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3

Let $s$ integrable and $\varepsilon$ such that $s\leqslant|f|$ on $X$ and $\displaystyle\int_X|f|\leqslant\varepsilon+\int_Xs$. Then, for every measurable $E\subseteq X$, $|f|-s\geqslant0$ on $X\setminus E$ hence $\displaystyle\varepsilon\geqslant\int_X|f|-s=\int_E|f|-s+\int_{X\setminus E}|f|-s\geqslant\int_E|f|-s$ , which implies ...

0

The proposition is trivial if the function $f$ is bounded. So assume that $f_n(x) = n$ if $f(x) \leq n$ and $f_n(x) = 0$ otherwise. Then each $f_n$ is bounded and $f_n \to f$ pointwise so by the Monotone convergence theorem $\int_E f_n \to \int_E f$. So given $\epsilon > 0$ there exists an $N$ such that $\int_E f - \int_E f_N < \epsilon/2$. Choose ...

0

Summary of comments Fix $a$ and let $T(x)=\sum_{k=1}^\infty a_kx_k$. For $n=1,2,\dots$ define linear functionals $T_nx = \sum_{k=1}^n a_kx_k$. These are bounded on $\ell^2$ and satisfy $\sup_n |T_n(x)|<\infty$ for every $x$, because the series $\sum_{k=1}^\infty a_kx_k$ converges. By the uniform boundedness principle, also known as the ...

2

And another proof: Let $\epsilon>0$ and choose $a$ large enough so that $\|f 1_{[a,\infty)} \|_p < {\epsilon \over 2}$. Now choose $L \ge a$ large enough so that $|\frac{1}{x^{1-\frac1{p}}}\int_0^a f(t)\,dt | < {\epsilon \over 2}$ whenever $x \ge L$. Then we have $|\frac{1}{x^{1-\frac1{p}}}\int_0^x f(t)\,dt | = |\frac{1}{x^{1-\frac1{p}}}\int_0^a ... 4 Use an approximation argument: it is true when$f$is a simple function (linear combination of characteristic function), and for any$s$simple and each positive$x$, $$x^{-(1-1/p)}\int_0^x|f(t)|\mathrm dt\leqslant \lVert f-s\rVert_p+x^{-(1-1/p)}\int_0^x|s(t)|\mathrm dt.$$ 0 When we consider integrability, estimates within a constant factor are enough. Then the sum of two positive numbers is as good as their maximum, because $$\max(a,b)\le a+b\le 2\max(a,b)$$ So, replace$x^2+y^a$by$\max(x^2,y^a)$. Then we only need to integrate $$\iint \chi_{\{x^2<y^a\}} y^{-a} \,dx\,dy + \iint \chi_{\{x^2>y^a\}} x^{-2} \,dy\,dx$$ ... 1 Because$L^{p}$spaces expose the subtle nature of arguments. You have reflexive, non-reflexive, separable, non-separable, algebra, Hilbert, Banach, etc.. And, interpolation works between such spaces because of the exponent. They're good spaces for testing conjectures. They're the original spaces that firmly established the need to separate a space from its ... 2 For$1<p<\infty$, strict convexity of the norm implies that the only shortest path is the line segment between these points. For$p=1$, the length of a path$(x(t),y(t))$is just the sum of the lengths of one-dimensional paths$x(t)$and$y(t)$. Both of those must go from$0$to$1$. They will have length$1$if and only if the function is ... 2 Instead of$(0,T)\times\Omega$work on$I=[0,1]. Let \begin{align*} f_n(x)=\begin{cases} x^{-2} & x\in[1/n;1]\\ 0 & x\in [0;1/n] \end{cases} \end{align*} Then|\{|f_n|>k\}|\leq |\{x^{-2}>k\}|=k^{-1/2}$but$\lim_{n\to\infty}\int_0^1f_n(x)dx=\infty$. 2 (Partial answer) For your question that there does not exists a function$g$increasing on$[0,1]$such that for all$0\leq a<b\leq 1$we have $$g(b)-g(a)\geq \frac{\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}}$$ you can argue as follows. First, changing$g$to$g(x)-g(0)$if necessary, we can suppose that$g(0)=0$. Then if$b>a=0$, we get$g(b)\geq 1$. Now ... 3 I don't see any use of any Hardy-Littlewood inequality here. You have a measure space$(Q,\delta^\alpha \,dx\,dt)$, which has finite total measure. I will denote the measure by$\mu$for simplicity. The assumption (2.7) says that$u$is in the weak$L^{\hat q}( d\mu)$space. Then it's just a matter of interpolation to get that$u\in L^q( d\mu)$for every ... 0 The following is not a complete answer. In the following let$f$be some candidate for a counterexample. First observation:$f$cannot be monotone. If it were, say, increasing, then we would have $$\int_0^x |f'| dx \leq f(x)-f(0)$$ by the Lebesgue decomposition theorem. Similarly$\int_0^x |f'(x)| dx \leq f(0) - f(x)$if$f$is decreasing. Second ... 2 If$p\gt 1$, we define the$L^{p,\infty}$semi-norm by $$\lVert f\rVert_{p,\infty}^p:=\sup_{t\gt 0}t^p\lambda\{s, |f(s)|\gt t\}$$ (this is equivalent to a norm, namely,$\sup_{A,\lambda(A)\in (0,\infty)}\mu(A)^{1/p-1}\int_A|f|\mathrm d\lambda$). If we define$x:=k^{1/m}$and if we use the inequality, we obtain $$x^{2m\frac{N+1}N}\lambda\{|u|\lt ... 1 Lemma. If g\colon\mathbf R^n\to\mathbf R is continuous and g(x)\leqslant 0 for each x\in A, then g(x)\leqslant 0 for each x\in \overline A. We use this with g(x):=|f(x)|-K and A:=\Omega\setminus N. To show the lemma, take x in the closure of A and (x_n)_{n\geqslant 1} a sequence of elements of A converging to x. Then ... 1 Well the biggest problem is that it is a classical result of analysis/topology that the space of continuous functions is closed under the L^\infty topology i.e. the topology of a.e. uniform convergence (in the case of continuous functions we take a continuous representative, and a convergent sequence in uniformly Cauchy, thus converges uniformly to a ... 1 Taking for simplicity \Omega compact, if C^\infty_c(\Omega) were to be dense in (L^\infty,\|\cdot\|_\infty) then having that C^\infty_c(\Omega) is separable because every differentiable function can be approximated by rational polynomials, so would have to be L^\infty(\Omega). 1 The compactly supported part is most important. Try to L^\infty-approximate the constant function g \equiv 1 by a compactly supported function f. There will be a set of positive measure where f is zero, so \| f - g \|_\infty will always be at least 1 no matter what f we pick. 0 Alternatively you can use Hölder's Inequality to find that$$ \lVert x \rVert_r \le \lVert x \rVert_p \tag{1} $$and then you can proceed by letting \epsilon > 0 and putting \delta = \epsilon so that$$ \lVert x - y \rVert_p < \delta \implies \lVert i(x) - i(y) \rVert_p < \epsilon \implies \lVert i(x) - i(y) \rVert_r < \epsilon $$Hint for ... 0 We have \|x\|_r \le \|x\|_p (from Jensen's inequality, using \lambda[0,1] = 1). Let i:L^p[0,1] \to L^r[0,1] be the injection. Then \|i(x)-i(y)\|_r = \|x-y\|_r\le \|x-y\|_p, hence it is Lipschitz continuous. This nesting is true more broadly for L^p(X,\mu), where \mu X < \infty (except the Lipschitz constant changes). 1 Consider h_n = f \cdot \chi_{\Bbb{R}^n \setminus B_{1/n}(\gamma_0)} and use dominated convergence. This implies that h_n \to f in L^p and each h_n is constant on the ball B_{1/n}(\gamma_0) with radius 1/n around \gamma_0. EDIT: Ok, to get a solution to the precise formulation of your problem, take$$ h(x) = (f(x) - f(\gamma_0)) \cdot ... 1 The duality$\ell^p(X)^*=\ell^q(X^*)$for$1<p<\infty$holds for every Banach space. Indeed,$c_{00}(X)$, the space of finitely supported sequences, is dense in$l^p(X)$. Therefore, every linear functional on$l^p(X)$is determined by its values on sequences with one nonzero element. This identifies such a functional with an$X^*$-valued sequence ... 2 The sum must extend over all integers, by the way, otherwise, it would be finite (namely$0$) for a constant function whose value is less than$1$in modulus. Then, with$B_n = \{x : 2^n < \lvert f(x)\rvert \leqslant 2^{n+1}\}$, we have $$A_n = \{ x : 2^n < \lvert f(x)\rvert\} = \bigcup_{k=n}^\infty B_k,$$ and the union is disjoint. Thus ... 3 Hint: Holder's inequality with$f$and$g(x)=1$. 1 Suppose that$L^p$is closed in$L^1$: in order to prove the existence of$C$, we only need to show that$L^1\subset L^p$, then we use the closed graph theorem. Let$f\in L^1$; define$f_n:=f\chi_{\{|f|\leqslant n\}}$. Then the sequence$(f_n)_{n\geqslant 1}$is an element of$L^p$, which converges in the$L^1$norm to$f$. By closeness,$f$belongs to ... 3 If$u\in\mathbb L^p$for some$p>1$, then take$u_n:=u\chi_{\{|u|\leqslant n\}}$. If$u$does not belong to any$\mathbb L^p$space for any$p>1$, then it is not possible: if$\lVert u_n-u\rVert_1\to 0$and$(u_n)_n$is bounded in$\mathbb L^p$, then extract a subsequence$(u_{n_k})_{k\geqslant 1}$which converges almost everywhere to$u$. Then using ... 3 Hint:$\Omega=(0,1)$,$f(x)=\log x$. 0 Since my attempt to close the question as a duplicate failed, I'll post an answer: the statement follows from a general theorem on the continuity of Nemytskii operator, which is stated and proved here. 3 Suppose$u_n \stackrel*\rightharpoonup u$in some$X^*$. Given$\epsilon > 0$choose some$x\in X$with$\|x\| = 1$and$|u(x)| \ge \|u\|-\epsilon$. We have $$\lim |u_n(x)| = |u(x)| \ge \|u\| - \epsilon$$ and on the other hand $$\lim |u_n(x)| \le \liminf \|u_n\|\|x\| = \liminf \|u_n\|$$ So $$\|u\| - \epsilon \le \liminf\|u_n\|$$ for each ... 2 This is the answer I've unravelled: Let$z \in (\ell^p)^{\ast\ast}$. I want to prove that exists$x \in \ell^p$such that$\langle z,f\rangle=\langle f,x \rangle$for every$f \in (\ell^p)^\ast$. I know that there are the isomorphisms:$j_p: \ell^q \rightarrow (\ell^p)^\ast$and$j_q: \ell^p \rightarrow (\ell^q)^\ast$Now, fix$z \in ...

0

It is not true that $$x=\sum_{n=1}^{\infty} x_ne_n$$ for $x\in\ell^{\infty}$, because convergence does not necessarily occur in the $\ell^{\infty}$-norm. Indeed, if $x=(1,1,1,\ldots)$, then, for any $N\in\mathbb N$, $$\left\|x-\sum_{n=1}^{N}x_ne_n\right\|_{\infty}=\|(\underbrace{0,\ldots,0}_{N\text{ times}},1,1,1,\ldots)\|_{\infty}=1,$$ so $x$ cannot be the ...

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Note that "$x = \sum_n x_n e_n$" does not hold in norm if $x \in \ell^\infty \setminus c_0$, neither does it weakly, as for $x \not\in c_0$ there is some $x^*\in (\ell^\infty)^*$ with $x^*|_{c_0} = 0$ and $x^*(x) = 1$. Then $$x^*\left(\sum_{n=1}^N x_ne_n\right) = 0 \not\to 1 = x^*(x)$$ It holds only weakly$^*$, as given $y \in \ell^1$, we have $xy\in ... 0 The sum you give is not norm convergent, unless$x\in c_0$. (It is weakly$\!^*$convergent, but that is beside the point.) You have here a Schauder basis for$c_0$, a (comparatively) tiny subspace of$l^\infty$. 1 No. If$f_\epsilon(x)>f(x)$, then$-f_\epsilon(x)<-f(x)$, so for any$f$, either$f$or$-f$is a counterexample. The convergence is decreasing when$f$is subharmonic. This is because$\frac{d}{d\epsilon}f_\epsilon$is the convolution of$f$against a radially symmetric function that is negative on small radii and positive for large ones. The ... 1 So you have the canonical maps: $$\begin{cases} f_q: \ell^p\to (\ell^q)^* \\ f_p: (\ell^q)^*\to (\ell^p)^{**}\end{cases}$$ which are the isomorphisms between$\ell^p$and$(\ell^q)^*$and$\ell^q$and$(\ell^p)^*$respectively. Then just write$f_p\circ f_q=j_p:\ell^p\to (\ell^p)^{**}$. 1 Here's another explanation of the failure of inequality $$\tag{1} \|f\ast g\|_{L^p(\mathbb{R}^N)}\le C \|f\|_{L^p(\mathbb{R}^N)}\|g\|_{L^p(\mathbb{R}^N)},\qquad f, g\in \mathcal{S}$$ for$p>1$. (The OP is about$N=1$but there is no added difficulty in considering the general case). It is a routine application of the so-called scaling argument. Assume ... 2 The inclusion stated in the title follows from the fact that$\mathcal{S}\subset L^1$. But the inequality would involve the$L^1$norm of$f$, not its$L^p$norm. (Namely, Young's inequality for convolution.) The point is, smoothness is irrelevant to$L^p$norm estimates of this sort. To see why you can't have$\|f\|_{L^p}$, consider$f=\chi_{[0,M]}$... 0 If$1<p<2$, it holds. Note that$f \in C(R)$implies$f \in L^{\infty}$. Since$f$is in$L^p \cap L^\infty$, we get$f \in L^2$. Thus we get the desired result. For$f \in L^p$with$ 2<p<\infty$, Fourier transform of$f$is not defined unless$f \in L^q$for some$1 \leq q \leq 2$. 0 This is false. Consider, for example,$f_n=n^{3/4}\chi_{(0,1/n)}$on$D=[0,1]$. Then$f_n\to 0$in$L^1$, but the sequence does not converge weakly in$L^2$(it would be bounded in norm if it did). The problem with your argument is that you are implicitly assuming that$\|f_n\|_2$is bounded; otherwise, there's no obvious way to carry out the approximation ... 3 By Jensen's inequality$\int |f|^2 \log |f|=\int |f|^2 \cdot \frac{1}{p-2}\log |f|^{p-2} = \frac{1}{p-2}\cdot\int |f|^2\log |f|^{p-2} \leq \frac{1}{p-2}\log (\int |f|^{p-2}\cdot |f|^2) = \frac{1}{p-2}\cdot \frac{p}{2}\log (\int|f|^p)^\frac{2}{p}=\frac{1}{p-2}\cdot \frac{p}{2} \log ||f||_p^2$because$\frac{1}{p-2}=\frac{n-2}{4}, ...

1

The denseness of $C_c^\infty(\def\R{\mathbb R}\R)$ in $L^p(\R)$ does the trick. Just use Hölder's inequality. For $f \in L^p(\R)$, choose $f_n \in C^\infty_c(\R)$ with $\def\norm#1{\left\|#1\right\|}\norm{f_n -f}_p \to 0$. Then $$\left|\left<f_n - f, g\right>\right| \le \norm{f_n - f}_p \norm g_q \to 0$$ Addendum: We will show $\sup B = \alpha$. ...

1

The inclusion $$(C([0,1]), || \cdot ||_{\infty}) \longrightarrow (L^2([0,1]),|| \cdot ||_2)$$ is continuous since for all $f \in C([0,1])$ $$|| f ||_2^2 = \int_0^1 f^2 \leq \int_0^1 ||f||_{\infty}^2 = ||f||_{\infty}^2$$ If you restrict the inclusion to $M$ $$i: (M, || \cdot ||_{\infty}) \longrightarrow (M,|| \cdot ||_2)$$ is continuous and ...

0

Define $\iota\colon (M,\lVert \cdot\rVert_2)\to (C[0,1],\lVert\cdot\rVert_\infty)$ by $\iota(f)=f$. We have to show that $\iota$ is continuous. Since $M$ and $C[0,1]$ endowed with their respective norms are complete, we have to show that if $\lVert f_n-f\rVert_2\to 0$ and $\lVert f_n-g\rVert_\infty=0$ then $f=g$. To see this, notice that $f_n\to g$ a.e. ...

1

Yes it is true because $C^\infty_c(\mathbb{R})$ is dense in $L^p(\mathbb{R})$ for any $1\leq p<\infty$ in the strong topology and is dense in $L^\infty$ in the weak-$\star$ topology. Edit: $L^\infty(\mathbb{R})$ is the dual of $L^1(\mathbb{R})$ and $C^\infty_c(\mathbb{R})$ is dense in $L^\infty(\mathbb{R})$. For any $f\in L^1(\mathbb{R})$ then the scalar ...

1

Adding to martini's answer: From $$\|f\|_p \le \|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^\theta$$ one finds using Young's inequality in the form $$ab \le \theta a^{\frac1\theta} + (1-\theta)b^{\frac1{1-\theta}}$$ the estimate $$\|f\|_p \le \|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^\theta \le (1-\theta)\|f\|_{p_0} + \theta\|f\|_{p_1}.$$ Using the definition of $p$ it ...

1

There is some convexity, don't know if this helps you, but: Let $p_0, p_1 \in [1,\infty]$, $\theta \in [0,1]$, $\frac 1p = \frac{1-\theta}{p_0} + \frac{\theta}{p_1}$. Then for $f \in L^{p_0}\cap L^{p_1}$, by Hölder \begin{align*}\def\norm#1#2{\left\|#1\right\|_{#2}}\def\abs#1{\left|#1\right|} \norm fp &= \norm{\abs{f}^{1-\theta}\abs{f}^\theta}p \\ ...

2

One can forget $p(s)$, $r(s)$ and the rest and simply try to show that, for every $a\lt b$, $$\|u\|_a\leqslant\|u\|_b.$$ To wit, considering $v=|u|^a$ and $p=b/a\gt1$, note that Hölder inequality yields $$\int |u|^a=\int v\leqslant\left(\int v^p\right)^{1/p}=\left(\int |u|^b\right)^{a/b},$$ that is, $$\left(\int |u|^a\right)^{1/a}\leqslant\left(\int ... 5 No. Nonlinear transformations and weak convergence go together like drinking and driving. For example, let r_k be the kth Rademacher function on [0,1], that is r_k = \operatorname{sign}\sin ( 2^k \pi x) . Then 2^p r_k \rightharpoonup 2^{p-1}\mathbf {1} in L^1, where \mathbf{1} is the constant function equal to 1. On the other hand, ... 0 Both assumptions imply convergence in the sense of distributions: that is, for every smooth compactly supported function \varphi we have$$ \int \varphi u_n\to \int \varphi u,\qquad \int \varphi u_n\to \int \varphi v \tag1$$So,$$\int \varphi u = \int \varphi v\tag2 for every such $\varphi$. This means exactly that $u=v$ in the sense of distributions. ...

1

Looking at non-negative unsigned Lebesgue integrable functions on $\mathbb{R}$, step function: taking finitely many values on finitely many intervals. simple function: taking finitely many values on finitely many Lebesgue measurable sets. True: Both step functions and simple functions are dense in $L^1$. They can approximate any Lebesgue integrable ...

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I expect you mean to ask if you can find a monotonically increasing sequence of step functions approximating the given function from below? The answer is a resounding no. For a simple example, consider the characteristic function of a fat Cantor set.

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