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6

Since your question is not very precisely formulated, I will interpret it as follows: Is there a (purely) measure theoretic proof of $\int_1^\infty 1/x \,dx =\infty$, which in particular avoids the use of antiderivatives. Indeed, there is. Assume towards a contradiction that $\int_1^\infty 1/x\,dx <\infty$. Define $$f_n = \frac{1}{n} 1_{(1,n)}.$$ I ...

4

$$\int_E \lvert f \rvert^p \cdot 1 \leqslant \left(\sup_E \lvert f \rvert^p \right) \int_E 1 = \lVert f \rVert_{\infty}^p m(E),$$ since $\lVert \lvert f \rvert^p \rVert_{\infty} = \lVert f \rVert_{\infty}^p$. Therefore, taking $p$th roots, $$\lVert f \rVert_{p} \leqslant m(E)^{1/p}\lVert f \rVert_{\infty}$$ It has to be this because the equality case ...

4

The fundamental result you need is that if $f \in L^1[a,b]$, then the measure $\mu A = \int_A |f(x)| dx$ is absolutely continuous with respect to the Lebesgue measure. This means that for all $\epsilon>0$ there is some $\delta >0$ such that if $m A < \delta$ then $\mu A < \epsilon$.

3

For any sequence $y\in c_0$ we have $$\|x-y\|_\infty = \sup_n|x_n-y_n|\ge \limsup_{n\to\infty} |x_n-y_n|=\limsup_{n\to\infty} |x_n|$$ This gives a lower bound on the distance. To get the matching upper bound, let $y_n=x_n$ when $n\le N$, and $y_n=0$ otherwise. This is an element of $c_0$, and $$\|x-y\|_\infty = \sup_{n>N}|x_n| ... 3 You are dealing with the convolution of f with$$ \phi_n(x) = \frac{1}{n}\chi_{[-n,0]}(x). $$That is,$$ (f\star\phi_n)(x)=\int_{-\infty}^{\infty}f(t)\phi_n(x-t)dt=\frac{1}{n}\int_{x}^{x+n}f(t)dt = f_n(x) $$Therefore \|f_n\|_1 \le \|f\|_1\|\phi_n\|_1 = \|f\|_1. If g \in \mathcal{C}_c^{\infty}(\mathbb{R}) (i.e., ... 3 You could use the equivalent definition of the Schwartz space:$$f\in\mathcal{S}(\mathbb{R})\iff \sup_{x\in\mathbb{R}}|x^iD^jf(x)|<\infty\iff\sup_{x\in\mathbb{R}}|(1+|x|)^nD^kf(x)|<\infty$$For i,j,k\in\mathbb{N}_0, n\geq k. See here or Folland's Real Analysis. From this it follows readily. Alternatively, ... 3 A precise way to present this problem would be as follows: The linear functional f\mapsto f(0), defined on the linear subspace L^2(\mathbb{R})\cap C(\mathbb{R}), is not continuous in the L^2 norm. (Hence, it cannot be extended to a continuous linear functional on L^2, no matter how we try to make sense of f(0) there.) The proof of the ... 3 Let \lambda denote the Lebesgue measure. If \lambda(A) = \infty, then both a_f and a_g are zero. Indeed, for a>0,$$ \lambda(\{x\in A: |f(x)|> a/2\})\le 2 a^{-1}\int_{A} |f(x)|\, dx<\infty, $$therefore,$$ \int_{A} |f(x)-a|\, dx\ge \frac{a}2\lambda(\{x\in A: |f(x)|\le a/2\}) = \infty. $$(Indeed, |z-a|\ge \frac{a}2\mathbf{1}_{|z|\le ... 3 Lemma. Let (X,M,\mu) be a \sigma-finite measurable space and p\in[1,+\infty). Then for measurable non-negative function f and measurable set A\in M we have$$ \int\limits_A f(x)^p d\mu(x)=p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt $$where F_{f,A}(t)=\mu(\{x\in A:f(x)>t\}) Proof #1. Using Fubini theorem for positive functions we ... 3 A direct proof, from which a proof by contradiction can be derive if necessary. Let A_n=\{x \in \mathbb R \ : \ \vert f(x) \vert^p \ge n\} for n integer and b = \int \vert f(x) \vert^p dx. We have$$0 \le n \mu(A_n) \le b \tag{1}$$where \mu is the measure. Hence \mu(A_n) \le \frac{b}{n} for all n \in \mathbb N.$$C=\{x \in \mathbb R \ : \ ...

3

The answer is no. Since $\ell_1$ and $\ell_p$ ($p\in (1,\infty)$) are totally incomparable (every operator from one space to the other is strictly singular), by a result of Edelstein and Wojtaszczyk (Theorem 3.5 here) infinite-dimensional, complemented suspaces of $\ell_1\oplus \ell_p$ are the obvious ones, that is, they are isomorphic either to $\ell_1, ... 2 I bet that this is the vector-valued$L^p(Y)$-space, so$L^p_n(Y) = (L^p(Y))^n$. And the divergence is meant in a weak sense: $$\int_Y q^\top \, \nabla\varphi \, \mathrm{d}x = 0$$ for all$\varphi \in C_0^\infty(\hat Y)$, where$\hat Y$is the interior of$Y$. For functions$q \in H^1(Y)^n$, this is equivalent (integration by parts) to$\mathrm{div} q = 0$... 2 By Sobolev embeeding, from$\sqrt\rho\in H^1(\mathbb R^3)$it follows that $$\sqrt \rho \in L^6(\mathbb R^3),$$ which implies $$\rho \in L^3(\mathbb R^3).$$ 2 The Fourier transform is a linear map, you only have to check that $$\forall f\in L^1(\mathbb R), \mathcal F(f)=0 \Rightarrow f=0.$$ Let$f\in L^1(\mathbb R)$such that$\mathcal F (f)=0$. Hence$\mathcal F(f)$is$L^1(\mathbb R)$since its the zero function and therefore its Fourier inverse exists. So $$f=\mathcal F^{-1}(\mathcal F(f))=\mathcal ... 2 If \mu(X)=\infty, the result is not true in general. Let X=[1,\infty) with \mu Lebesgue measure and let f_n be the characteristic function of the interval [n,2\,n]. Then f_n\in L^p, \|f_n\|_\infty=1 for all n and f_n(x) converges point wise to f(x)=0 for all x\in X. Let g(x)=1/x. Then g\in L^q for q>1. We have$$\int_X ... 2 Use (Riesz-Thorin) interpolation. Show that$\|K\|_{L^1(\Bbb R^d) \to L^1(\Bbb R^d)} \le c$using$\sup_y \int |k(x,y)|\, dx \le c$, and show that$\|K\|_{L^\infty(\Bbb R^d)\to L^\infty(\Bbb R^d)} \le c$using$\sup_x \int |k(x,y)|\, dy \le c$. These imply$\|K\|_{L^p(\Bbb R^d) \to L^p(\Bbb R^d)} \le c$for all$1 < p < \infty$by interpolation. This ... 2 Later hint,$p=2$: Let$\epsilon >0.$For$0<x<\epsilon$we have $$|f(\epsilon)-f(x)| \le \int_x^\epsilon |f'(t)|\, dt \le (\int_x^\epsilon|f'(t)|^2t\, dt)^{1/2}(\int_x^\epsilon t^{-1}\, dt)^{1/2}$$ $$\le (\int_0^\epsilon|f'(t)|^2t\, dt)^{1/2}(\int_x^\epsilon t^{-1}\, dt)^{1/2}.$$ Note the first integral on the last line$\to 0$as$\epsilon \to ...

2

Since $\frac{p}{q}+\frac{q-p}{q}=1$, apply Holder's inequality. $$\left(\int_X |f|^p\,d\mu\right)^{1/p}\leq \left[\left(\int_X |f|^{q}\right)^{p/q}\mu(X)^{(q-p)/q}\right]^{1/p}=\left(\int_X|f|^q\right)^{1/q}\mu(X)^{(q-p)/(pq)}.$$ Note Holder's inequality only applies if the condition $\mu(X)<\infty$ holds. The inequality you ask for will hold ...

2

I'm not sure about better, but here is an alternate solution based on Holder's inequality: since $p_1 < p < p_2$ you have $\frac 1{p_2} < \frac 1p < \frac 1{p_1}$ so there exists a constant $0 < \alpha < 1$ with $\frac 1p = \frac{\alpha}{p_1} + \frac{1-\alpha}{p_2}$. Then $1 = \frac{p\alpha}{p_1} + \frac{p(1-\alpha)}{p_2}$ and by Holder's ...

2

A function in $\mathcal{S}$ decays faster than the reciprocal of any polynomial at infinity and is bounded. (This is usually built into the definition.) So its magnitude is bounded by some $M_1$ on the ball of some radius $R$ centered at the origin, and by $M_2/|x|^N$ outside this ball. Choose $N$ sufficiently large (depending on the dimension and $p$) to ...

2

$U$ is some other normed vector space. In this case $L^2([0,T];U)$, sometimes lazily written as $L^2(0,T;U)$, consists of functions $f$ from $[0,T]$ to $U$ such that $\int_0^T \| f(t) \|^2 dt<\infty$, where $\| \cdot \|$ is the norm on $U$. This notation is used in, for instance, Partial Differential Equations by Evans. Most commonly the $U$ in question ...

2

So you have a set of the form $$C=\{x\in\ell_2; |x_n|\le a_n\}$$ where $a_n$ is a series of positive real numbers such that $\sum a_n^2<+\infty$. You want to show that this space is totally bounded. Let $\varepsilon>0$. We want to show that there exist finitely many points in $C$ such that each point of $C$ is within the distance $\varepsilon$ from ...

2

For any $p \geq 1$, we have $$|x+y|^p \leq 2^p (|x|^p+|y|^p),$$ and therefore \begin{align*} \mathbb{E}(|X_n-X|^p \mid \mathcal{F}) &\leq 2^p \mathbb{E}(|X_n|^p \mid \mathcal{F}) + 2^p \mathbb{E}(|Y|^p \mid \mathcal{F}) \\ &\leq 2^p \mathbb{E}(|X|^p \mid \mathcal{F}) + 2^p \mathbb{E}(|Y|^p \mid \mathcal{F}). \end{align*} This shows that ...

2

Let $f \in L^{p_1}(E)$ bounded. Then there exists $M\geq 0$ such that $\sup_{x \in E} |f(x)| \leq M$ and $\|f\|_{p_1}^{p_1}=\int_E |f(x)|^{p_1}\,dx < \infty$. Case 1: $p_2<\infty$ For $p_2 > p_1$, we have \begin{align} \|f\|_{p_2}^{p_2}& =\int_E |f(x)|^{p_2}\,dx \\ &= \int_E |f(x)|^{p_2-p_1}|f(x)|^{p_1}\,dx \\ & \leq \left(\sup_{x ...

2

For the first statement: as mentioned by Giovanni in the comments, you need to use Fubini's theorem, after a change of variables. After the change of variable $y=x-t$ in the inner integral, you have $$\|f_n\|_1= \int_{\mathbb R} \left| \frac1n \int_x^{x+n}f(t)\,dt\right|\, dx \leq \frac1n \int_{\mathbb R} \int_0^n |f(x-y)|\,dy\,dx,$$ and now by Fubini you ...

2

Fact. If $\,f\in L^1[a,b]$, then for every $\varepsilon>0$, there exists a $\delta>0$, such that $\mu(I)<\delta$ implies that $\int_I\lvert\, f\rvert \,dx<\varepsilon$. Proof. If we set $E_M=\{x\in[a,b]: \lvert\, f(x)\rvert\ge M\}$, then $\lim_{M\to\infty}\int_{E_M}\lvert\, f\rvert\,dx=0$. To see this, let $$f_M(x)=\left\{\begin{array}{ll} f(x) ... 2 By divergence theorem$$ \int_{\partial \Omega }u\cdot n \ \phi=\int_\Omega div(u\phi) = \int_\Omega \phi div(u) + \nabla \phi \cdot u. $$This gives$$ \left|\int_{\partial \Omega }u\cdot n \ \phi\right|\le \|\phi\|_{L^2(\Omega)}\|div(u)\|_{L^2(\Omega)} + \|\nabla \phi\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)} \le \|\phi\|_{H^1(\Omega)}\|u\|_{H(div,\Omega)}. $$2 This is not true in general. Let X = \ell^2, and T \in L(\ell^2) given by$$ Tx = \left(\frac{x_n}{n+1}\right)_n $$Then T is one-to-one, but for g = e_n \in \def\ran{\operatorname{im}}\ran T we have that the inverse image is f = (n+1)e_n, and$$ \|f\| = \|(n+1)e_n\| = (n+1)\|e_n\| \stackrel ?\le C \|e_n\| $$Hence, such a C cannot exist. 2 Your idea is not wrong, it just needs to be written down more carefully. I think there are a few points that you are slightly misunderstanding. To prove that your subspace M is closed we need to prove that every convergent sequence in M converges to a point of M. So, as you do, let's take a sequence X_k = \left((x_n^k)_{n \in \mathbb N}\right)^{k ... 1 Let$$f_n = n\chi_{\left[0,\frac{1}{n^{p+1}}\right]}$$and$$F(x) = x^{\frac{p+1}{p}}. Then $\|f_n\|_{L^p} = n^{-\frac{1}{p}}\rightarrow 0$, but $\|F(f_n)\|_{L^p} = 1$ for all $n$.

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