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The even $p$ norms corresponds directly to the $p$th moment of the random variable. $$E[X^p] = ||X||_p^p$$ The association for odd $p$ is not as clear because the definition of the norm requires taking the absolute value of the outcomes. At best you can note: $$E[|X|^p] = ||X||_p^p$$ In this way, for a symmetric distribution, you can ...

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Hints: Fix $\epsilon>0$. Fix $k \in \mathbb{N}$. Using Markov's inequality, show that $$A_N^k := \left\{x; \exists n \geq N: |f_n(x)-f(x)| \geq \frac{1}{k} \right\}$$ satisfies $$m(A_N^k) \leq k \sum_{n=N}^{\infty} \|f_n-f\|_{L^1}.$$ Conclude from the first step that there exists $N=N(k)$ such that $m(A_{N(k)}^k) \leq \epsilon 2^{-k}$. Set $$A := ... 1 There may be a misunderstanding here; I don't see how one would use Jensen's inequality to control L^p norm by L^1 norm. Anyway, the proof of$$ \lim_{n\to\infty}\|f-h_n*f\|_p =0 $$is essentially the same for 1<p<\infty as it is for p=1. For continuous functions g we have h_n*g\to g uniformly: just split the integral in ... 1 This is a broad question with no definitive answer. You are asking for a description of characteristic functions \chi_S that are Fourier multipliers bounded on L^p, or C^\alpha, etc. Certainly, this is so when S finite or \mathbb{Z}\setminus S is finite. The best known nontrivial case is S=\{n\in \mathbb{Z}:n\ge 0\} which is directly related to ... 1 The argument is the following: Let v_n be a sequence in W^{1, p}_0(\Omega) for some \Omega \subset \mathbb R^N and p >N. If v_n \to v weakly in W^{1,p}_0(\Omega), then ||v_n||_{1, p} is uniformly bounded. By the Sobolev Embedding (Theorem 7.17) and the fact that$$C^{0, \alpha}(\overline \Omega) \to C(\overline \Omega)$$is compact, there ... 2 This is not true. In fact, Kolmogorov constructed (1923) an example of a L^1 function whose Fourier series diverges almost everywhere (later improved to everywhere divergent). Om the other hand, if f\in L^p for some p>1, it's a deep theorem by Carleson (the p=2 case) and Hunt (p>1) that the Fourier series of f converges pointwise almost ... 0 If f \in L^{2} is supported in [-R,R], then it is also in L^{1} because of Cauchy-Schwarz:$$ \int_{-R}^{R}|f|dx \le \left(\int_{-R}^{R}|f|^{2}dx\right)^{1/2} \left(\int_{-R}^{R}1^{2}dx\right)^{1/2} $$Therefore, f_{R} = \chi_{[-R,R]}f \in L^{1}\cap L^{2} for all 0 < R < \infty. And f_{R}\rightarrow f in L^{2} as ... 2 In the case q<p^* the Rellich–Kondrachov theorem applies, if your domain \Omega has the required properties. If q=p^*, the L^q norm is still controlled by the W^{1,p} norm; and since a weakly convergent sequence is norm-bounded, L^q norm is bounded. Moreover, you will have weak convergence in L^q, since linear maps preserve weak ... 3 Yes, it's complete. First, observe that your norm is comparable to the simpler norm$$ \|f\| = \sup_{n\in\mathbb{Z}} \int_n^{n+1}|f(y)|\,dy \tag{1} $$Indeed, \|f\|\le |f|\le 2\|f\| because every interval of length 1 is contained in some interval of the form [n,n+2]. The space with (1) is just the direct sum \bigoplus_\infty X_n of Banach spaces ... 2 Note that this is a special case of the general theorem of means. If a, b are non-negative and 0 < x, y < 1 with x + y = 1 then a^{x}b^{y} \leq ax + by. Equality occurs if a = b. Here in the current question a = t, b = 1, x = \theta, y = 1 - \theta. An easy proof of the general theorem mentioned above is based on Mean Value Theorem. ... 1 This is indeed a version of the Dominated Convergence Theorem. Let's first assume that \mu(X) < \infty. Fix \epsilon > 0 and set$$X_n(\epsilon) := \left\{ x \in X \, : \, \vert f_n(x) - f(x) \vert \geq \epsilon \right\} \quad \text{for $n \in \mathbb N$} \; .$$Show that \vert f_n \vert \xrightarrow{\mu} \vert f \vert and \vert f_n \vert ... 1 I think it's enough to use the sub-sub-sequences lemma. (I don't remember the true name and I'll look for a reference tomorrow morning). But more or less this lemma say to you that if every subsequence of your sequence has a sub-sub-sequence converging (all of them to the same limit) then your entire sequence converges to that limit. 0 The set \mathbb L^1(\mathbf R)\cap\mathbb L^2(\mathbf R) contains the collection of linear combination of indicator functions of sets of finite measure (denoted S), which is dense in \mathbb L^2(\mathbf R). Indeed, assume that f belongs to \mathbb L^2(\mathbf R) and 0\leqslant f(x)\leqslant M. Then we can find a sequence (s_n)\subset S such ... 2 If we take g:\theta\to t^{\theta} we are just stating that the graphics of g on the interval (0,1) lies below the line through (0,1) and (1,g(1)). However, that is trivial, since g(\theta) is a convex function due to:$$ g''(\theta) = t^{\theta}\log^2 t \geq 0.$$4 For t > 0, write$$t^\theta = t^\theta\cdot 1^{1-\theta}.$$Taking the logarithm of both sides, what you need to show becomes$$\theta\log t + (1-\theta) \log 1 \leqslant \log (\theta\cdot t + (1-\theta)\cdot 1),$$which follows from the concavity of \log. 1 This is true in even weaker hypothesis, i.e. only weak convergence. In fact, taking \phi \in C_c^{\infty}(\mathbb R), we have$$ \begin{matrix} \int f'_n \phi & = & - \int f_n \phi'\\ \downarrow & & \downarrow\\ \int g \phi & & -\int f \phi' &= \int f' \phi \end{matrix} $$therefore f'=g a.e.. 1 Follow your idea, s = \|x\|, Consider A = \{s < 2\}, B = \{s > 2\}. On A, we have$$f\le \dfrac{C}{s}$$for some constant C. because$$\inf_{0 \le s < 2} s^{1/2}\log s$$is bounded.$$\int_A |f|^p\le \int_{A} \dfrac{C}{s^p} dxdy = \int_{0}^{2\pi}\int_{0}^{2}\dfrac{C}{s^p}s ds d\theta = 2\pi\int_0^{2}\frac{C}{s^{p-1}}ds$$it is ... 4 Since$$ \lim_{h\to0}\int_0^x\left|\frac{f(t+h)-f(t)}{h}-g(t)\right|^2\,\mathrm{d}t=0\tag{1} $$Hölder's Inequality says$$ \lim_{h\to0}\int_0^x\left|\frac{f(t+h)-f(t)}{h}-g(t)\right|\,\mathrm{d}t=0\tag{2} $$There is no way to derive the continuity of f since changing f on a set of measure 0 will not affect (1). However, the Lebesgue Differentiation ... 3 I elaborate on a remark in one of the other answers. Consider a measure space (X,\mathcal{M},\mu) which has sets of arbitrarily small measure and sets of arbitrarily large measure. (Here I think of the Lebesgue measure on the real line.) Let 1 \leq q < \infty and choose$$g \in L^q \setminus \bigcup_{p \in [1,\infty), p \neq q} L^p.$$This is ... 2 I think part of it is because of Holder's inequality, which dictates that$$ |fg|_{1}\le |f|_{p}|g|_{q} $$But the other reasons goes into this is much deeper. For example, this duality is "structural" as it holds regardless of the measure space is \sigma-finite. And this has to do with concepts like uniform convexity. I used to wondering the same thing ... 3 To expand a bit on the standard answer ("p=2 is a Hilbert space and Hilbert spaces are nice"), let's talk about why Hilbert spaces are nice. One of the key feature of Hilbert spaces is that they are self dual. This is expressed through the Riesz representation theorem, which says that if I want to define a bounded, linear map \phi:L^2\rightarrow ... 0 It's the only Hilbert space. That is, the norm comes from an inner product. Hilbert spaces have a lot of nice properties. For instance they are self-dual. They have a concept of orthogonality and angle. 0 (a) For the inclusion, use and show |f(x)|^r\leqslant |f(x)|^p +1. For strictness: x^{-1/p}. (b) If f belongs to L^\infty, f is almost surely bounded by a constant, hence so are all the powers of f. Constants on finite measure spaces are integrable. For strictness: f(x)=\log x. 1 If such a function exists then$$\tag{1} f(t)=\frac1\pi\sum_{n=1}^\infty B_n\sin(nt). $$But then we would have: \begin{eqnarray} \infty>\|f\|^2&=&\langle f,f\rangle=\frac{1}{\pi^2}\sum_{m,n}B_mB_n\int_{-\pi}^\pi\sin(mt)\sin(nt)\,dt=\frac{1}{\pi^2}\sum_{m,n}\pi B_mB_n\delta_{m,n}\\ &=&\frac1\pi\sum_{n=1}^\infty ... 0 In the Fourier transform represented by C_n terms used by engineers this would more or less be equal to i d(t) where i is an imaginary unit and d is a delta function. I conclude that the function in the coordinate domain was a flat offset with a phase factor. 1 As Farnight suggested, it is better to use \sup. I also prefer to work on a more concrete level: with numbers, not with sets of numbers. So I would write: "for n\in \mathbb{N} ,$$|v(n)|=\left|\frac{n\sin{(n!)}}{n^2+1}\right|\le \frac{n }{n^2+1}\le \frac{n^2}{n^2+1} <1 $$Therefore, the sequence v is bounded." Note that your write up leaves the ... 2 It is certainly not true that |f|^2 < |f|^4 (think of the case where |f|<1). Hint: Cauchy-Schwarz on \int |f|^2 = \int g h where g = |f|^2 and h = 1. 2 Write |f|^2 = |f|^2 \cdot 1 and apply the Cauchy-Schwarz inequality to get \||f|^2\|_1 \le \||f|^2\|_2 \|1\|_2. Simplify the inequality. Note if f \in L_4([0,1]), then the inequality \|f\|_2 \le \|f\|_4 implies \|f\|_2 < \infty and hence f \in L_2([0,1]). 1 From Minkowski,  \| \sum_1^n |f_k|\|_p \le \sum_1^n \|f_k\|_p \le \sum_1^\infty \|f_k\|_p. Now raise to the pth power. 1 Triangle inequality then dominated convergence theorem. Edit: By triangle inequality, we have,$$\int G^p= \lim \int G_k^p=\lim \int (\sum_{k=1}^n |f_k|)^p=\lim ||(\sum_{k=1}^n |f_k|) ||^2\leq \lim \sum_{k=1}^n ||f_k||^2$$By DCT, we then conclude the given inequality. 1 Another proof. More elementary than Urban's proof. (But less general.) Consider the subspace c \subseteq l^\infty consisting of the real sequences that converge. So c is a closed linear subspace of l^\infty. Let L \;:\; c \to \mathbb R be the "limit" linear functional. That is, for x = (x_1,x_2,\cdots), define$$ L(x) := \lim_{n\to\infty} ...

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This is not true because of the following Theorem $\textbf{Theorem:}$ If $X$ is a normed space such that its dual $X'$ is separable, then $X$ itself is separable. So, if $(l^{\infty})' =l_1$, then it will follow that $l^{\infty}$ is separable which is not true.

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It suffices to show that $$\sum_{k=0}^\infty \int_{A_{2^k}} |f|<\infty \tag{1}$$ because $\bigcup_{k=0}^\infty A_{2^k}$ covers the plane except for a subset of finite measure (and on a subset of finite measure $L^{3/2}\subset L^1$). Naturally, Hölder's inequality comes into play: $$\int_{A_{2^k}} |f| \le \left(\int_{A_{2^k}} |f|^{3/2}\right)^{2/3} ... 0 Since constant functions have zero H^{1/2} seminorm, it follows that$$\vert T u \vert_{H^{1/2} (\partial M)} = \vert (T u) -c \vert_{H^{1/2} (\partial M)} = \vert T (u -c) \vert_{H^{1/2} (\partial M)}\tag{1}for every c\in\mathbb{R}. (Trace operator commutes with adding a constant, because the trace of a constant function is that constant function.) ... 2 Why is it important that L^p spaces be complete? Among other things, because we can use it to prove existence of solutions for some problems. For instance, we use (among other things) the completeness of L^p to define a certain Banach space X and to prove that the operator \begin{align*} A&=\begin{pmatrix} 0 & 1 & 0 & 0 & 0 ... 3 It is very difficult to do analysis on a space if it is not complete. Consider for instance \mathbb{Q}. The function f\colon[0,2]\cap\mathbb{Q}\to\mathbb{Q} defined as f(x)=\frac{1}{x^2-2} $$is continuous, but is not bounded. An important result in Banach spaces (and in particular in L^p) that depends on completeness is: Theorem. If X is a ... 3 In one dimension, consider in order$$\chi_{[0,1]}, \chi_{[0,1/2]}, \chi_{[1/2,1]},\chi_{[0,1/3]},\chi_{[1/3,2/3]},\chi_{[2/3,1]}, \dots $$This sequence \to 0 in L^1, and pointwise nowhere. 0 We will give an example in \mathbb{R}^1. The analagous construction in \mathbb{R}^d is similar. Fix k \in \mathbb{Z}. Take some enumeration \{I^k_n\}_{n \in \mathbb{N}} of all subintervals of [k,k+1] which have the form [\frac{p}{q},\frac{p+1}{q}] for some integers p,q. Notice that m(I^k_n) \to 0 as n \to \infty, because for any \epsilon ... 1 A C^*-algebra A is isometric to an L_1-space (even as Banach space!) iff it is one dimensional. Assume A is isometric to L_1 space and \operatorname{dim}(A)>1, then A is weakly sequentially complete. By result of Sakai (proposition 2), this is possible only if A is finite dimensional. By classification theorem for C^* algebras we know ... 2 In what follows, we assume that  G  is a locally compact Hausdorff group that does not have to be abelian. You can certainly define an involution  ^{*}  on  {L^{1}}(G)  by$$ \forall f \in {L^{1}}(G), ~ \forall x \in G: \quad {f^{*}}(x) \stackrel{\text{df}}{=} \overline{f(x^{-1})} \cdot \Delta(x^{-1}),  where $\Delta$ denotes the modular ...

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Let $(f_n)_{n\geqslant 1}$ be a sequence with converges almost everywhere to $0$.  In this thread, it is shown that if $\sup_n\lVert f_n\rVert_p$ is finite, then $f_n\to 0$ weakly in $L^p$. Since boundedness in $L^p$ is a necessary condition for the weak convergence in $L^p$, boundedness in $L^p$ is a necessary and sufficient condition for an almost ...

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If you didn't use any specific properties of the measure, then yes, since the sequence spaces are, as you note, also $L^p$ spaces. If you use specific properties of the measure, you need to check whether the counting measure has these properties. – Daniel Fischer

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You approach looks correct to me. Notice that an application of the inequality in this question with $p=2$ does not give the result you want. Therefore, I am not sure there is a faster solution.

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