# Tag Info

5

No. Nonlinear transformations and weak convergence go together like drinking and driving. For example, let $r_k$ be the $k$th 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, ...

4

Rewrite the integral as follows $$\int_0^1 |f(x)|^p|g(x)|dx = \int_0^1 \int_0^{|f(x)|} pt^{p-1}|g(x)| dt dx$$ Switching the order of integration we obtain $$\int_0^\infty \int_{|f(x)|>t} pt^{p-1}|g(x)| dx dt = \int_0^\infty pt^{p-1}\int_{|f(x)|>t} |g(x)| dx dt$$ Now note that $\int_{|f(x)|>t} |g(x)| dx \leq \min(\frac{3}{t^2},|g|_{L^1})$ Hence the ...

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 $x$, $$x^{-(1-1/p)}\int_0^x|f(t)|\mathrm dt\leqslant \lVert f-s\rVert_p+x^{-(1-1/p)}\int_0^x|s(t)|\mathrm dt.$$

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 ...

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 $\frac{1}{p-2}=\frac{n-2}{4}, ... 3 Let$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$, which implies ... 3 If$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$. 3 Suppose$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 ... 3 Hint: Holder's inequality with$f$and$g(x)=1$. 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 ... 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$f=\chi_{[0,M]}$... 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 $$x^{2m\frac{N+1}N}\lambda\{|u|\lt ... 2 For 1<p<\infty, strict convexity of the norm implies that the only shortest path is the line segment between these points. For p=1, the length of a path (x(t),y(t)) is just the sum of the lengths of one-dimensional paths x(t) and y(t). Both of those must go from 0 to 1. They will have length 1 if and only if the function is ... 2 This is the answer I've unravelled: Let z \in (\ell^p)^{\ast\ast}. I want to prove that exists x \in \ell^p such that \langle z,f\rangle=\langle f,x \rangle for every f \in (\ell^p)^\ast. I know that there are the isomorphisms: j_p: \ell^q \rightarrow (\ell^p)^\ast and j_q: \ell^p \rightarrow (\ell^q)^\ast Now, fix z \in ... 2 The sum must extend over all integers, by the way, otherwise, it would be finite (namely 0) for a constant function whose value is less than 1 in modulus. Then, with B_n = \{x : 2^n < \lvert f(x)\rvert \leqslant 2^{n+1}\}, we have$$A_n = \{ x : 2^n < \lvert f(x)\rvert\} = \bigcup_{k=n}^\infty B_k,and the union is disjoint. Thus ... 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 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 ... 2 To prove the first inequality it suffices to prove that (x+y)^{p}\le x^{p}+y^{p} for x,y\ge0 and p\in(0,1). Notice that we can use this as follows: \vert x\rvert^{p}\le\lvert x-y\rvert^{p}+\lvert y\rvert^{p} then reversing the roles of x and y. Notice that this Lemma can be proven by reducing to the case when y=1 then looking at ... 2 For the first problem, use that any L^2-convergent sequence has a subsequence that converges almost everywhere. Alternatively, you can use that any finite-dimensional subspace of any normed vector space is closed. This essentially follows (in your case), because\Bbb{R} \rightarrow \Bbb{R}_+, x \mapsto \Vert x \cdot \chi_{[0,1]} \Vert_{L^2}$$gives a ... 2 One can forget p(s), r(s) and the rest and simply try to show that, for every a\lt b,$$\|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 ... 1 The inclusion $$(C([0,1]), || \cdot ||_{\infty}) \longrightarrow (L^2([0,1]),|| \cdot ||_2)$$ is continuous since for allf \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 ... 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 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 ... 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 ... 1 For the constants: prove that the subspace is complete. In an Hilbert space, this is equivalent to being closed, and in this case it is easier. For the null integrals subspace: show that$\int f_n\to \int f$. 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. 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 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

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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