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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 ...



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