New answers tagged

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 $L^1$.


4

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


0

We assume that $\mu$ is a measure on $X$. Note that is the only standing assumption here; no norms or measures are assumed finite below except where explicitly stated. The question as stated is unclear. The question at the linked post is clear, but slightly silly because the hypotheses are obviously redundant. Here are the facts, or at least some of the ...


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 ...


0

For any subset: $$A\subseteq\mathcal{H}:\quad A\cap A^\perp=\{0\}$$ As well as it is: $$A\subseteq\mathcal{H}:\quad\overline{\langle A\rangle}^\perp=A^\perp=\overline{\langle A^\perp\rangle}$$ Moreover it holds: $$A\subseteq\mathcal{H}:\quad A^{\perp\perp}=\overline{\langle A\rangle}$$


0

I'm not following you here. So let $H$ be a Hilbert space and $V$ a (not necessarily closed) subspace, then $V^{\perp}=\left\{w\in H\mid \left\langle v,w\right\rangle=0 \mbox{ for all }v\in V\right\}$ is the orthogonal complement. Using the continuity of the inner-product you can show that $V^{\perp}$ is a closed subspace of $H$. It follows that $V^{\perp\...


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 \...


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 \...


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]\...


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 ...


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 ...


0

Consider a sequence of Borel subsets of the unit interval $\left(A_n\right)_{n\geqslant 1}$ such that for each Borel subset $B$ of the unit interval, $\lim_{n\to +\infty}\lambda\left(A_n\cap B\right)=\lambda(B)/2$. For example, we can choose $A_n=\bigcup_{j=0}^{2^n-1}\left[j2^{-n},(j+1)2^{-n}\right)$. Then define $f_n(x):=2\mathbf 1_{A_n}(x)-1$. We have the ...


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 ...


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)...


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*} ...


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{...


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

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 ...


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

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

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_{\...


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.


0

This is a fairly standard result, here is a proof that uses the Borel Cantelli lemma. A slightly more general result is true, and the more general result hints at a direction of proof. It is not difficult to show that if $f_n\to f$ (in $L^1$) then $f_n$ converges to $f$ in probability, that is for all $\epsilon>0$, $\lim_n \lambda \{ x | |f_n(x)-f(x)| \...


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 $L^2(\mathbb R)$.


5

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


0

Here's the issue with your boundedness proof. Your argument boils down to the following calculation: \begin{align*} \|Bf\|_{L^2}&=(\int_0^\infty\left|\frac{1}{x}\int_0^xf(t)dt\right|^2dx)^{1/2}\\ &\leq(\int_0^\infty\left(\frac{1}{x}\int_0^\infty|f(t)|\mathbf1_{[0,x]}dt\right)^2dx)^{1/2}\\ &\leq(\int_0^\infty\frac{1}{x^2}\|f\|_{L^2}^2xdx)^{1/2}\\ &...


0

The 2nd and 3rd of the facts you cite give the result. Indeed, you have proved that for any $\epsilon>0$ there is $N_1$ such that $\max_{\overline{\Omega}}|\rho_n *u-u|<\epsilon$ whenever $n>N_1$. Also, you know that there exists $N_2$ such that $\xi_n\equiv 1$ on $\overline{\Omega}$ whenever $n>N_2$. Conclusion: $\max_{\overline{\Omega}}|(\...


2

The following two theorems (see Partial Differential Equations (chapter 5) by Evans) can answer your question:


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 ...


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 $C > 0$...


2

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


0

Answer to both questions is yes. Bounded sequences in $ L^2$ have weakly converging subsequences. By uniqueness of the almost everywhere pointwise limit, these weak limits coincide with your $ f $. Furthermore, the norm of the weak limit of a subsequence is smaller than the limit inferior of the norms of the elements of the sequence. Using a bit less ...



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