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0

Well my solution looks basically the same as that of A.G; perhaps it's a little more elementary. I'll include it since I just spent some time on it! Suppose $$\sum_{k=1}^{\infty}|a_k| < \infty, \sum_{k=1}^{\infty}|b_k|^2 < \infty.$$ Let $f(z) = \sum_{k=1}^{\infty}a_kz^k, g(z) = \sum_{k=1}^{\infty}b_kz^k.$ Then in the open unit disc, $$f(z)g(z) = ...


1

We need to prove that $$ \sum_{k=1}^\infty\left|\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}\right|^2\frac{1}{\lambda^2}<+\infty. $$ Denote $x_k=\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}$, which gives $$ x_{k+1}=\frac{1}{\lambda}x_k+y_{k+1}, \qquad x_0=0,\ k\ge 0.\tag1 $$ Do $z$-transform of the equation, i.e. multiply by $z^{k+1}$ and ...


2

Given a measure space $(X,\mathcal M,\mu)$, we define $$L^1(X) = \left\{f : X\to\mathbb C \;\mid\; \int_X |f|\mathsf d\mu < \infty\right\}$$ and $$\mathcal L^1(X) = L^1(X) / \sim $$ where $\sim$ is the equivalence relation on the bracketed set defined by $f\sim g$ iff $f=g$ a.e. That is, $$\mu(\{x\in X : f(x)\ne g(x)\}) = 0. $$ In other words, we identify ...


1

After you use Holder, it is $\|f'\|_{L_2([0,x])}$ and it goes to $0$ when $x$ goes to $0$, because $f'$ is in $L_2([0,1])$.


4

Do what you did. Except don't plug in the $L^2$ norm: $$\int_0^x|f'(t)|\,dt\le\sqrt x\left(\int_0^x|f'(t)|^2\,dt\right)^{1/2}.$$ Now (prove and) use the fact that $\int_0^x|f'|^2\to0$ as $x\to0$.


1

This is not quite what was asked for, but I thought it worth mentioning: Below is (the non-trivial) Proposition 11.1.9 in Kalton and Albiac's Topics in Banach Space Theory: $\ \ \ $(i) For $1\le p\le2$, $L_q[0,1]$ embeds in $L_p[0,1]$ if and only if $p\le q\le2$. $\ \ \ $(ii) For $2< p<\infty$, $L_q[0,1]$ embeds in $L_p[0,1]$ if and only if $q=2$ ...


2

Here are some interesting facts about the relations between different $L^p$-spaces over the same measure space $(X,\Sigma,\mu)$ (based on Section 6.1 of Folland, 1999): If $0<p<q<r\leq\infty$ and if $f\in L^q$, then there exist $g\in L^p$ and $h\in L^r$ such that $f=g+h$. If $0<p<q<r\leq\infty$, then $L^p\cap L^r\subseteq L^q$. If ...


3

Here are some examples of relations between the spaces: If $p,q\in[1,\infty)$, there is a bijection $L^p\to L^q$, namely $L^p\ni f\mapsto |f|^{p/q-1}f\in L^q$. If $A$ has finite measure, then $p\geq q$ implies $L^p(A)\subset L^q$. If $f\in L^p$ and $g\in L^q$ so that $1/p+1/q=1/s$ (assuming $p,q,s\in[1,\infty]$), then the pointwise product $fg$ is in ...


3

Corollary 3 of Chapter 7 of Royden: If $E$ is measurable with finite measure and $1\leq p_1< p_2\leq \infty$, then $L^{p_2}(E)\subseteq L^{p_1}(E)$. Furthermore, $||f||_{p_1}\leq c||f||_{p_2}$ for all $f\in L^{p_2}(E)$ where $c=[m(E)]^{\frac{p_2-p_1}{p_1p_2}}$ if $p_2<\infty$ and $c=[m(E)] ^{\frac{1}{p_1}}$ if $p_2=\infty$. Royden remarks in an ...


1

Actually, I made a mistake: The question is not an exercise in the book, but rather a complement and a rather famous result in functional analysis, known as the Dunford-Pettis theorem (see Uniform Integrability Wiki). The proof can be found in several textbooks and in a short research note here.


3

Normal Human already gave a good answer, but I came up with another one that I would like to share. It is based on the coarea formula and I tried to make it feel natural, with few arbitrary choices. As has been noted, it suffices to consider $L^1$ functions, so let $f\in L^1$. (For $f\in L^p$ we have $|f|^p\in L^1$.) This argument works for any ...


4

As felipeh noted, the problem reduces to $p=1$ by replacing $f$ with $g = |f|^p$. (The case $p=\infty$ should be treated separately.) Also, the annulus can be replaced by the sphere $S^{n-1}$. Indeed, define a function $h$ on the unit sphere by $h(\xi) = \int_r^1 g(t\xi)\,dt$. Then $h\in L^1(S^{n-1})$, with a norm comparable to $\|g\|_{L^1(A)}$ because ...


2

The ordinary heat equation is $$ \frac{\partial F}{\partial t}=\frac{\partial^{2}F}{\partial x^{2}},\\ F(0,x)=f(x). $$ The initial heat distribution is $f$. The ordinary heat kernel is the Guassian, and the resulting time evolution operator $T(t)=e^{tL}$ is a constractive $C_0$ semigroup on every ...


2

The functions $g_m(t) = H_m(t)\,e^{-t^2/2} $ just give an orthogonal base of $L^2(\mathbb{R})$, since: $$ \int_{-\infty}^{+\infty} f_m(t)^2\,dt = 2^m m! \sqrt{\pi}. $$ An orthonormal base is given by: $$ f_m(x) = \frac{1}{\sqrt{2^n n! \sqrt{\pi}}}\,H_m(x)\, e^{-x^2/2}. $$ We may notice that if $m$ is odd then $\int_\mathbb{R}f_m(x)\,dx = 0$, while: $$ ...


8

It is certainly not dense. The linear functional $$ T(x_n) = \sum_{n=0}^{\infty} \frac{x_n}{n} $$ Is continuous on $l^2$. Thus, the set $T^{-1}(0)$ must be a closed subspace of $l^2$. If it were dense, we'd have $l^2 = T^{-1}(0)$. But the sequence ${\frac{1}{n}}$ obviously isn't in this space.


1

Hint: try the Radon-Nikodym theorem.


2

You know (why?) that $L^2 \neq L^{3/2}$, so choose an $f \in L^{3/2}\setminus L^2$, and consider the map $$ L^3 \mapsto \mathbb{C} \text{ given by } g \mapsto \int fgd\mu $$ This is a bounded linear map. Suppose it is the restriction of a bounded linear map on $L^2$, then $\exists h \in L^2$ such that $$ \int fgd\mu = \int hg d\mu \quad\forall g\in L^3 $$ ...


2

If by “die down,” you mean “converge to $0$”, then the answer is no. Consider the following function: \begin{align*} f(x)\equiv \begin{cases} 0&\text{if $x\in[0,1)$,}\\ 1&\text{if $x\in[1,1+1/1^2)$,}\\ 1&\text{if $x\in[2,2+1/2^2)$,}\\ 0&\text{if $x\in[2+1/2^2,3)$,}\\ 1&\text{if $x\in[3,3+1/3^2)$,}\\ 0&\text{if $x\in[3+1/3^2,4)$,}\\ ...


0

We can find that \begin{align*} &\big \vert G_\lambda f (s,x_1 + h) - G_\lambda(s,x_1) - G_\lambda f (s,x_2 + h) + G_\lambda(s,x_2)\big \vert = \vert G_\lambda'(s,\tilde{x}_1)h - G_\lambda'(s,\tilde{x}_2)h\big \vert\\ &\leq h C_{\lambda,q}\|g'_\lambda\|_q \|f - \tau_{\tilde{x}_2 - \tilde{x}_1}f\|_p \end{align*} And consider the following as a ...


1

For any $s>1/2$ the norm of $H^s$ controls the $L^\infty$ norm. Indeed, when the series of Fourier coefficients is absolutely convergent, their sum bounds $\sup|u|$. By the Cauchy-Schwarz inequality, $$ \|u\|_{L^\infty}\le \sum_{n \in \mathbb{Z}} |\hat u_n| \le C(s,P) \left( \sum_{n \in \mathbb{Z}} \bigg(1 + \frac{4 \pi^2 n^2}{P^2}\bigg)^{s} ...


0

There are several ways to construct $L^1(\mathbb{R})$. One way is to let $L^1(\mathbb{R})$ be the smallest Banach space containing $C_0^\infty(\mathbb{R})$ under the norm $\|\cdot\|=\|\cdot\|_{L^1}$. For other definitions of $L^1$ this is also true (because it is the same space), however then this would need a proof based on the particular construction.


0

Let $f \in L¹(\mathbb{R})$. Given an $\epsilon > 0$ you can find a simple compactly supported function $\phi$ such that $\|f - \phi\|_{L^1} < \epsilon/2$. Now you just have to find a smooth and compactly supported function near $\phi$. One way to do this is by doing a convolution against a mollifier. There are other ways to prove this. Another proof ...


1

By the inequality $|Af| \leq A|f|$, we may assume that $f$ is non-negative. Then from the Tonelli's theorem (a.k.a. Fubini's theorem for non-negative functions), $$\| Af \|_2^2 = \int_0^1 \int_0^1 f(y)f(z) \left( \int_0^1 \frac{dx}{|x-y|^{\alpha}|x-z|^{\alpha}} \right) \, dydz $$ and we may try to estimate the following function $$ k(y, z) = \int_0^1 ...


2

HINT: The Cauchy-Schwarz Inequality reveals that $$\begin{align} \left|Af(x)\right|^2 &= \left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2\\\\ &\le \int_0^1\left|f(y)\right|^2\,dy\,\int_0^1\frac{1}{|x-y|^{2\alpha}} dy \end{align}$$ And thus, the square of the operator norm is $$\begin{align} ||A||_2^2&=\sup_{f\in \mathscr{L}^2} ...


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 $p>0$, ...



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