# Tag Info

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As Frank provided a counterexample the inclusion in general doesn't hold. In fact if your measure is finite the inclusion is vice versa because via Hölder you get $\|f\|_q \leq \mu(X)^{\frac{1}{p} -\frac{1}{q}} \cdot \|f\|_p.$ Maybe you are talking about the sequence spaces $\ell^p$ because there the inclusion holds as every function which is in ...

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Let $f(x):= x^{-1/p} \mathbf{1}_{(1,\infty)}$. Then $f^q$ is integrable because $\frac{q}{p} > 1$. But $f^p = \frac{1}{x}$ (on $(1,\infty)$) is not. That is, $f \in L^q(\mathbb{R})$ but $f \notin L^p(\mathbb{R})$.

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Yes, it is. Let $M=\{x\in X: y^*(x)=0 \text{ for all }y^*\in Y^*\}$. This is sometimes denoted ${}^\perp Y^*$ and called the pre-annihilator of $Y^*$. Every $y^*\in Y^*$ induces a linear functional on $X/M$ in a natural way: $y^*(x+M)=y^*(x)$ is well-defined. Conversely, if $\phi$ is a linear functional on $X/M$, then its composition with the projection ...

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Your set $A_\sim$ is the projection onto the second coordinate of the set $$\{(X,Y)\in \mathrm{SB}^2 \mid (X,Y)\in A\times \mathrm{SB} \text{ and } X\sim Y\},$$ and $A\times \mathrm{SB}$ is the preimage of $A$ by the first projection, so both sets are analytic.

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First note that the way you are defining $T$ does not give you guarantee that $T$ will be linear. To make sure that $T$ is linear you need $\{x_n \mid n \in \mathbb{N}\}$ and $\{y_n \mid n \in \mathbb{N}\}$ to be linearly independent, which you can assume without loss of generality. Because if they are not linearly independent then you can choose a maximal ...

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"Naturally identified" leaves enough wiggle room to identify an infinite subset $X$ of $\mathbb N$ with the sequence in $\mathcal N$ that enumerates the elements of $X$ in increasing order. The set of all these increasing sequnces is a closed subset of $\mathcal N$.

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Let $U$ be a subspace of a Banach space $V$. Suppose $x\in U$ is an interior point of $U$, which means that $$B(x,\varepsilon)=\{y:\|x-y\|<\varepsilon\}\subseteq U$$ for some $\varepsilon>0$. Since $U$ is a subspace, also $$B(0,\varepsilon)=-x+B(x,\varepsilon)$$ is contained in $U$. Now, for every $v\in V$, $v\ne0$, the vector ... 1 This might seem redundant but also seems correct to me. If we take the sequence (S_n) to be the constant sequence of \textbf{Identity operators} then this sequence converges strongly to the Identity operator. Since (T_n) converges to T weakly and in this case the sequence T_nS_n is nothing but the sequence (T_n), which clearly fails to converge ... 1 Hint: Try to prove the following result \textbf{Result}: If for a sequence (x_n) in a Banach space X, the sequence (f(x_n)) is bounded for all f\in X^*, then the sequence (\|x_n\|) is bounded. Now suppose that you have proved the result. Suppose (T_n) converges to T in weak operator topology, i.e., \begin{align*} |f(T_nx)-f(Tx)| ... 2 The category of Banach spaces is locally \aleph_1-presentable. 2\langle x+\alpha y,x+\alpha y\rangle\geq\langle x,x\rangle\forall \alpha\\ \langle x,x\rangle+2\alpha\langle x,y\rangle+\alpha^2\langle y,y\rangle\geq\langle x,x\rangle\forall\alpha\\ 2\alpha \langle x,y\rangle+\alpha^2\langle y,y\rangle \geq0\forall\alpha$$But the left-hand side equals zero if \alpha=0 and if \alpha=-2\langle x,y\rangle/\langle ... 0 I just did |T'(X)| = |2014 cos(2014x)| < sup (2014 cos(2014x)) and it was not smaller than 1 so T is not a contraction is that right? 1 I think that you should add \alpha < 1 in your statement. So, you just have to show that there exists at least one x\in[0, 2\pi] such that this inqequality doesn't hold. Do you have any ideas? Just take the derivative of the \sin(2014x). 1 First of all you should understand that what is meant when one says that "\textit{Not all normed linear spaces are inner product spaces.}" Here is the explanation. An \textit{inner product space} is a vector space with an inner product defined on it. An inner product on X defines a norm on X given by \|x\| = \sqrt{\langle x,x ... 4 The space of continuous functions on [0,1] with the norm \|x\|= \sup\{|x(t)| : t\in[0,1] \} is a Banach space. To justify that claim, you would need to know that all continuous functions on [0,1] are bounded, and that Cauchy sequences in this metric space actually converge, and those take some work. One way to prove that this norm does not come from ... 0 There are at most 2^{\aleph_0} subsets of a standard Borel space, or of a separable metric space. The same bound applies to coanalytic subsets. But usually, you have a lot more of non Borel subsets (for instance, in the case your family {G} is uncountable). 0 Here's a proof that T is continuous in the strong operator topology. As noted in the question, for each x\in X, the map t\mapsto T_tx is right-continuous under the weak topology, and norm-bounded on each bounded interval. To complete the proof given in the question for an arbitrary Banach space, we just require the following lemma. Lemma: Let X ... -1 every finite dimensional Banach space is a reflexive bot it need not be uniformly convex .X=\mathbb{R}^n , n\geq2 with the norm \Vat x\Vat_{1}=\sigma {i=1}^\n\vert x{i}\vert 2 Yes, it follows from the following general theorem (see Dixmier "C^*-algebras and representations", 2.10.2): Every irreducible representation \rho of a C^*-subalgebra C of a C^*-algebra A can be continued to an irreducible representation \pi of A in a possibly larger Hilbert space. You can also apply the Lemma 2.10.1 from Dixmier directly. ... 0 Well, it is enough to show that the unit ball is compact with respect to some locally convex topology, that is weaker than the norm topology. See this article. For example, if X is a reflexive space with a basis, then the unit ball of \mathscr B(X), the space of bounded linear operators on X, furnished with the weak operator topology is compact. This ... 3 You assumed that T_\lambda is surjective (when applying bounded inverse theorem). It is not true. In fact a small modification of your second argument shows precisely that T_\lambda cannot be surjective. 0 What about the following generalisation (it works, but is it useful?) ? Let's deal with polynomials of the form P(A,B,z)=A-Bz first and let's give a modified definition to the "generalised resolvent set" \rho(A,B): \left\{z\in \mathbb{C}\left| \right. \left(A-zB\right)^{-1}\text{,}B\left(A-zB \right)^{-1}\text{exist on a dense (common) ... 2 If \epsilon>0, choose a step function \tau such that \sup_{x\in[a,b]}\|f(x)-\tau(x)\|<\frac\epsilon4. Now, since \tau is a step function, there is a partition \{I_k\}_{k=1}^n of [a,b] such that \tau is constant on each interval I_k. Let M denote the set of endpoints of the intervals I_k. This set is finite. Now, suppose that x\not ... 0 Let \{x_n\}_{n\in \mathbb{N}} be a set with dense linear span and T:X\rightarrow Y be an isomorphism. Put y_n=Tx_n. Then \{y_n\}_{n\in \mathbb{N}} has dense linear span in Y. Now let (a_n) be a sequence such that \sum_{n=1}^\infty a_n x_n exists. For every N the following inequality is true:$$\|\sum_{n=1}^N a_n x_n\|=\|T^{-1}\sum_{n=1}^N a_n ...

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Q1. Non-zero constant functions DO NOT BELONG to $W_0^{1,2}(\Omega)$. Q2. The space $W_0^{1,2}(\Omega)$ is defined as the «completion» of $C_0^\infty(\Omega)$ with norm $$\|u\|_{W^{1,2}_0(\Omega)}=\left(\int_\Omega\big(u^2(x)+\big|\nabla u(x)\big|^2\big)\,dx\right)^{1/2}.$$ Thus, indeed, every element of $W_0^{1,2}(\Omega)$ can be ...

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In terms of getting a proof done, it really is just Cauchy-Schwarz: $$\left | \int_0^t f(s) ds \right | = (f,1)_{L^2} \leq \| f \|_{L^2} \| 1 \|_{L^2} = t^{1/2} \left ( \int_0^t f(s)^2 ds \right )^{1/2}.$$ Cauchy-Schwarz can be proven in the abstract setting of a general inner product space, and then it follows in your case from the fact that the $L^2$ ...

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Try applying Cauchy-Schwartz to $|f|$ and $1$.

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You can characterize $W^{1,2}_0$ as "the 'compactly supported' $L^2$ functions that have $L^2$ weak derivatives." Since the weak derivatives of a constant function are all zero, hence $L^2$, the answer to your question is the answer to, "are constant functions compactly supported $L^2$?" This is never true unless the constant is zero.

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Just take the union of $0$ and a sequence of disjoint closed annuli around $0$ of decreasing sizes. It's still closed and absorbing, but not convex, and it includes no ball around 0. Edit: the above is wrong. The definition of absorbing set is equivalent to be a neighbor of 0 in every intersection with a one-dimensional subspace. An example of a radial set ...

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Let $y \in V \setminus W$ with $\def\norm#1{\left\|#1\right\|}\norm y = 1$. Then, as $W$ is closed $d := \def\dist{\mathop{\rm dist}}\dist(y, W) > 0$. Choose $\delta > 0$ such that $\frac{d}{d+ \delta} > 1-\epsilon$. There is some $w \in W$ with $\norm{y-w} < d + \delta$. We let $v := \frac{y-w}{\norm{y-w}}$. Then $\norm v = 1$ and $$\dist(v,W) ... 1 if x \in V is non-zero then ||x|| \gt 0, so set:$$ x' = \frac{x}{||x||} $$it is now evident that ||x'|| = 1 1 Let (g_{n}) be a sequence in K satisfying \left\|f-g_{n}\right\|_{p}\rightarrow\delta:=\inf_{g\in K}\left\|f-g\right\|_{p}. It follows from the triangle inequality that the g_{n} are bounded in norm, and therefore lie in a closed, convex subset E\subset K, which is weakly compact by Alaoglu's theorem. Whence (g_{n}) has a subset ... 2 For L^1, take K to be the set of functions for which \int h \mathrm{d}x = 0. Let f be the characteristic function of the interval [0,1]. Note$$ 1 = \int h - f \mathrm{d}x \leq \|h - f\|_1 $$Note that this minimum is achieved for any h of the form: h = +1 on a measure 1/2 subset of [0,1], h = -1 on its complement in [0,1], and h = 0 ... 0 It means closed under isomorphism. 1 Hint:$$\|f\|^2 = \sum_{k=1}^n \|f^{(k)}\|_2^2 \ge \|f^{(k)}\|_2^2$$1 One example might be M(K), the space of all regular Borel measures on K of finite variation, where K is compact space. This space arises as the dual of C(K). 3 No, it cannot. For any x_1, \ldots, x_n \in X, the linear span V of x_1, \ldots, x_n is finite-dimensional, so it is not all of X, and by Hahn-Banach there is a nonzero continuous linear functional f such that f = 0 on V. Take y \in X with 0 < \|y\| < r_0 so that f(y) > r \|f\|. Now$$\|y - x_j\| \ge \|f\|^{-1} |f(y - x_j)| = ...

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That $I - A$ is both injective and surjective follow from the fact that it is invertible. Indeed, let $B:X \to X$ be any invertible operator; then we have an operator $B^{-1}$ such that $BB^{-1} = B^{-1}B = I. \tag{1}$ To see that (1) implies $B$ is injective, suppose that $Bx_1 = Bx_2 \tag{2}$ for some $x_1, x_2 \in X$. Then from (1) $x_1 = Ix_1 = ... 2 Hint Prove that$Id+A+A^2+..+A^n+...$defines an operator which is the inverse of$Id-A$. 1 You know that$B(X)$, the space of bounded linear operators on$X$is a Banach space. Consider the series $$\sum_{n=0}^{\infty} A^n$$ Since$\|A\| < 1$, it converges absolutely, and thus converges in$B(X)$to an operator$B$. Now check that$B(I-A) = (I-A)B = I$and so$(I-A)$must be surjective. 0 Given$\varepsilon > 0$, there is$n_0 \in \mathbb{N}$such that: $$m,n \geq n_0 \implies \Vert f_m - f_n \Vert < \frac{\varepsilon}{4}$$ Then, we have:$|f_m(0) - f_n(0)|< \varepsilon /4\sup \left\{ \frac{|(f_m-f_n)(t)-(f_m-f_n)(s)|}{|t-s|^p}: t\neq s, \quad t,s\in [0,1] \right\} < \varepsilon /4$Given$t\in]0,1]$, we have for all ... 1 In fact all normed spaces are subspaces of some function spaces. This could be the reason why functional analysis have its name. 1$\mathbb{R}$is an infinite dimensional vector spaces over$\mathbb{Q}$, but it is not an Hilbert space. You can see that all the axioms for a vector space are verified if you define the sum of two ''vectors" as the usual sum of real numbers and the product for a scalar$q \in \mathbb{Q}$as the usual product. This space has an infinite dimensional Hamel ... 1 Nevermind, I answered it myself using an elementary argument. Thanks anyway! EDIT: Here is the argument, in brief. Consider the real Banach spaces$\widetilde{X}=X\oplus_{\ell_1}X$and$\widetilde{Y}=Y\oplus_{\ell_1}Y$. It is a simple exercise to show that$T\oplus 0$and$0\oplus T$both lie in$\mathcal{FSS}(\widetilde{X},\widetilde{Y})$, and hence so ... 1 Relative directions of elements of vector spaces can be expressed via the scalar product: If we have a (pre-)Hilbert space$\cal H$with a scalar product$\langle \cdot, \cdot \rangle$, then we say that two elements$f,g \in \cal H$are orthogonal (i.e. their relative angle is 90°) if$\langle f, g \rangle = 0$. If you take a (say, finite) set of vectors in ... 0 In my opinion direction can be defined only fo Hilbert spaces ( sice these are the spaces where we cand define an angle). If$V$is an Hilbert space than, given two vectors$v,w$we can say that they have the same direction ad are equi-oriented iff$ \langle v,w\rangle=||v||\cdot||w||$and have the same direction but are opposite iff$ \langle ...

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Well we already seem to have a structure that encompasses length and direction in the notions of $\mathbb{R}^2$ and $\mathbb{R}^n$. Describe a vector in $\mathbb{R}^2$ by $(r,\theta)$ where $r$ is the length of the vector and $\theta$ is the direction of the vector. In $\mathbb{R}^n$ we can do the exact same thing, just using the $n$-tuple ...

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Note: Here are two examples of partial converses which could be convenient. Partial converse in section 27 of A Hilbert Space Problem Book by Paul R. Halmos: Every bounded subset of a Hilbert Space is weakly bounded. We can read in section $27$: Uniform boundedness The celebrated principle of uniform boundedness (true for all Banach ...

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Let $(u,v)\neq(0,0)$. $D_{(0,0)}f(u,v)=\displaystyle\lim_{\epsilon\to0}\dfrac{f\big((0,0)+\epsilon((u,v))\big)-f(0,0)}{\epsilon}=\lim_{\epsilon\to0}\dfrac{f(\epsilon u,\epsilon v)}{\epsilon}=\lim_{\epsilon\to0}\dfrac{\epsilon^4 u^3v}{\epsilon(\epsilon^4 u^4+\epsilon^2 ... 1 Being able to extend an isometry to an isometry is rare. I don't think any nontrivial Banach space has this property. Here is a proof for$\ell^1$and$\ell^\infty$, the two spaces of main interest to you. In both cases$e_1,e_2,\dots$is the standard set of$0-1$vectors, e.g.$e_2=(0,1,0,0,\dots)$. The case of$\ell^1$Let$A=\{0,e_1,e_2\}\$. Define ...

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