# Tag Info

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The fact that there is an $\epsilon>0$ such that $B_{\epsilon}(x)\subset \overline{B_r(a)}^c$ does not imply $||x-a||\geq r+\epsilon>r$. This implication is not justified by the open-ness of $\overline{B_r(a)}^c$. You want this to say that $x$ is at least $\epsilon$ distance away from $\partial \overline{B_r(a)}$. This is the intuitive idea from ...

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Hint If every term of a sequence $(b_n)$ is in the interval $[0,1]$ and $b_k=1$ for some $k\ge 2$, then the sequence is not in $M$, but it is in $\overline M$ (this needs to be proved).

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We denote the closure of a set $A$ by $\bar A$ and the interior by $\mathring A$. Hint: Consider the set $B_1(0)$ where $B$ denotes the open ball and $0$ denotes the sequence $(0,0,\dots)$. What is the closure of this set? What is the boundary of this set? How does this set relate to $M$?

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Let $x = \{x_k\}$ denote an arbitrary sequence in $\ell^p$. For $j \in \Bbb N$, let $x^{(j)} = \{x^{(j)}_k\}$ denote the sequence given by $$x^{(j)}_k = \begin{cases} x_k & k \leq j\\ 0 & k > j \end{cases}$$ Note in particular that $x^{(j)} \in V$ for all $j$. Claim: In the space $\ell^p(\Bbb N)$, $x^{(j)} \to x$ as $j \to \infty$. Proof: We ...

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Given a continuous mapping $F:X\rightarrow Y$ it induces continuous map $F^{**}:X^{**}\rightarrow Y^{**}$ given by the formula $$[F^{**}(x^{**})](y^*):=x^{**}(F^*(y^*)).$$ This assigment is functorial, i.e. it has two properties: $(id_X)^{**}=id_{X^{**}}$ and $(G\circ F)^{**}=G^{**}\circ F^{**}.$ Additionaly for every normed space $X$ we have continuos ...

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Given a bijective linear isometry $T:X\rightarrow Y$, the dual map $T^*:Y^*\rightarrow X^*$ is also a bijective linear isometry. (This follows from the fact that $T^*$ has inverse $(T^{-1})^*$ and $\|T^*\|=\|T\|$.) From this, we have that $T^{**}:X^{**}\rightarrow Y^{**}$ is a bijective linear isometry as well. Let $J_X:X\rightarrow X^{**}$ is the canonical ...

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It seems the following. Let $S$ be a linear subspace of the linear space $E$ of codimension $2$. Take an arbitrary vector $x\in E\setminus S$. Let $S’=\langle \{x\}\cup S\rangle$ be a linear hull of the set $\{x\}\cup S$. Then $S$ is a subspace of codimension $1$ of the space $S’$ and $S’$ is a subspace of codimension $1$ of the space $E$. So if $S$ in ...

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Your proof is fine. You constructed a countable subset of $E\subset X$, then proved that each $x\in X$ has a sequence $(y_n)\subset E$ such that $y_n\rightarrow x$ in your topology. You use the linearity of $X$ and the separability of $B_X$ to construct $E$ and $y_n$, and the Hausdorff condition (implicitly) to prove that $y_n\rightarrow x$. There's ...

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The answer is "no". The unit sphere is norm closed in the unit ball under the (norm) subspace topology. But in an infinite dimensional normed space, the weak closure of the unit sphere is the unit ball. See this post for a proof of this. So, in the space $B(X, {\rm weak})$, the closure of the unit sphere is again all of $B_X$ (in general if $A$ is a ...

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Let $\Bbb{F}$ denote the base field of $V$. If we assume that $\star$ is a binary operation on $V$ that turns $V$ into an $\Bbb{F}$-algebra, i.e., Left distributivity: $x \star (y + z) = x \star y + x \star z$ for every $x,y,z \in V$, Right distributivity: $(x + y) \star z = x \star z + y \star z$ for every $x,y,z \in V$, and ...

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Edit: As OP points out, this solution only works if $X$ is second-countable. Let $f, g: \mathbb{R} \to X$ be Borel measurable and $h(s) = f(s) + g(s)$. Consider mappings \begin{align} \mathbb{R} \ni s & \stackrel{h_1}{\mapsto} (s, s) \in \mathbb{R}^2 \\[1ex] \mathbb{R}^2 \ni (s, t) & \stackrel{h_2}{\mapsto} (f(s), g(t)) \in X^2 \\[1ex] X^2 ... 1 Suppose f open. The image of the unit open ball is open. In particular contains an open ball centered on the origin of radius 2\alpha. Therefore contains all the vectors \alpha f_1,\dots,\alpha f_n where (f_1,\dots,f_n) is a basis of F of vectors having norms equal to 1. Hence f is surjective as supposed to be linear. Conversely, suppose f ... 0 See the method described in the answer to your other question here. In particular, take x = (1,0) and y = (0,1). 1 Since A is convex, for all x,z\in A, \theta \in [0,1], x + \theta(z-x) \in A \|y - (x + \theta(z-x))\|^2 = \|y - x\|^2 - 2\theta \langle y-x,z-x\rangle + \theta^2\|z-x\|^2 $$So if \langle y-x,z-x\rangle \le 0 for all z\in A, take \theta=1 in the above equation to show \|y-z\|^2 \ge \|y - x\|^2. On the other hand, if for some z\in A ... 1 First of all: since all norms on a finite-dimensional space are comparable, the condition could be simpler stated as$$ C_1\|x-y\| \leq d(x,y) \leq C_2\|x-y\| $$where \|\cdot \| is a norm of our choice, e.g., Euclidean. Second: the answer is negative, for example$$d(x,y) = |x-y|+\min(|x-y|,1)$$is a translation-invariant metric on \mathbb{R} that ... 1 I have found the following somewhat messy estimates. First define r=1/p, so that -1<r-1<\nu<r<1. Then notice that the potentially problematic points are -\infty,0,1 and +\infty. We consider 1 first: Let |t-1|<1/2. Then by Taylor's theorem and the triangle inequality, for some \tau between 1 and t, ... 0 Let$$ S = \{(y_n) \in \ell^{\infty} : \exists C > 0 \text{ such that } |y_n| \leq C/n\quad\forall n\in \mathbb{N}\} $$If (x_n) \in \ell^{\infty} then T(x_n) \in S with c = \|(x_n)\|_{\infty}, and conversely, if (y_n) \in S, then the sequence (x_n) defined by$$ x_n := ny_n $$is in \ell^{\infty} and satisfies T(x_n) = (y_n). Hence, R(T) = ... 0 Let us denote two norms as \mu(x) = \|x\|_1  and \nu(x) = \|x\|. From the definition of \|x\|_1  we have \mu(x) = \nu(x) + |f(x)|. Two norms \mu and \nu are equivalent if there exist two constants c,C\in\mathbb{R} such that \forall x\in X c\nu(x) \leq \mu(x) \leq C\nu(x). First, since \mu(x) = \nu(x) + |f(x)| and |f(x)|>0 we ... 1 The answer is AFFIRMATIVE First suppose that the two norms are equivalent and a and b be such that a\|x\|\leq \|x\|_1\leq b\|x\| for all x\in X. Then |f(x)|=|\|x\|_1-\|x\||\leq \|x\|_1+\|x\|\leq (1+b)\|x\| for all x\in X. Therefore, f is continuous. Conversely, let us assume that f is continuous. It is clear from the definition of ... 1 It is clear that \|x\| \le \|x\|_1. If f is continuous, it is a bounded operator and so there is some M such that |f(x)| \le M\|x\|. Then \|x\|_1 \le (1+M) \|x\|, and so \|\cdot\|_1 and \|\cdot\| are equivalent. If the two norms are equivalent, there is some L such that \|x\|_1 \le L \|x\|, and so |f(x)| \le (L-1) \|x\|. Hence f is ... 4 If \{ x_{n} \} converges to x in \|\cdot\|_1, then it converges in \|\cdot\| because$$ \|x-x_n\| \le \|x-x_n\|_1. $$Conversely, if \{ x_n \} converges to x in \|\cdot\|, then \{f(x_n)\} converges to f(x) because f is continuous; therefore, \{ x_{n} \} converges to x in \|\cdot\|_1. 1 Hint. It suffices to prove the following:$$ \|x_n-x\|\to 0 \quad\text{iff}\quad \|x_n-x\|_1\to 0. $$1 You are correct in stating that M is both convex and complete. The proof of this is a fairly straightforward application of the definitions. The second part doesn't require any particularly high-powered tools for functions of multiple complex variables. Hints: \|x\|_\infty is minimized with \xi_i = \frac{z}{n} Use Hölder's inequality with the ... 4 In case that W is closed, there is indeed a canonical choice. It is called the quotient norm and defined by \begin{equation*} \| [\hat v] \|_{V / W} := \inf_{v \in [\hat v]} \| v \|_V . \end{equation*} This also leads to \|\pi\| = 1 (in case W \ne V). As already mentioned by matt biesecker, you always have \operatorname{ker}(\pi) = W, since this ... 2 Consider the subset consisting of the e_n's. It's bounded, because each sequence has norm 1. Since the distance between any two distinct sequences is 2, any Cauchy sequence is constant and thus converges, so the subset is closed. However, it's not compact because, it does not have a convergent subsequence. The reason your sequence doesn't have a ... 3 The distance from a closed subspace is a continuous function and D is the inverse image of a closed set, hence topology gives that the claim is trivial. 2 There are two understandings of the notation U \oplus V. The first interpretation means that if you write U \oplus V, then U and V are both subspaces of a common inner product space and orthogonal to each other (i.e. \langle x,y\rangle =0 for x\in U,y\in V and U \oplus V=\{x+y\mid x\in U, y \in V\}. In this interpretation, you have ... 3 1. T is surjective because of the Fundamental Theorem of Calculus. The reason that T^{-1} does not exist is that T is failing to be injective. T is not injective because any two functions that differ by a constant will get mapped to the same function, e.g., x and x+c, c\in \mathbb{R} both get mapped to the constant function 1. 2. Since ... 1 Let \mathbf{x}=(x_n) be any sequence in l^{\infty}, then it can be easily seen that give any \epsilon>0 you can find a natural number N such that \frac{\|\mathbf{x}\|}{N} < \epsilon. So, for all n\geq N you can see that |x_n|\leq\|\mathbf{x}\| and \frac{1}{n}\leq \frac{1}{N}, so, |\frac{x_n}{n}| \leq ... 2 By Fundamental Theorem of Calculus you can see that the range of T is R(T) =\{f\in C^1[0,1]\mid f(0)=0\} where C^1[0,1] is the space of all continuously differentiable functions. T^{-1} is linear is easy to see. But T^{-1} is not bounded. To see this first of all notice that T^{-1}(x(t)) = x'(t), i.e., T^{-1} is the differentiation operator ... 0 Question 1: T^{-1} does not exist because T is not injective. E.g. Let x(t)=t , y(t)=t+2. Then Tx(t)=Ty(t)=1. Observe that T is surjective by the fundamental theorem of calculus (which can also be applied to question 3). 0 Consider the fact that if (f_n) is a Cauchy-series in C^1[a, b] with respect to the ||.||_{C^1}-norm, then it also a Cauchy-series with respect to ||.||_\infty, and the series of functions (f_n') is a Cauchy-series with respect to the supremum norm. Now, consider the space C[a, b] = \{f: [a, b] \to \mathbb{C}, f continuous\}. It's easy to ... 1 First, X is "exactly" same as \mathbb{R}^n w.r.t euclidean norm. Let id be the transformation (a_1,\cdots,a_n)\mapsto \sum a_je_j. What you wrote in your question, proves the continuity of id. To prove continuity of id^{-1}, it's sufficient to prove that the set \{\frac{\|x\|_e}{\|x\|_2}\;;x\neq0\} is bounded. Suppose S^{n-1} is the unit ... 0 Ok, I've made a real mess of this, but I've figured it out. Simple is good. Let T : (a_1,\ldots, a_n) \to \sum a_j e_j. It is obvious that T is linear. We explore its continuity at zero. Let \|\sum a_je_j\|_1 = \sum |a_j| < \epsilon for any \epsilon > 0. Then, \left(\sum |a_j|\right)^2 < \epsilon^2, and hence \left(\sum ... 1 Suppose \|x \| = a; write x = \frac{a}{\delta} y, where \|y\| = \delta. You know that T(x) = \frac{a}{\delta} T(y), and \| T(y) \| < 1. So \|T(x)\|< |\frac{a}{\delta}|. The case where \| x \| < a is similar. 0 A Cauchy sequence (f_k) with respect to the norm \|\cdot\|_{C^1} obviously is a Cauchy sequence with respect to \|\cdot\|_\infty, and the sequence (f_k^\prime) as well is a Cauchy sequence with respect to that latter norm. Therefore we have a uniformly convergent sequence of differentiable functions such that the derivatives are uniformly convergent ... 1 If \|f\|_{C^1}=0, then \|f\|_{\infty}=0 and \|f'\|_{\infty}=0. In particular,$$0=\|f\|_{\infty}=\sup_{x\in\left[a,b\right]}|f\left(x\right)|$$and thus f is identically zero. Then, for all f,g\in\mathcal{C}^{1}\left(\left[a,b\right];\mathbb{R}\right), we have ... 1$$||f+g||_{C^1} = ||f'+g'||_{\infty} + ||f+g||_{\infty} \le \le ||f'||_{\infty} +||g'||_{\infty} + ||f||_{\infty} +||g||_{\infty} = ||f||_{C^1} + ||g||_{C^1}$$when the inequality comes from triangle inequality applied to ||\cdot ||_{\infty} 2 1. If T is bounded then it is not hard to see that T maps bounded sets to bounded sets. Conversely, let us assume that T maps bounded sets to bounded sets. In particular T(B_X) will be bounded, where B_X is the closed unit ball of X. So, there exists c>0 such that \|Tx\|\leq c\|x\| for all x\in B_X. Now, let y\in X be any arbitrary ... 1 T is obtained from continuous functions by composition and pairing and hence is continuous. To be more precise, T is:$$X_{\neq0} \stackrel{\Delta}{\longrightarrow} X_{\neq0} \times X_{\neq0} \stackrel{\langle\|\cdot\|, \mbox{id}\rangle}\longrightarrow \mathbb{R}_{\neq0} \times X_{\neq0} \stackrel{\langle(\cdot)^{-1}, \mbox{id}\rangle }{\longrightarrow} ...

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The map $T$ takes any $x \in X \setminus \{0\}$ to the unit sphere in $X$. Furthermore, it maps the entire ray $\{ r x \mid r>0 \}$ to the same point $T(x)$. It is easy to see that the preimage of any open set in $X$ will either be the empty set or it will be a union of open wedges, so $T$ is continuous.

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Suppose $(x_n,y_n)$ is a sequence in $U$ and $(x_n,y_n)\to (a,b).$ We want to show $(a,b)\in U.$ As we know, $(x_n,y_n)\to (a,b)$ implies $x_n\to a, y_n \to b.$ By the product rule for limits, $x_ny_n \to ab.$ But $x_ny_n \in (-\infty, 1]$ for all $n.$ Because $(-\infty, 1]$ is closed in $\mathbb {R},$ the limit of $x_ny_n$ is in $(-\infty, 1].$ I.e., ...

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Another proof. More elementary than Urban's proof. (But less general.) Consider the subspace $c \subseteq l^\infty$ consisting of the real sequences that converge. So $c$ is a closed linear subspace of $l^\infty$. Let $L \;:\; c \to \mathbb R$ be the "limit" linear functional. That is, for $x = (x_1,x_2,\cdots)$, define $$L(x) := \lim_{n\to\infty} ... 1 This is not true because of the following Theorem \textbf{Theorem:} If X is a normed space such that its dual X' is separable, then X itself is separable. So, if (l^{\infty})' =l_1, then it will follow that l^{\infty} is separable which is not true. 1 Like you did, consider U^c and we'll show it's an open set. take (x,y) \in U^c such that xy= 1+\delta Now, as y\neq 0, x= \frac{1+\delta}{y} Consider the norm \|(x,y)\| = |x|+|y| Take (x',y') such that \|(x',y')\|< \epsilon$$c = (x+x')(y+y') = xy + xy'+x'y+x'y' 1+\delta - \epsilon |x| - \epsilon |y| + \epsilon^2 \leq c $$... 1 Choose a point (x,y)\in \mathbb{R^2} such that xy \gt 1. Then Take \epsilon=\frac{1}{2}min\{|x-\frac{1}{y}|,|y-\frac{1}{x}|\}. Then you will be done. 1 A few comments: 1) you don't need "contradiction". Your argument (if right) shows that if g\in X^*, then g is in the span of F_N. 2) I don't see why you claim that \cap_{f\in F_N}\ker f\subset\ker g implies g\in \text{span}\,F_N. 3) You seem to prove that X=\text{span}\,\{F_N:\ N\} without closure, which looks suspicious to me. 4) Here is ... 0 I assume you've proved \|\,\| is a norm on V. Suppose f_n is Cauchy in this norm. Fix x. Then$$|f_n(x) - f_m(x)| = |f_n(x) - f_m(x)- (f_n(a) - f_m(a))| \le \text {Var} (f_n-f_m). Hence $f_n(x)$ is Cauchy in $\mathbb {R}$ and thus converges. Thus $f_n \to$ some $f$ on $[a,b]$ pointwise everywhere. Now $(f_n)$ Cauchy in this norm implies \$\sup_n ...

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