# Tag Info

0

The $1$ in parentheses is useless; it doesn't scale if you multiply $u$ by a constant. So your inequality can hold for all $u\in H^1$ if and only if $$\|u\|_{2^*}\leq \epsilon (\|\nabla u\|_2+\|u\|_p^{\frac{p}{2}})+C_\epsilon \|u\|_2$$ holds. And here the scaling of the variable $u_t = u^{n/2^*}u(tx)$ becomes an issue: namely, $\|u_t\|_{2^*}$ and $\|\nabla ... 0 You have that if$a \in l_q\big(\Bbb N\big)$then$a \in l_q\big(\Bbb Z\big)$since a sequence of natural numbers is a sequence of integers. You also have if$a$and$b$are two sequences of natural numbers then so is$ a*b$, since the convolution only involves multiplications and additions. So the equality you wrote holds for sequences of natural numbers ... 2 I can't bring myself to write$fg$for the convolution of$f$and$g$. So I'm going to write$f\mapsto f'$for the involution, so I can write$f*g$for the convolution. Are you certain you got the definition of$f'$straight? What would make much more sense to me would be $$f'(t)=\overline{f(-t).}$$ That seems to me is the "standard" involution on$L^1$. ... 2 For any sequence$f_n \to f''$in$L^p$, there exists a subsequence such that$f_{n(k)} \to f''$almost everywhere as$k \to \infty$. There are (at least) two possibilites to prove this statement: In the proof of the Riesz-Fischer theorem (which states that$L^p$is a complete space), one usually constructs such a sequence (see e.g. René Schilling: ... 0 Note that if$\mu(X)=\infty$, then your series condition doesn't necessarily imply$f\in L_{\mu}^{p}(X)$. For example, take$X=\mathbb{R}$with Borel$\sigma$-algebra and Lebesgue measure. If$f=1/2\chi_{[0,\infty)}$, then clearly$f\notin L^{p}$; however,$\lambda_{f}(n)=0$for$n\geq 1$. Assume that$\lambda_{f}(0)<\infty$. Observe that for$p>0$, ... 1 It holds true that$q>p\implies \ell^p\subseteq \ell^q$Indeed, let$(a_n)_{n\in\mathbb N}$such that$\sum_n |a_n|^p<+\infty$Since$q>p$, as soon as$|a_n|<1$, which holds from a certain$N$onwards,$|a_n|^q<|a_n|^p$. So$\sum_{n\in\mathbb ...

1

Fix $y \in [0,\infty)$. Let $f_N(x,y) := \sum_{k=1}^N e^{-y\sqrt{\lambda_k}} (u,\varphi_k) \varphi_k(x)$ be the partial sum, and let $g_N := f-f_N$ be the tail of the series. Consider the integral$\int_{\Omega } g_N^2$: writing $g_N^2$ as a product of two sums and multiplying out the terms, then only the diagonal terms survive after integrating because the ...

4

No, this is not correct. Just consider $((0,1],\mathcal{B}((0,1]),\text{Leb})$ and $$X_n(\omega) := n^2 \cdot 1_{(0,1/n)}(\omega), \qquad \omega \in (0,1].$$ Then $X_n \to 0$ almost surely (hence, in particular, $X_n \to 0$ in probability), but $\mathbb{E}X_n = n$ is not bounded. Another counterexample: $$X_n(\omega) := \frac{1}{\omega} ... 0 Split into small x and large x: |x|\le 2. Nothing to discuss here: the function is continuous. |x|>2. Then u(x)\le \dfrac{1}{|x| \ln^p |x|}. By symmetry, it's enough to consider positive x. Integrate using substitution y=1/x. This is one of standard improper integrals that just barely converge (thanks to p>1). 1 Hint: Write u=(u_1,\ldots,u_n) and v=(v_1,\ldots,v_n). By Young's inequality,$$|u_i \cdot v_i| \leq \frac{|u_i|^p}{p} + \frac{|v_i|^q}{q}. \tag{1}$$Rewrite this inequality using the definition of u and v. Sum (1) over i=1,\ldots,n. Deduce that$$\sum_{i=1}^n |u_i v_i| \leq 1.$$Conclude. 1 This is more of a problem relating to weak convergence in L^{2}. Since f_{n}\to f weakly in H^{1}\left(\Omega\right), we know g_{n}:=\nabla f_{n}\to\nabla f=:g weakly in L^{2}\left(\Omega\right). Now, we require the following fact: If x_{n}\rightharpoonup x weakly and y_{n}\to y in norm, then \left(x_{n}\right)_{n} is bounded (lets say by ... 1 I think this does it: if a sequence converges in L^1, then a subsequence converges a.s. to the same limit. That subsequence converges in probability by Egorov's theorem. So the L^1 limit of an L^1-convergent sequence must be in an L^0-closed set. The only catch here is that this assumes that you are dealing with a subset of L^1 which is closed in ... 0 HINT: Once you have a continuous g that is a good approximation to the generic f\in L^2(-1, 1), take a \delta>0 and do a linear fit between the points (-1, \alpha) and (-1+\delta, g(-1+\delta)). That is, consider the function$$ g_\delta(x)=\begin{cases} \text{linear}, & x\in[-1, -1+\delta) \\ g(x), & x\in [-1+\delta, 1] \end{cases} $$... 1 Generally, if T:X\to Y is a surjective linear map, then X/\ker T is isomorphic to Y. Applied to your case, X=L^p, Y=\mathbb{R}, T=\int f, this yields L^p/L^p_0  being isomorphic to \mathbb{R}. (Or \mathbb{C} if you use complex scalars) Bonus content: If M is a closed subspace of Banach space X, the inclusion i:M\to X has adjoint ... 0 Writing l(x) for |\log|x||. Hints: First, all that matters is what happens for, say, |x| > 10 and what happens for |x|<1/10. For |x|>10, 1+|x|^\alpha \sim |x|^\alpha and 1+l(x)^\beta\sim l(x)^\beta. After that "substitution", write the integral of the p-th power in polar corrdinates; you get \int_{10}^\infty t^\gamma ... 2 Let me give you a hint: First, let us define$$v_\epsilon:= \epsilon^{-\frac{n}{p^*}}u\left(\frac{x}{\epsilon}\right) $$Can you compute \|v_\epsilon\|_{L^p} and \|\nabla v_\epsilon \|_{L^p} in term of u? Try to write it down explicitly. Then you will know why it is bounded in W^{1,p} Secondly, the fact that v_\epsilon has no convergent ... 1 Well first there's a typo in the question; you mean to ask how to prove ||f*g||_p \le||g||_p, right? And there's some funny business in the question itself; smoothness and compact support are irrelevant. (And they don't make it any easier as far as I can see.) All that matters is f\ge 0 and \int f=1. That means if you define a measure \mu by ... 1 The first question is, in what sense do you want to understand convergence of the sum? The correct notion here is unconditional convergence, i.e. we show that there is some h \in H such that for every \epsilon > 0, there is some finite subset J_\epsilon \subset I with \Vert h - \sum_{j\in J} a_j x_j \Vert < \epsilon for all finite sets ... 2 I am not sure about how \tau is a translation. But the existence of the required functional \Psi can be proved in a more general setting. It will cover your case if we assume that \{nq-1\}_{n=1}^\infty is a strictly increasing sequence of positive integers. This is my proof: Consider any strictly increasing function ... -1 Looks like a job for the Schauder fixed-point theorem. EDIT: I'm assuming q is an integer > 1, so n \in \mathbb N \implies qn - 1 \in \mathbb N. The unit ball B of (\ell^\infty)^* with the weak-* topology is compact. The intersection K of the subsets \{\psi \in B: \liminf x \le \psi(x) \le \limsup x \} is again compact and convex (and ... 0 It seems the following. I will consider that s\in\Bbb R, not in \Bbb R^n, because I don’t understand what is sx\in\Bbb R^n The correctness of the definition of the operator imposes restrictions on the set U: U+a\subset U or sU\subset U. Or we may for each function f\in L^p(U) consider its extension in L^p(\Bbb R^n) such that f|(\Bbb ... 2 Use that \chi_{(a,b]} = \chi_{(0,b]} - \chi_{(0,a]} for a<b. This will show that your first assumption yields$$ \int f_n \cdot \chi_{(a,b]} \to \int f \cdot \chi_{(a,b]} $$for all a<b. By linearity, we get the same convergence for arbitrary Riemann step functions, i.e. for step functions with respect to intervals. Now, one possibility is to ... 2 Let S\equiv\sup_{x\in\mathbb R^n}|f(x)|. By definition, the essential supremum norm is defined as follows:$$\|f\|_{\infty}=\inf_{c\geq 0}\big\{\lambda(\{x\in\mathbb R^n\,|\,|f(x)|>c\})=0\big\}.$$In words, \|f\|_{\infty} is the infimum of such non-negative numbers above which the function |f| takes values only on a measure-zero set. Intuitively, ... 1 Since the Lebesgue measure is \sigma-finite we have that$$\| {f}\|_{\infty}=\inf\left\{M\geq 0:\,\lambda(\left\{x: |f(x)| > M\right\})=0 \right\}$$Clearly \lambda(\left\{x: |f(x)| > \sup_{x \in \mathbb{R}^{d}}|f(x)|\right\})=0 Now suppose there is a constant M < \sup_{x \in \mathbb{R}^{d}}|f(x)|, that satisfies the above critera. It should ... 0 For c), apply Fatou's lemma to 2^{p-1}(|f|^p + |f_k|^p) - |f - f_k|^p, giving$$\liminf_{k \to \infty} \int_X 2^{p-1}(|f|^p + |f_k|^p - |f - f_k|^p)\, d\mu\ge 2^p \int_X |f|^p\, d\mu.$$Thus$$2^{p-1}(\|f\|_p^p + \liminf_{k\to \infty} \|f_k\|_p^p) - \limsup_{k\to \infty} \|f - f_k\|_p^p \ge 2^p\|f\|_p^p.$$Since \|f_k\|_p^p \to \|f\|_p^p, we have ... 2 You haven't got the correct negation of the limit hypothesis. Remember that the opposite of f(x) \to b as x \to \infty if for every \varepsilon>0 there is an n such that \lvert f(x)-b \rvert < \varepsilon for all x>n is f(x) \not\to b as x \to a if there is an \varepsilon>0 such that for every n there is an x>n ... 3 Use Holder to see$$(1)\,\,\,\,|\int_x^{x+1}f\,| \le \int_x^{x+1}|f| = \int_x^{x+1}|f|\cdot 1 \le (\int_x^{x+1}|f|^p)^{1/p}\cdot (\int_x^{x+1}1)^{1/q}=(\int_x^{x+1}|f|^p)^{1/p}.$$Because f\in L^p, \int_x^{x+1}|f|^p \to 0 by the dominated convergence theorem. That shows (1) \to 0 as desired. 4 Thanks to zhw for catching an oversight. Write \mathbb{R} = \bigcup_n [n,n+1) and then$$\int_\mathbb{R} |f|^p = \sum_n \int_{[n,n+1)} |f|^p,$$and hence$$\int_{[n,n+1)} |f|^p \stackrel{n\to\infty}{\longrightarrow} 0.$$If p=1, we are finished, otherwise suppose p>1. Since$$\left(\int_{[a,b)} |f|\right)^p = \left(\int_{[a,b)} |f|1\right)^p \le ...

1

The idea to use $f_k=k^\alpha \chi_{(0,k^{-1})}$ is good. The sequence converges to $0$ a.e., so if there is any $L^p$ limit at all, it has to be zero (recall that $L^p$ convergence implies a.e. convergence for a subsequence). Therefore, if the sequence of norms stays constant, the sequence does not have a limit in $L^p$. Choose $\alpha$ so that these ...

1

This was getting too long for a comment. Analysis is not my thing, but I think you've reduced the problem to showing that for a continuous function $f\in \mathcal{C}^0(\left[0,1\right])$ and $\epsilon > 0$, you can find $g\in E_\alpha$ such that $\left|f-g\right|_2<\epsilon$. So, given a continuous $f$, let $g$ be equal to $f$ except on a super tiny ...

2

This follows from the triangle inequality of the$\ell^2$ norm, since $x,y \in \ell^2$ then $\|x\|_2 <\infty$ and $\|y\|_2 <\infty$ thus $$\sum_{n=0}^\infty |x_n + y_n|^2 =\| x+ y\|_2^2 \leq (\|x\|_2+\|y\|_2)^2 < \infty$$ Hence, indeed $x+y \in \ell^2$

3

Absolutely cloddish inequality: If $a,b \ge 0,$ then $(a+b)^2 \le 4a^2 + 4 b^2.$ Proof: If $a\le b,$ then the left side is $\le (2b)^2 = 4b^2,$ same idea of course if $a\ge b.$ So $$\sum (x_n+y_n)^2 \le \sum (|x_n|+|y_n|)^2 \le \sum (4|x_n|^2 + 4|y_n|^2)$$ and that does it.

1

Hint: multiply out the terms in the sum for $||x + y||$, and use Cauchy-Schwarz to find a bound for $\sum x_iy_i$ in terms of $||x||$ and $||y||$.

2

$(\Rightarrow)$ By the reverse triangle inequality $$| \|g_n\| -\|g\| | \leq \| g_n -g \|$$ Since $\| g_n -g \| \to 0$ then clearly $\|g_n\| \to \|g\|$ The converse is also true as long $g, g_n \in L ^1$ and $g_n \to g$ a.e. In that case you can use the next argument: $(\Leftarrow)$ First note that since $|g_n -g| \leq |g_n| + |g|$ then $|g_n -g| ... 0 It looks like your reasoning is fine, except that you may want to add some more detail as to how $$\left|\, \int_{X} g_n \, dm - \int_{X}g\, dm \,\right|$$ is related to the (absolute value of) the difference of the norms. Specifically, I’d write something like $$\left|\, ||g_n||_{1} - ||g||_1 \,\right| = \left|\, \int_{X} |g_n| \, dm - \int_{X} |g| \, dm ... 0 In the case p = 2, we first compute T^*. Noting that$$ \langle Tx, y\rangle = \sum_{k} \sum_{j = 1}^k x_j \bar{y}_k / k = \sum_{j} \sum_{k = j}^\infty x_j \bar{y}_k / k $$we have that$$ (T^*y)_j = \sum_{k = j}^\infty \frac{y_k}{k} $$So we have that$$ (T T^* y)_\ell = \sum_{j = 1}^\ell \sum_{k = j}^\infty \frac{y_k}{k \ell} $$and$$ (T^* T ... 5 The dual space of$L^\infty$is the space of "finitely addditive" signed measures and not the space of finite signed measures. The usual Radon-Nikodym theorem does not apply. 0 The$\xi$-integral can be computed as $$\int e^{i(x-y)\xi}e^{-|\xi|^2/n}\mathrm{d}\xi = e^{-n|x-y|^2}\int e^{-(\xi/\sqrt{n}-i\sqrt{n}(x-y))^2}\mathrm{d}\xi = (\pi n)^{3/2} e^{-n|x-y|^2} ,$$ and hence $$I_n(x)=\int\chi_{B(n)}(y)f(y)(\pi n)^{3/2} e^{-n|x-y|^2}\mathrm{d}y = \int f_n(y)g_n(x-y)\mathrm{d}y ,$$ where $$f_n(y) = \chi_{B(n)}(y)f(y) ,$$ and ... -1 Let$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \sin (2\pi i \,n x) + b_n \cos (2\pi i \,n x) \in L^2[0,1]$then we have Bessel inequality: $$\frac{a_0}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2) < \bigg|\bigg|\int_0^1 f(x)^2 \, dx\bigg|\bigg|^2 < \infty$$ Then the Fourier coefficients tend to zero$|a_n| =|\langle f ,\sin (2\pi i \,n x)\rangle | \to ...

0

Suppose that $\mu(B_{n})=0$. That would imply that the $a$ in the definition of $\Vert f\Vert_{\infty}$ satisfies $a\leq \Vert f\Vert_{\infty}-\frac{1}{n}$. Since we assumed that $\mu(\{|f(x)|>\Vert f\Vert_{\infty}-1/n\})=0$. But this is non-sense since it implies \begin{align*} \Vert f \Vert_{\infty} &\leq \Vert f\Vert_{\infty}-\frac{1}{n} \\ ...

1

If both $f$ and $g$ are continuous with compact support, then so is their convolution. And continuous functions with compact support are dense in $L^p$ for $1<p<\infty$. So, you need to do the following: Given $f\in L^p$ and $g\in L^q$, approximate them with $C_c$ functions $f_n$ and $g_n$ in the respective norm. Show that $f_n*g_n$ is a Cauchy ...

0

Indeed this is just Holder's inequality: Pick $V \subset U$ with $|V|<\infty$, then $$||u||_{L^1(V)}=\int_V |u| dx\le \sqrt{\int_V 1^2 dx}\sqrt{\int_V |u|^2 dx} = \sqrt{|V|}\ ||u||_{L^2(U)}$$ then $u \in L^1_{loc}(U)$.

1

Try $X = \mathbb R$ with the usual topology, and counting measure on the rationals. The only continuous function in $L^1$ is $0$.

2

For the case $1\leq p_1 < p_2 < \infty$, let $p=p_2/p_1 >1$. Then take any $f \in L^{p_2}(E)$, this means $$\int_{E} |f|^{p_2} dm < \infty$$ Now, using that $p=p_2/p_1$ we observe that $$\int_{E} |f^{p_1}|^{p} dm = \int_{E} |f|^{ p_1 \cdot p} dm =\int_{E} |f|^{p_2} dm < \infty$$ Thus indeed $f^{p_1} \in L^p(E)$. Now for the rest, note ...

1

Given $f \in L^{p_2}$, $f^{p_2} \in L^1$. So since $p_2 = p_1p$, $(f^{p_1})^p\in L^1$, i.e., $f^{p_1}\in L^p$.

2

Let $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. If $X$ is a set, $\mathcal{S}$ a sigma algebra on $X$ and $\mu : \mathcal{S} \to \mathbb{R}^+$ a measure, then, by definition $$L^p(X, \mathcal{S}, \mu) = \left\{ f:X\to \mathbb{F} : f \text{ is \mathcal{S} measurable and } \int_X |f|^p d\mu < \infty \right\}$$ In that case, if $f \in L^p(X, \mathcal{S}, ... 0 Actually,$f$is not necessarily bounded.$f(x)=\frac{\chi_{(0,1]}(x)}{\sqrt{x}}$is an unbounded (Lebesgue) integrable function (its integral is$2$). Hint: try to argue that$\int_{-\infty}^n f(x) dx$converges to$\int_{-\infty}^\infty f(x) dx$by using the dominated convergence theorem. Then notice that$\int_{-\infty}^\infty f(x) dx - \int_{-\infty}^n ...

0

Well the part $\int_{-\infty}^nf(x)dx$ must converge to $\int_{-\infty}^{\infty}f(x)dx$ so the part $\int_n^{-\infty}f(x)dx$ must go to zero. Just write the whole integral as a sum of those two parts and take the limit.

1

This is true provided that $C_b \cap L^1$ is dense in $L^1$, which holds if $m$ is a finite measure, or if $X$ is locally compact and $m$ is Radon (I imagine many other conditions are possible). In this case, we can use a standard triangle inequality trick. Given any $h \in L^1$ and any $\epsilon > 0$ we can find $g \in C_b \cap L_1$ such that ...

0

Even in the finite measure case this seems false. Take $f_n(x)=\mathrm{sign}(2^n x)$ in $L^\infty(0,2\pi)$. Then for all $n \neq m$, $\mathrm{Leb}\{ x , |f_n(x)-f_m(x)|>1\}=\pi$.

Top 50 recent answers are included