x^2} 2}. Note, ... 2 I wouldn't know about the proof in the book, but here's a proof. It could probably be streamlined some - you should see what it looked like a few days ago. Going to change some of the notation; this is going to be enough typing as it is. Going to assume we're talking about real-valued functions, so that for every f there exists E with \left|\int_E ... 2 Use the fact that$$ 0 \leq 2^p (|f_n|^p +|f|^p) -|f_n-f|^p. $$Hence by Fatou's lemma,$$ 2^{p+1} \int_\Omega |f|^p \leq \liminf_{n \to +\infty} \int_\Omega \left( 2^p (|f_n|^p +|f|^p) -|f_n-f|^p \right) = 2^{p+1} \int_\Omega |f|^p - \limsup_{n \to +\infty} \int_\Omega |f_n-f|^p. $$Remark that this holds true also for p=1. 2 If (\Omega,\mathcal{T},\mu) is a measured space such that \mu(\Omega)<\infty then you have$$\forall (p,q)\in[1,\infty], p\leq q\Rightarrow L^q\subset L^p \textrm{ (with continuous injections)}$$This doesn't apply here as \mu(\mathbb{N})=\infty if \mu is the usual counting measure. However, denoting ... 2 Consider first the functions$$h_1(x)=\frac{|\pi-x|}{2},\,\,h_2(x)=-2\ln\left|\sin\frac{x}{2}\right|.$$It is shown here that the Fourier series of h_1 on (-\pi,\pi) is \sum_{j=1}^{\infty}\frac{\sin jx}{j}, and the Fourier series for h_2 on (-\pi,\pi) is \sum_{j=1}^{\infty}\frac{\cos jx}{j}. Therefore, the Fourier series of the function \sin ... 2 Note that \begin{equation*} \mathbf{f}=(1,\frac{1}{2},\frac{1}{3},\cdots ) \end{equation*} is contained in l^{2}. Let \begin{equation*} \mathbf{x}=(x_{1},x_{2},\cdots )\in l^{2} \end{equation*} Then for \mathbf{x}\in H \begin{equation*} (\mathbf{x},\mathbf{f})=1 \end{equation*} I think it is useful to switch to \begin{equation*} ... 2 I will answer a bit more than what the question actually asks for, since the question itself doesn't really give the full picture. A function can be in L^p but not L^q for q<p if it has a long tail which larger powers shrink. For instance, for r<0, x^r 1_{[1,\infty)}(x) is in L^p if and only if pr<-1, i.e. p>-1/r. A function can ... 2 Since f_n\xrightarrow{L^1(\Omega)} f, there is a subsequence (f_{n_k}) of (f_n) such that f_{n_k} \rightarrow f pointwise a.e. on \Omega. By Fatou's lemma,$$\|f\|_{L^2}^2 \le \varliminf_{k\to \infty} \|f_{n_k}\|_{L^2}^2\le \varliminf_{k\to \infty} M = M.$$2 By assumption p\neq q. Case 1: p,q\in(1,+\infty). Assume we have a surjection T:\ell_p\to\ell_q, then T^*:\ell_{q'}\to \ell_{p'} is an embedding. Since p',q'\in(1,+\infty), we get a contradiction because by corollary of Pitt's theorem theses spaces are totally incomparable. Thus for this case a desired surjection doesn't exists. Case 2: q=1, ... 1 I would rather say that the opposite is true, and that it comes from Jensen's inequality: since x \mapsto x^{1/p} is concave for p>1, one has:$$ \left( \int g \right)^{1/p} \geq \int \left(g^{1/p}\right).$$Then you apply this result to the function g = \sum_k \vert f \vert^p. 1 In general, this is false (see below), but if (f_n)_n is bounded,it is true, since we have for every f \in L^2 that$$\langle f_n , f\rangle = \langle f_n g_n , f\rangle + \langle (f_n (1-g_n), f\rangle ,$$where the first summand on the right converges to \langle h,f\rangle and where$$ |\langle f_n (1-g_n), f\rangle \leq \|f_n\|_2 \|(1-g_n)f\|_2 \to ...

1

Counterexamples: $(i)$ $$f(x)=\begin{cases}e^{-x},& x\notin \mathbb N\\1,&\text{otherwise}\end{cases}$$ $(ii)$ $$f(x)=\begin{cases}{1\over x},& x\neq 0\\1,&\text{otherwise}\end{cases}$$

1

For $(i)$: Take an integrable bump function $\phi$ and form $$f(x) = \sum_{k=0}^\infty \phi(2^k(x-k)).$$

1

(i) We can even find some $f \in L^1[0,\infty)$ that is continuous and does not satisfy $\lim\limits_{x\rightarrow\infty} f(x)=0$ Consider $\phi$ a positive bump function and still continuous, with support in $[-1/2,1/2]$ and $\int_\mathbb{R} \phi(x) dx=1$. We can also assume $\phi(0)=1$. Then $\int_\mathbb{R} \phi(nx) dx=n$ for $1/n \in \mathbb{N}$ and ...

1

Continuous representatives are unique whenever they exist. That is just because if you modify a continuous function at one point, to have the result also be a continuous function, you need to make a modification on an interval, which will leave the equivalence class you started in (since intervals have positive measure). Another way of putting it is that ...

1

Apparently this is not true, according to link: If $\Omega = (0,1)$ and $f(x) = g(x) = x^{-1/3}$, we have $\|fg\|_2 = \infty$ but $\|f\|_2 = \|g\|_2 < \infty$.

1

$$X_{n+1} = X_n + T_{n+1}$$ where $T_{n+1}$ is independent of $X_1,\ldots,X_n$ and $$\Pr(T_{n+1}\in A) = \int_A \frac{dt}{(2+t^2)^{3/2}}$$ for every measurable set $A$. (Thus $T_{n+1}$ has a t-distribution with $2$ degrees of freedom.)

1

We have to prove finiteness of $$I:=\int_{[0,+\infty)}\int_{[0,+\infty)}|f(t)|e^{x-t}\mathbf 1\{t\geqslant x\}\mathrm dm(t)\mathrm dm(x).$$ Using Fubini-Tonnelli's theorem (that is, switching the integral when the integrand is a non-negative function), we infer that $$I=\int_{[0,+\infty)}e^{-t}|f(t)|\int_{[0,t)}e^x\mathrm dm(x)\mathrm dm(t).$$

1

The best intuition I can get for your result comes from one dimension. Think of $u$ as being $1$ on an interval $(a,b)$ and then diminishing down to $0$ outside a bigger interval $(c,d)\supset (a,b)$. As you exponentiate $u$ to higher and higher powers, it becomes dominated by what happens on $(a,b)$ and goes to zero outside $(a,b)$. Also, no matter how ...

1

We will use the following inequality: $$\forall a,b\ge 0, s\ge 1: \, (a+b)^s\leq 2^{s-1}(a^s+b^s)$$ This is easily proved by noting that the function $f(t)=t^s$ is convex for $t\ge 0$ and therefore you have $$f(\frac{1}{2}a+\frac{1}{2}b)\leq \frac{1}{2}f(a)+\frac{1}{2}f(b)\Leftrightarrow \big(\frac{a+b}{2}\big)^s\leq \frac{1}{2}a^s+\frac{1}{2}b^s$$ ...

1

Given $f\in\mathcal{S}(\mathbb{R}^d)$ and $a>0$ let $f_n(x)=n^{-a}\,f(x/n)$. Then $f_n$ converges uniformly to $0$ but $\|f_n\|_p=n^{d/p-a}\,\|f\|_p$ does not converge to $0$ if $1\le p\le d/a$. If $f_n$ converges to $0$ in the metric of $\mathcal{S}(\mathbb{R}^d)$, then $$\lim_{n\to\infty}\sup_{x\in\mathbb{R}^d}(1+|x|)^{d+1}\,|f_n(x)|=0.$$ From this ...

1

Hint: Think about $$\sum_{n=2}^{\infty}\frac{1}{n^a (\ln)^b}$$ for $a, b> 0.$ This diverges for all $a, 0< a <1,$ but converges for $a= 1,b>1.$

1

Consider that if $a_n<p$ for each $n$ with $\lim_{n\to \infty}a_n=0$, and $x=(x_n)_{n\in N}=(n^{-1/(p-a_n)})_{n\in N}$ then $x\notin l_q$ for any $q<p.$ I tried putting $(x_n)^p=n^{-1}(\log n)^{-2}$ for all sufficienly large $n$, to get $x\in lp.$ For $n>1$, I got $a_n=p y_n/(1+y_n)$ where $y_n=2(\log (\log n))/\log n.$ And $n\geq 4\implies ... 1 Consider$x=(x_1,x_2,...,x_n,...)\in \bar {H}$. T.P.$x\in H$. as$x\in \bar H$there exists seq say$x^n=(x^n_1,x^n_2,x^n_3,....)\in H$converging to$x$.$\implies\|x^n-x\|_2< \varepsilon\implies$on simplifying we will get$|x_k^n-x_k|< \varepsilon\forall n \geq n_0$,$\forall k$. So$(x^n_k)_{n\in\mathbb{N}}\to x_k\$ for each k. Now ...

1

Note that $$\int_X|f|=\int_X\chi_E|f|+\int_X\chi_{E^c}|f|.$$

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