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Note that $$f_x = \frac{(1 + x^2) \frac 1 2 x^{-1/2} - x^{1/2} \cdot 2x}{(1 + x^2)^2} = \frac{\frac 1 2 x^{-1/2} - \frac 3 2 x^{3/2}}{(1 + x^2)^2}$$ For large values of $x$, we have the estimate (where $\lesssim$ means "up to some constants") $$|f_x(x)| \lesssim \frac{x^{3/2}}{x^4} = x^{-5/2}$$ For small values of $x$, we have the similar estimate ...

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Hint: you can show that $l^1$ is separable, and $l^\infty$ is not.

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The following gives the main idea. \begin{eqnarray*} |F(k)| &\leqslant &\frac{1}{(1+k^{2})^{2}}\Rightarrow F(k)\in L^{1}\cap L^{\infty } \\ |kF(k)| &\leqslant &|k|\frac{1}{(1+k^{2})^{2}}\Rightarrow H(k)=kF(k)\in L^{1}\cap L^{\infty } \end{eqnarray*} It follows that $g(x)\in L^{1}$ and $g(x)\in L^{2}\cap L^{\infty }$ so $% g(x)\in L^{1}\cap ... 1 The inclusion$L^2\subset L^1_w$does not hold: take$f(x):=\frac 1{\sqrt x\log x}\chi_{(1,\infty)}(x)$. The converse reduces to ask whether$g\in L^1$implies$x\mapsto xg(x)^2\in L^1$. Define $$g(x):=\sum_{j=1}^\infty c_j\cdot\chi(j^2-a_j,j^2+a_j)(x),$$ with the$a_j$small enough. Then$g$is integrable if and only if$\sum_{j=1}^\infty c_ja_j$is ... 0 By a change of variables$u=\frac{y}{x}$, $$\int_0^a e^{\frac{p}{x}} d\lambda(x)=y\int_{y/a}^\infty \frac{e^u}{u^2} d\lambda(u)$$ But$e^{u}/u^2>\frac{1}{u^2}+\frac{1}{u}+\frac{1}{2}$is not integrable at$\infty$. 1 This can be generalized: this is a notion called uniform integrability. It gives nice convergence properties in mesure theory (in particular a generalization of the dominated convergence theorem). 1 I hope I didnt miss something, but this should work, For$p=1$the inequality trivialy holds and notice that$\forall p>0$we have, $$|a-b|^p\leq|a+b|^p$$ We have two cases to consider, (1)$p>1$: Note that$h(x)=x^p$is convex and monotone for$p>1$. Hence, ... 0 Given a measurable space$X$equipped with a measure$\mu$, a function$f : X \to \mathbb{C}$which is defined almost everywhere (that is, up to a set of measure$0$) is said to be an element of$L^1$if $$\int_X |f| d\mu < \infty$$ More properly, we have to identify functions which are equal almost everywhere, so the elements of the Lebesgue space ... 0 As @John pointed out in the comment, your apporach may not work, because$u'_n$does not need to be dounded in$C^0$. One way to approach this problem is the following. We will use Theorem 4.26. from Brezis book. To this end, extend all functions$u_n$to$\mathbb{R}$by using Thereom 8.6. of the same book and note that the extended sequence still bounded. ... 1 An$L^1$functional from a space$X$to$\mathbb{R}$is an$\mu$-measurable function such that $$\int_{X} |f|\,d\mu < \infty.$$ 1 Yes, your proof is good. In terms of writing, you start using a fixed$x\in\ell^1$without saying so. Also, the sentence that says "For that we will show..." doesn't really make sense; I understand it because I know how to prove the density, but otherwise it looks hard to understand. 2 Write$s = tp+(1-t)q$. Consider$1/p'+1/q'=1$. $$\int |f|^{s} = \int |f|^{tp+(1-t)q} \le \left\{\int |f|^{tpp'} \right\}^{1/p'} \left\{\int |f|^{(1-t)qq'} \right\}^{1/q'}$$ Now you can choose $$tpp'=p, (1-t)qq'=q\iff p'=\frac 1t, q'=\frac 1{1-t}$$ because $$\frac 1{p'}+\frac 1{q'} = t+(1-t)=1$$ Then $$\int |f|^{s} \le ||f||_p^{p/p'} ||f||_q^{q/q'} ... 1 Hint: It follows as a corollary of the following statement: Let \{f_i:i\in I\} be a family of functions with f_i\in L^{p_i}(\Omega) and \dfrac{1}{p}=\sum \dfrac{1}{p_i}\leq1. Then \prod f_i\in L^p(\Omega) and$$\|\prod f_i\|_{L^p(\Omega)}\leq\prod\|f_i\|_{L^{p_i}(\Omega)}$$This is the interpolation inequality (H. Brezis "Functional Analysis, ... 1 Here's an idea that might be helpful: that L^p measures spikiness or broadness of a function in some sense. In particular, L^{\infty} consists of functions with no spikes whatsoever, while L^1 functions can't be too broad; the higher p is, the more control you have over spikes, and less control over broadness. Since f \in L^p \cap L^q, we've ... 1 If a function u is actually a radially symmetric function, then we have the integration formula$$ \int_{B(0,R)} u(x)\, dx = |\mathbb{S}^{n-1}| \int_0^R u(r)r^{n-1}\, dr, $$where n is the dimension of the space where u is defined. In your case, u(x)=|x|^{-\lambda}, and it is straightforward to apply the formula and check the integrability of u. 2 Given f \in L^\infty, you can find a sequence f_n \in C_c^\infty such that \|f_n\|_\infty \le \|f\|_\infty, and f_n \to f in L^1, that is, \|f-f_n\|_1 \to 0. 2 If (g_n) is a sequence in C_c^{\infty} and \|g_n-f\|_{L^p}\rightarrow0, then (g_n) is a Cauchy sequence and the limit g\in L^p exists. It can easily be shown that f=g almost everywhere. 1 EDIT: the fact that the sequence is unbounded for y = 0 is not a problem since we don't care about sets of measure 0. Clearly$$\lim_{n \to \infty} \frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big) = \frac{1}{1 + x^2}f(0).$$Moreover we can find N such that if n \ge N then \Big|f\Big(\frac{x}{\sqrt{n}}\Big) - f(0)\Big| \le 1, then in particular ... 1 A is closed, but not open. The sequence x=(1,1,1,...) is not an interior point. If (x_n) is a convergent sequence in A with the limit, then it also converges pointwise. But since 0\le x_{n,m}\le 1, we also have 0\le \lim \limits_{n\to \infty} x_{n,m}\le1, so \lim \limits_{n\to \infty}x_n\in A. B is open, but not closed. If f\in B, not that ... 2 A is not closed. Take any function f\in L^1 such that f is not in L^2, then approximate it by functions in A. I.e. f(x) = \chi_{(0,1)}(x) x^{-1/2}, then f\in L^1, f\not\in L^2. Define f_n(x) = \min(n,f(x)). Then f_n\to f in L^1, \|f_n\|_{L^2}\to\infty. As to your second question: sets like \{f\in L^1: \ f\ge 0 \} are closed as ... 3 From the inclusions of sets$$\ell_p\subset c_0\subset \ell_\infty,$$where c_0 denotes the set of convergent sequences and the separability of (c_0,\lVert \cdot\rVert_\infty) is separable we conclude that (\ell_p,\lVert\cdot\rVert_\infty) is separable. \ell_p endowed with the supremum norm is not a closed subspace of \ell_\infty because its ... -2 Generally, to prove that A=B for sets A,B one shows that if a \in A then also a \in B (i.e. A \subseteq B) and also if b \in B then also b \in A (so B \subseteq A). Let A = L^2(0,\pi) and let B = \mathrm{span}\left(\left\{\sin kx\right\}_{k=0}^\infty\right). How do you proceed to apply this to your problem? 1 Unfortunately, a sequence can be weakly convergent without being pointwise convergent (f_n(x))=(\sin(n\pi x)) on the unit interval (which is not convergent except at 0). However we can show that for each simple function g (linear combination of characteristic functions of measurable sets),$$\lim_{n\to +\infty}\int_{(a,b)} f_ngdx= \int_{(a,b)} ... 1 Consider the unit interval$[0,1]$and $$f_\epsilon(x) = \frac 1 {\sqrt x} \chi_E = g_{\epsilon}(x)$$ where$E = [\epsilon, 1]$and$\chi_E$is a characteristic function. Then$f_{\epsilon}g_{\epsilon} \in L^1$, and we have $$\|f_{\epsilon}\|_1 = \int_{\epsilon}^1 \frac{1}{\sqrt x} dm(x) \le \int_0^1 \frac{1}{\sqrt x} dm(x) = 1$$ and likewise for ... 4 In fact, we can derive that$\,\dim S \le c^2$. Let me describe the proof as it is very elegant. Assume that$v_1,\ldots, v_n\in S$are orthonormal functions, i.e.,$\int_0^1 v_iv_j\,dx=\delta_{ij}$, and for a fixed$a=(a_1,\ldots,a_n)\in\mathbb R^n$define$\varPhi_a :\mathbb R^n\to \mathbb R$as $$\varPhi_a(x)=\sum_{j=1}^n a_jv_j(x).$$ Then $$... 2 Here is a counterexample. Let \phi\colon\mathbb{R}\to\mathbb{R} be C^\infty supported on [0,1], positive with \int_0^1\phi(x)=\int_0^1\phi(x)^2\,dx=1. Let$$ f_n(x)=\sum_{k=1}^n2^{k}\phi(2^{2k}(x-k)). $$Then$$ \int_{\mathbb{R}}f_n(x)\,dx=\sum_{k=1}^n2^{k}2^{-2k}<1. $$Moreover f_n(x)\le f_{n+1}(x) for all x\in\mathbb{R}. It follows that ... 3 No, you can't expect. You can approximate any f\in L^1 by functions f_n in the Schwartz space \mathscr S (even by smooth functions with compact support). Then \hat f_n \in \mathscr S \subseteq L^1 but there is no reason to expect convergence of \hat f_n. 8 You can apply a theorem of Grothendieck to the closure of S in L^2 which is (as you show) contained in C([0,1]) \subseteq L^\infty. Grothendieck's theorem says that every closed subspace of L^p(\mu) (where \mu is a probability measure on some measurable space and 0<p<\infty) which is contained in L^\infty(\mu) is finite dimensional. A ... 4 The statement is false, as I discovered here. Since not everybody has access to the paper, let me provide a summary of the argument: Let R=\mathbb C[t], L the left shift operator, and view \ell^p as an R-module by defining t\cdot x=Lx. Let X=\sum \ker L^i \subset \ell^p be the subspace of eventually-zero sequences. Lemma: Given a PID P, a ... 3 There is no reason to expect that the g_n converge, the analogous implication does also not hold in \mathbb{R}. For an - admittedly boring - explicit example, take f_n = \chi_{[0,2]} for all n (where \chi_A is the characteristic function of the measurable set A\subset \mathbb{R}), and g_n = \chi_{[n,n+1]}. Evidently the constant sequence f_n ... 0 We are given that \lVert f_n\rVert_p converges to 0. Good. But it can converge be arbitrarily slowly. However, given a sequence (c_k)_k of real numbers, we can find a subsequence (f_{n_k})_k such that \lVert f_{n_k}\rVert_p\leqslant c_k for each k. In particular, we obtain for each k,$$\mu\{x,|f_{n_k}(x)|\geqslant k^{-1}\}\leqslant kc_k.$$... 0 In a general context, note that L^p convergence \implies convergence in probability \implies almost sure convergence (up to an extraction!). edit: for a simple example, consider$$ f(x) = 1_{nx<1} \text{ if$n$is even}\\ f(x) = 1_{n(1-x)<1} \text{ if$n$is odd} $$and the subsequence \phi(n) = 2n. 1 If you are working in \mathbb{R}: Without loss of generality assume that f,g \geq 0. For each n \in \mathbb{N} let A_n \subseteq \mathbb{R} such that 0<\mu(A_n)<\infty and f>n on A_n, where I use \mu to denote Lebesgue measure on \mathbb{R}. Let g_n:= 1_{A_n}/\mu(A_n) \in L^1. Then$$\|fg_n\|_1 = \frac{1}{\mu(A_n)}\int_{A_n} ... 0 If$L^{1}=L^1(\Omega)$where$\Omega$is$\sigma$-finite, then$(L^{1})^*=L^\infty$, and if your claim didn't hold, there would be an$M>0$such that for all$g\in L^1$,$\|fg\|_{L^1}<M$, and so$g\mapsto\int fg$would be in$(L^1)^*$, which is impossible. So under the assumption that your measure space is$\sigma$-finite, we can actually find a ... 2 Hint: given any continuous function on$[0,1]$, you can approximate it in the$L^1$norm by continuous functions that vanish at$0$. 1 Just as you suspect,$\left\|\cdot\right\|_{p,q}$is the norm$L_p \rightarrow L_q$. In this context$\left\|\cdot\right\|_{2}$is just$\left\|\cdot\right\|_{2,2}$and so on. 3 Note that for every$t\ge 0$, we have that $$t^h\le t^p+t^q.$$ To see this consider the possibilities$t\le 1$and$t>1$. Hence $$\lvert f(x)\rvert \le \lvert f(x)\rvert^{p/h}+\lvert f(x)\rvert^{q/h}=g_1(x)+g_2(x)=g(x),$$ and thus $$\|f\|_h\le \|g\|_h\le \|g_1\|_h+\|g_2\|_h=\left(\int_E g_1^h\right)^{1/h}+\left(\int_E ... 1 Hint 1 I believe$$f \in L^p(0,\infty) \Leftrightarrow \int_0^\infty |f(x)|^p dx < \infty.$$So you need to find such a p that$$ \int_0^\infty \frac{dx}{\left( x^\alpha + x^\beta \right)^p} < \infty. $$How would you proceed to do that? Can you get an idea about how to evaluate that integral? Hint 2 A possibly useful transformation is$$ ... 1 If$K$consists of finitely many atoms and a set of measure zero, then your set of functions is finite-dimensional, and compactness follows from the compactness of closed bounded subsets of$\mathbb R^n$. But then$X$is not such an interesting space to look at. If$K$contains a set$A_0$of positive measure with no atoms, then no. By a theorem of ... 1 This is a very standard and yet elegant proof. We shall create a bounded linear functional on$\ell^\infty(\mathbb N)$, which is not realized by any element of$\ell^1(\mathbb )N $, using the Hahn-Banach. Let first$c(\mathbb N)$be the set of converging sequences. Clearly$c(\mathbb N)$is a closed subspace of$\ell^\infty(\mathbb N)$. Let ... 3 If$(x_k)$is a sequence of vectors in$\ell_q$, each with$p$-norm at most one, that converges to$x$in$\ell_q$, then$(x_k)$converges to$x$coordinatewise. For each$k$and$n$,$\sum\limits_{i=1}^n |x_k(i)|^p\le 1$. Fix$n$and let$k\rightarrow\infty$to deduce that$\sum\limits_{i=1}^n |x(i)|^p\le1$for each$n\$. Now let ...

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