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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}. $$We may notice that if m is odd then \int_\mathbb{R}f_m(x)\,dx = 0, while:$$ ...

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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.

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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 $$\int fgd\mu = \int hg d\mu \quad\forall g\in L^3$$ ...

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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 $x\in[3+1/3^2,4)$,}\\ ...

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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 ...

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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 ...

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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 $L^1$. ...

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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: ...

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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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