# Tag Info

5

Since $f(x) = \dfrac 1x$ does not belong to $L^1((0,1])$, your example shows that the statement is false.

4

Triangle inequality $$|\, ||f_n ||_2 -|| f ||_2\, |\leq ||f_n-f||_2$$

4

The vanishing integral over the small ball is enough to get a Poincaré-type estimate. Let $B = B(0,1)$ and $\Omega = B(0,r)$. We define $$\|u\|_\star := \big|\int_B u \, \mathrm dx\big| + \|\nabla u\|_{L^p(\Omega)}.$$ It is clear that $\|\cdot\|_\star$ is a norm on $W^{1,p}(\Omega)$ and that $\|u\|_\star \le C \, \|u\|_{W^{1,p}(\Omega)}$ for some $C > 0$...

4

First, note that we have$$|\mathcal{F}f(\xi)| = \left|\int_{\mathbb{R}^d} f(x)e^{-ix \cdot \xi}dx\right| \le \int_{\mathbb{R}^d} |f(x)|\,dx = 1 = \mathcal{F}f(0),$$where the final equality comes from the fact that $f \ge 0$. Since $f$ is real-valued, we may decompose $\mathcal{F}f(\xi)$ into its real and imaginary parts as$$\mathcal{F}f(\xi) = \int_{\mathbb{... 3 L^2 spaces should not be sensitive to the topology or shape of whatever underlying space you're working with. Indeed, given a "manifold" (a generalization of circles and surfaces), one way of defining an L^2 space on it is to pick a chart D^n \to M, where D is the unit disc, that is injective except at a set of measure zero, and then pull back ... 3 Arguing by contradiction, suppose such a measure, say \nu, does exist. The main idea is that we can place continuum many disjoint balls, or I would rather say cubes, inside a ball. Hence, there are infinitely many of them that have measure greater than some constant \epsilon > 0. Let us denote by B_r(x) a ball in \ell_\infty with center x = (x_i)... 3 analysis would benefit from some Banach space X^{p,q} that lets you measure the function locally in p and globally in q. Such spaces do exist, in several flavors. Interpolation spaces The spaces L^p\cap L^q and L^p+L^q sometimes appear in PDE and frequently in interpolation theory. In L^p\cap L^q, the smaller exponent controls the global ... 3 f=g a.e. as L^1 convergence implies almost everywhere convergence to f for a subsequence (you can find this as part of the proof of completeness of L^1 in any textbook). Since you know f_n already converges pointwise to g, you must have that the same subsequence as above converges to g, and hence f=g almost everywhere. 3 Let E_\infty = \{x : f(x) = 0\}. Then clearly \int_{E_\infty} |f| = 0. Since f is a measurable almost everywhere finite function, we have$$\mathbb{R}^d = \left(\bigcup_{n \in \mathbb{Z}} E_n\right) \cup E_\infty \cup F,$$with m(F) = 0. By the monotonicity property of Lebesgue integration, we have$$\int_{\mathbb{R}^d} |f| = \sum_{n \in \mathbb{Z}} \...

2

Let $r > 0$. Then $$\int_{0}^{\infty}e^{-rx}e^{-isx}dx=\frac{1}{r+is}.$$ The function $f(x)=e^{-rx}\chi_{[0,\infty)}(x)$ is in $L^1$, but $\hat{f}(s)=\frac{1}{\sqrt{2\pi}(r+is)}$ is not in $L^1$.

2

A less functional-analytic argument is through Fatou's lemma (applied to $|f_n|^2$), since a subsequence converges pointwise a.e.

2

This is just two changes of variables (assuming your definition of $\mathbb N$ starts at $1$). \begin{align*} \int_0^{2\pi}f(nx) \, dx=\int_0^{2\pi n}f(x)\frac{dx}{n}=\frac{1}{n} \sum_{k=1}^n \int_{2\pi(k-1)}^{2\pi k} f(x) \, dx=\frac{1}{n} \sum_{k=1}^n \int_0^{2\pi} f(x+2\pi(k-1)) \, dx\\ =\frac{1}{n} \sum_{k=1}^n \int_0^{2\pi}f(x) \, dx=\frac1n\cdot n\...

2

The typical proof of this uses $$L^p \text{ convergence } \,\, \implies \text{ convergence in measure } \,\, \implies \text{ subseq. converges pointwise a.e.}$$ For the latter, if $f_n \to f$ in measure, then for each $k \in \mathbb N$ we can find $n_k \in \mathbb N$ so that $$\mu \left( \left\{ \lvert f_{n} - f \rvert > \frac{1}{k} \right\} \right) \le \... 2 Of course as Hilbert spaces L^2(S^1) and L^2\bigl([0,1]\bigr) are isomorphic, and you could also say that L^2\bigl([0,1]\bigr) is the prime example of a Hilbert space arising from Lebesgue theory. But note that L^2(S^1) is one of the most important Hilbert spaces in the world, and there definitively is an essential difference between L^2\bigl([0,1]\... 2 Hint: C_c(\Bbb R) (the space of continuous functions of compact support) is \| \cdot \|_1-dense in L^1(\Bbb R). Try to prove it for functions in this space. Given that, proceed as follows: Let f \in L^1(\Bbb R) and \eta > 0. Then, there is \varphi_{\eta} \in C_c(\Bbb R), such that \| f - \varphi_{\eta} \|_1 < \eta /3. Denote by \tau_{\... 2 Consider the sequence \mathbf x^{(n)}\in\ell^1\subset\ell^2 defined by$$ x^{(n)}_k=\frac{1}{k}\textrm{ for } k\leq n,\quad x^{(n)}_k=0\textrm{ for } k > n. $$Then \mathbf x^{(n)}\overset{d_2}{\to} \mathbf x\in\ell^2\setminus\ell^1, with$$ x^{(n)}_k=\frac{1}{k}\quad \forall k. $$So \ell^1 is not d_2-complete. 2 You are correct. Polynomials are not square-summable over the real line (or a half-axis), and thus are not (elements of the equivalence classes which are) members of L^2(\mathbb R). 2 Suppose f\ge0 to save typing. Let q be the conjugate exponent. Now$$\begin{aligned}\int_0^1\int_n^{n+n^{-\alpha}}f(x+y)\,dydx &=\int_n^{n+n^{-\alpha}}\int_0^{1}f(x+y)\,dxdy \\&=\int_n^{n+n^{-\alpha}}\int_y^{y+1}f(x)\,dxdy \\&\le \int_n^{n+n^{-\alpha}}\int_n^{n+2}f(x)\,dxdy \\&= n^{-\alpha}\int_n^{n+2}f(x)\,dx\end{aligned}.So if F(x)... 2 The following two theorems (see Partial Differential Equations (chapter 5) by Evans) can answer your question: 2 I am not sure about the inequality you mention, but since |g_k| \le g seems to be sufficient for your purposes, you can get this by \begin{align*} |g_k| &= |g_1+(g_2-g_1)+\dots+(g_k-g_{k-1})| \\ &\le |g_1|+|g_2-g_1|+\dots|+|g_k-g_{k-1}|\\ &=|g_1| + \sum_{j=1}^{k-1} |g_{j+1}-g_j| \\ &\le |g_1| + \sum_{j=1}^\infty |g_{j+1}-g_j|=g \end{align*} ... 2 You should define all your terms. I presume \cal F is the Fourier transform. The standard formula is that\cal F f(x) = \int_{\mathbb R} e^{-2\pi i x y} f(y) \, dy.$$Since 2\pi i x y is purely imaginary or zero, |e^{-2\pi i xy}| = 1. Apply the triangle inequality to get$$|\cal F f(x)| \le \int_{\mathbb R} |e^{-2\pi i x y} f(y)| \, dy = \int_{\...

1

I assume that $H$ is a positive measure. In this case, the function $x\mapsto 1/x$ will be integrable on $(0,1)$ for the measure $H$ as long as the series $\sum_{n=1}^{+\infty}2^n\cdot H\left(\left[2^{-(n+1)},2^{-n}\right)\right)$ converges. Indeed, we have the pointwise inequalities $$2^{n}\mathbf 1\left(\left[2^{-(n+1)},2^{-n}\right)\right)(x)\leqslant \... 1 Why not L^1: In studying elliptic equation, it is most convenient to consider L^p space for 1<p<\infty. The reason is that one does not have nice L^p-estimates$$\|u\|_{W^{2,p}(\Omega)}\le C (\|f\|_{L^p(\Omega)} + \|u\|_{L^p(\Omega)}$$for p=1 (Here we assume that p(x) is nice). Note that the above estimates is crucial in establishing ... 1 I figured it out while editing the image in. f_n\in L^\infty, and since the measure is finite this implies f_n\in L^p for all p\in[1,\infty]. By Fatou's lemma (poor Fatou, always forgotten), we have:$$\int|f|^p=\int\liminf|f_{n_k}|^p\leq\liminf\int|f_{n_k}|^p\leq\liminf\|f_n\|_\infty\leq1,$$hence f\in L^p for all p. But since the measure is ... 1 Since \{f_n\} converges, \{f_n\} is Cauchy. In probability theory, we have to deal with different types of convergence, and therefore - in order to avoid any confusion - it is always good to mention which kind of convergence you are talking about, e.g. "Since f_n converges in L^p(\mathbb{R}^n), \{f_n\} is an L^p-Cauchy sequence." Define a ... 1 Notice that you may expand h as h = \sum \limits _{n \in \Bbb Z} a_n \varphi_n, so that$$h (t-s) = \sum \limits _{n \in \Bbb Z} a_n \varphi_n (t-s) = \sum \limits _{n \in \Bbb Z} a_n \frac 1 {\sqrt {2 \pi}} \Bbb e ^{\textrm i n (t-s)} = \sum \limits _{n \in \Bbb Z} \sqrt {2 \pi} a_n \varphi_n (t) \overline {\varphi_n (s)} . Inputting this in the ...

1

Here is a proof inspired by @Bananach's answer: (I wanted to find a proof that didn't utilise pointwise limits to show equality.) The set $\bar{B}(0,C_0) \subset L^2[0,1]$ is weakly compact (by Banach Alaoglu), hence there is some $\tilde{f} \in \bar{B}(0,C_0)$ such that $f_{k_k}\overset{\text{weak}}{\to} \tilde{f}$ for some subsequence. To finish, we just ...

Only top voted, non community-wiki answers of a minimum length are eligible