# Tag Info

22

Because complex-valued functions are used. The square of a complex number need not be non-negative.

6

Besides the complex-valued case, I suspect it also has to do with the existence of other Lp spaces; Wikipedia gives the general definition as $$\|f\|_p = \left(\int_S |f|^p d\mu\right)^{1/p}$$ Since the absolute value symbols are redundant only for even integral $p$, omitting them disrupts the uniformity of the notation without buying a whole lot.

6

No, consider $u=1\mbox{}{}{}{}{}{}{}{}{}$.

5

Fix $\varepsilon>0$ and $g\in L^{\infty}([0,1])$. If $T$ is surjective, then $g=T(f)$ for some $f\in L^1([0,1])$, indeed. By the continuity of $T$ at $f$, there exists some $\delta>0$ such that \begin{align*} \left.\begin{array}{ll}\bullet\,h\in ...

3

By one hand, if $\Omega\subset \mathbb{R}^N$ is any open set then, $C_0^\infty(\Omega)$ is dense in $L^p(\Omega)$, as you can see, for example, in Brezis chapter 4. Once $$C_0^\infty(\Omega)\subset C^\infty(\Omega)\cap L^p(\Omega)\subset L^p(\Omega),$$ the result follows. On the other hand, if $-\infty<a<b<\infty$, you can see in chapter 8 of ...

3

Assume $f(x)=\sum _{n=0}^\infty a_n x^n$. Put $c_n=\sum_{m=0}^n a_m a_{n-m}$, so that $f(x)^2=\sum _{n=0}^\infty c_n x^n$. Then, invoking first the Monotone Convergence Theorem, then the Tonelli/Fubini Theorem and then removing vanishing terms, we see that $$\int f^2 dx=\lim_{k\rightarrow\infty} \int_{-k}^k f^2 dx=\lim_{k\rightarrow \infty} ... 3 It is very easy to see that f\circ g\in L^{p}\left(\mathbb{R}\right) holds for all g\in L^{p}\left(\mathbb{R}\right) if f is Borel measurable with \left|f\left(x\right)\right|\leq C\cdot\left|x\right| for all x\in\mathbb{R} and some (fixed) C\geq0, so I will only show the converse (and I will not go into Borel-measurability, because that is a ... 2 Let (X_n)_{n=1}^\infty be a sequence of disjoint sets of strictly positive finite measure. Set g_n = \mathbf{1}_{X_n}. For each non-empty set A\subset \mathbb{N} let$$f_A(x) = \sum_{n\in A}g_n(x).$$Then \|f_A - f_B\| = 1 for distinct subsets A,B\subseteq \mathbb{N}. The power-set of \mathbb{N} has cardinality continuum so L_\infty(\mu) is ... 2 Your argument is fine. You can also do a direct variant of it like this. If \{f_{n_k}\}_{k=1}^{\infty} were a Cauchy sequence, then \lim_{k \rightarrow 0} ||f_{n_{k+1}} - f_{n_k}||_{L^p[0,2\pi]} would be zero. But by Holder's inequality$$\int_0^{2\pi}|f_{n_{k+1}} - f_{n_k}|^2 \leq||f_{n_{k+1}} - f_{n_k}||_{L^p[0,2\pi]}\,\,\,\,||f_{n_{k+1}} - ...

2

It is not compact by the Khintchine inequality which tells you that the closed span of Rademacher functions is isomorphic to a Hilbert space in every $L_p$-space. The inclusion then takes a copy of a Hilbert space to another copy of a Hilbert, so it cannot be compact.

2

Three things to use: Triangle inequality for integrals: $\left|\int fg\right| \le \int |fg|$. This yields $\sup_g \left|\int fg\right| \le \sup_g \int |fg|$. Hölder's inequality: $\int |fg|\le \|f\|_p\|g\|_q$. This yields $\sup_g\int |fg|\le \|f\|_p$. Equality case of Hölder's inequality: you need only the easy direction, namely $\int fg= ... 2 It looks to me that$\xi\cdot\eta$has a finite$L^p$norm due to Young's inequality. For instance: $$\|\xi\cdot\eta\|_p^p = \mathbb{E}[|\xi|^p\cdot|\eta|^p]\leq\frac{1}{2}\left(\mathbb{E}[|\xi|^{2p}]+\mathbb{E}[|\eta|^{2p}]\right)=\frac{\|\xi\|_{2p}^{2p}+\|\eta\|_{2p}^{2p}}{2}.$$ 2 Take$f\in L^{2,n}(\Omega)$. Almost every point of$\Omega$is a Lebesgue point of$|f|^2$. At such points$x$, $$|f(x)|^2 = \lim_{r\to 0} \left( r^{-n} \int_{\Omega_r(x_0)}|f|^2\, dx \right) \le [f]^2_{L^{2,n}}$$ Therefore,$\|f\|_{L^\infty}\le [f]_{L^{2,n}}$. When the measure of$\Omega$is finite, the$L^\infty$norm controls the$L^2$norm and we get ... 1 Suggestions: Recall that every weakly convergent sequence is bounded in the norm. Think of what it means for$n_{k}u_{k}$to be bounded in the norm for every sequence$n_k$Conclude that your sequence$u_k$is very special. 1 The answer becomes yes if we consider$C_0$and assume$u\in L^1_{loc}$. Weak derivative being zero implies$u$has to be zero$a.e$see here 1 It is not true in general that convergence in$\ell^p$takes place: take$a_n^k=0$if$n\neq k$and$1$otherwise. However, if the remainders are uniformly convergent to$0$, that is, $$\lim_{N\to \infty}\sup_{k\geqslant 1}\sum_{n=N}^{+\infty}|a_n^k|^p=0,$$ then the convergence in$\ell^p$holds by an approximation argument. 1 You can take$\{\sin(nx)\}_{n\geq 1}$. It is bounded in$L^p$but you can't take a sub-sequence converging in$L^q$because$\sin(nx)$has no sub-sub sequences converging a.e. (Remind that strong convergence in$L^q$implies a. e. convergence up to taking subsequences). 1 EDIT: A lot of what I said in my original post is wrong, as it turns out :( The$L^1$example below shows that your formulation of the Fatou property is not as useful as it first appears. The Fatou property as you formulate it is not true in general. As a counterexample, consider $$C_0 := \{f : \Bbb{R} \to \Bbb{C} \mid f(x) \to 0 \text{ as } |x|\to\infty ... 1 Suppose g(x) = x and I_n = [2^n - 3^{-n}, 2^n]. Let I = \cup_{n=1}^{\infty} I_n and f(x) = g(x)\chi_{I}(x). Note that I_n has length 3^{-n}, and on this interval, g is bounded above by 2^n. Therefore,$$\int_{I_n} g(x) dx \leq \left(\frac{2}{3}\right)^n$$and so$$\int_{\mathbb{R}} f(x) = \sum_{n=1}^{\infty} \int_{I_n} g(x) dx \leq ... 1 Yes. Two steps: Step 1. using the inequality $$y>x>a\Longrightarrow \vert f(x)-f(y) \vert\leq\int_a^\infty\vert f'(t)\vert dt$$ we conclude that$f$satisfies Cauchy criterion as$x\to\infty$(because$f'\in L^1$,) and consequently the limit$\lim\limits_{x\to\infty}f(x)=\ell$does exist. Step 2. the limit$\ell$must be zero because otherwise$f$... 1$f'\in L^1(\Bbb R)$, then$\int_{-\infty}^0|f'(t)|dt$and$\int_0^{+\infty}|f'(t)|dt$converges, in particular$\int_{-\infty}^0f'(t)dt$and$\int_0^{+\infty}f'(t)dt$converges. The convergence of the integral$\int_0^{+\infty}f'(t)dt$implies that the function$x\to \int_0^{x}f'(t)dt$admit a finite limit$l$(as$x\to +\infty$). But ... 1 The given arguments are good. I agree with Golbez, the fact that if$(f_{n_k})_k$converge weakly to something, then it converges to$0$has to be detailed, for example noticing that$\int_{[0,2\pi]} f_{n_k}g\mathrm d\lambda\to 0$for each continuous function$g$and concluding by a density argument. 1 Reading JimmyK4542's comment made me want to delete this question as this is very much covered in that thread. But then again some poor soul like me might find this useful in future so here goes. let$x_1$be the maximum of the lot. And I take it out of the expression.$x_1(1 + (x_2/x_1)^p + (x_2/x_1)^p + (x_3/x_1)^p .. + (x_n/x_1)^p)^{1/p}$---- (1) ... 1 Try to adapt the following example: the functions$f_t(x) = \chi_{(-\infty,t]}(x)$all belong to$L^\infty(\mathbb R)$, but satisfy$\|f_t - f_s\|_\infty = 1$if$s \not= t$. 1 No, the condition$\int_{\mathbb R} \left(\int_{\mathbb R}|F(x,y)|^{p}dx\right)^{1/p} dy < \infty$is sensitive to the interchange of coordinates. For example, let$p=2$and $$F(x,y) = \begin{cases}y^{-1/2},\quad & 0\le x\le y\le 1 \\ 0 &\text{otherwise}\end{cases}$$ Then for$y\in [0,1]$we have $$\int_{\mathbb R}|F(x,y)|^{2}dx = y^{-1} ... 1 In the case of finite dimensional spaces, measurability can be reduced to the "ordinary" notion of measurability as follows: A function f : X \to \Bbb{R}^n is measurable if and only if for each i \in \{1, \dots, n\}, the component function f_i : X \to \Bbb{R} is measurable. This can be seen as follows: Measurability is preserved by continuous ... 1 As pointed out in the comments, this is easier to prove if you use the uniform boundedness principle. I will assume you meant that \frac1q+\frac1p=1. Now, since (L^q(\mathbb{R}))' is isometric to L^p(\mathbb{R}), we can interpret the sequence (f_n) as a sequence of functionals in (L^q(\mathbb{R}))'. Then the condition says that for each g\in ... 1 You cannot have a norm bound like the one you wanted, since adding a constant to a function doesn't change it's BMO-norm but does affect the L^p-norm. As for proving the non-quantitative inclusion one has to somehow manage to control the L^p-norm of a function with essentially an L^1-integral (or supremum of). Therefore something nontrivial must ... 1 Let (t_i)_{i \in \mathbb{N}} be a dense subset of [0,T] such that t_k = T for some k \geq 1. If we set \mathcal{H}_n := \sigma(B_{t_1},\ldots,B_{t_n}) and$$\mathcal{H}_{\infty}:= \sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{H}_n \right) \qquad \quad \mathcal{F}_T := \sigma(B_s; s \leq T),$$then$\mathcal{H}_{\infty} = \mathcal{F}_T\$. ...

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