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Sketch of proof. Suppose you have a Cauchy sequence $\{f_n\}\subset C_b^k(U)$, with respect to $\|\cdot\|_k$. Then, for every $\lvert\alpha\rvert\le k$, the sequence $\{D^\alpha f_n\}$ is uniformly Cauchy in $U$ and hence uniformly convergent to a continuous and bounded function, say $f^\alpha\in C_b(U)$ - All the $f^\alpha$'s are for the moment only ...

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This is obvious. Note that $m(r)$ is not only an infimum, but a minimum. This means that you can cover $B$ with $m:=m(r)$ suitably placed balls of radius $r$. When $r'>r$ then $m$ balls of radius $r'$ with the same centers will obviously cover $B$, but there might be a better covering with balls of radius $r'$. So $m(r')\leq m(r)$. (You cannot hope for ...

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Please refer to the exercise 7 on page 155 of Folland's real analysis. You will find more useful information. Let $X$ be a Banach space. If $T\in L(X,X)$ and $\|I-T\|<1$ where $I$ is the identity operator, then $T$ is invertible; in fact, the series $\sum_0^\infty(I-T)^n$ converges to $(I-T)^{-1}$. If $T\in L(X,X)$ is invertible and ...

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Visualizing such things might prove hard, since we're dealing with vectors in $n^2$ dimensions for some $n$. The theorems are important since they provide, along with the contraction principle, a very good proof of the inverse functions theorem which Rudin gives in full detail. To prove the first theorem, we start with $\lVert A-B\rVert \rVert ... 0 If you do not assume continuity, for what I have said in the comment, you can not put this definition, that is,$T$, not being bounded (on$X$) and discontinuous at each point in$||x|| \leq 1$, will not have in general the$\sup$. 1 If the space$X$is banach it is an easy consequences of the open mappig theorem. Anyway with the norm induced topology over$ X $you in fact are resizing a ball so it is a ball again and it is open by definition of the topology. So the map sends open ball in open ball therefore it is open. This reasoning heavily rely on the "absolute omogeneity" of the ... 1 Alternative proof by contradiction. Suppose for all$k$there is some$v_k\neq 0$such that$\|T v_k \| \ge k \|v_k\|$. Let$x_k = {1 \over k} { v_k \over \|v_k\|}$, and note that$\|T x_k \| \ge 1$, but$x_k \to 0$, which contradicts continuity. 0$T$is continuous. Let$\epsilon = 1$. We can find$\delta$so that$||Tv|| < 1$when$||v|| < \delta$. Now,$||Tv|| = || \frac{||v||}{\delta} T(\frac{\delta v}{||v||})|| \leq \frac{||v||}{\delta}$. Take$c = \frac{1}{\delta}$. 0 The fact that$\| \pi(x) \| \le \|x\|$for all$x \in X$tells you that$\|\pi\| \le 1$as$\|\pi\| := \sup\{\|\pi(x)\| : \|x\| \le 1 \}$. If you pick a point$z \in M$then$\|\pi(z)\| = \|z\|$and so$\|\pi\| = 1$. 0 You have$\pi^2 = \pi$, so$||\pi|| = ||\pi^2|| \le ||\pi||^2$, implying that$||\pi|| \ge 1$. Your statement implies that$||\pi|| \le 1$. 2 If$null(f)$is not dense in$X$, you can find$x\in X$and$r>0$such that$B(x,r)\cap null(f)=\varnothing$. if$y\in X$is such that$|f(y)|\geq|f(x)|$, then for some$\alpha$with$|\alpha|\leq 1$we have$f(\alpha y)=f(x)$, so$x-\alpha y\in null(f)$, hence$x-\alpha y\not\in B(x,r)$, so$\Vert y\Vert\geq\Vert\alpha y\Vert\geq r$. What this just said ... 1 Try this:$null(f)$is a subspace of$X$whose codimension is 1. Now if it is not dense, then$null(f)$is closed (because its closure is a subspace containing$null(f)$and it is not$X$). Then you show that it implies continuity. For example, as$f\neq 0$there is a$y\in X$such as$f(y) = 1$. $$\{x\in X| |f(x)| = 1 \}= (y + null(f))\cup (-y + null(f)) ... 0 You did the hard part, namely (ii) \Longrightarrow (iii). Now u is continuous means every sequence x_n \to x \in E must have u(x_n) \to u(x) \in F. Now use (iii) combined with linearity, (i) will be immediate. Part (b) is also not so hard if you use the fact that every norm on a finite dimensional vector space is equivalent. Try proving it ... 2 Fix a basis v_1,\ldots,v_m \in V_k. Then the map \sum \alpha_jv_j\to\left(\sum |\alpha_j|^2\right)^{1/2} defines a norm on V_k, and this norm is induced by the inner product \langle \sum \alpha_jv_j,\sum \beta_jv_j\rangle = \sum \alpha_j\overline{\beta_j}. In a finite-dimensional space all norms are equivalent, so the identity map becomes a ... 3 The Banach function spaces may be a better choice for your question. You can see here for the definition of a Banach function space or this book Function Spaces, Volume 1 (chapter 6). Answer to your question: The first thing is to introduce the definition of the Riesz-Fischer property. We say that a normed linear space (X,\left\|\cdot\right\|) has ... 3 Yes. The isomorphism is given by:$$\begin{array}{llll} \varphi:&C^0[a,b]&\longrightarrow&C^0[0,1]\\ &f(x)&\longmapsto&(b-a)f(a+(b-a)x) \end{array}$$0 Let S be the set of all tuples of scalars (\alpha_1,\dots,\alpha_n)\in K^n such that |\alpha_1|+\cdots+|\alpha_n|=1. Then the map f:S\rightarrow [0,\infty) given as$$f(\alpha_1,\dots,\alpha_n)=\|\alpha_1 u_1+\cdots+\alpha_n u_n\|$$is continuous. Because S is compact, the continuous function f attains its infimum on S at some x_0\in S, ... 3 For each k,$$\left\lVert\sum_{n=1}^{k}f_n\right\rVert \leqslant \sum_{n=1}^{k}\lVert f_n \rVert$$Then take k\to\infty. 1 The answer is yes. For every normed space X we can define as \alpha, the cardinality of the set of minimum cardinality which is dense in X. Clearly, \alpha\ge\aleph_0. For every such set D, described in the question, we shall show that \lvert D\rvert=\alpha. It is not hard to see that the set E=\bigcup_{n\in\mathbb N} 2^{-n}D is dense in ... 1 Let X, Y be Banach spaces and U\subset X open. A function u\colon U\to Y is said to be differentiable at a\in U if there exists a linear operator Du_a\colon X\to Y (the differential of u at a) such that$$ u(a+h)=u(a)+Du_a(h)+o(h). $$If u is continuous, then Du_a is also continuous. If u is continuous and differentiable at very point ... 0 I don't know if you can construct a sequence as you've proposed above, but I do know that the inequality you proposed is sharp. Proof of this is neatly presented in Becker's "Inequalites in Fourier Analysis'' (1975): http://www.jstor.org/stable/1970980 The sharp Young's inequality reads:$$\|f \ast g \|_r \leq \left( A_p A_q A_{r'} \right)^n ... 1 If$V$is not separable, then$L^p(X,\mu,V)$is not either. Take$\{v_j\}$an uncountable set in$V$without a limit point and$f_j\in L^p(X,\mu,V)$, such that$f_j(x)=\varphi_j(x)v_j$, where$\varphi_j\ne 0$scalar. Clearly, there is no limit point in$\{f_j\}$. If$X$is separable, then again it is not certain that$L^p(X,\mu,\mathbb R)$is separable. It ... 3 Complementing the very well detailed answer of Zev Chonoles, I think you've got it in the opposite direction.$T$as is defined in the title is the mapping$x\mapsto T(x)=x^T$, such that$T(x)(y)=x^Ty$is a linear transformation of the vector$y$. This means that the domain of$T$is$(\mathbb R^n, \Vert \cdot \Vert_\infty)$and its codomain is the dual ... 1 A pair$(V,\|\cdot\|)$denotes a vector space$V$over$\mathbb{R}$, together with a norm function$\|\cdot\|:V\to\mathbb{R}$. Thus,$(\mathbb{R}^n,\|\cdot\|_1)$and$(\mathbb{R}^n,\|\cdot\|_\infty)$mean "the vector space$\mathbb{R}^n$equipped with the$L^1$norm", and "the vector space$\mathbb{R}^n$equipped with the$L^\infty$norm", respectively. ... 0 I don't think you're going to get anywhere this way- it seems as though what you're trying to prove at the end there is stronger than what is really true! Let$Conv(A)$denote the convex hull of$A$, which for me will be the set of all finite convex combinations of elements of$A$. We need not worry about closures: the Hausdorff distance only sees closures, ... 0 I thought that I could use that, since all the norms are equivalent in any finite dimensional vector space, the convergence with the norm$\lVert.\rVert_\infty$is equivalent to the convergence with the norm$\lVert.\rVert$, defined in 1. This way, I could argue that the convergence with$\lVert.\rVert$is equivalent to the convergence of every addend in ... 0 This is a consequence of the Uniform Boundedness Principle, which hold for a Banach space, and not a normed one. However, you can think of$S$as a subset of$X^{**}$which is indeed a Banach space. 1 It is true if you assume that$T$is self-adjoint (i.e. symmetric), meaning that $$(Tx,y)=(x,Ty), \quad \text{for all}\,\, x,y\in H, \tag{1}$$ and assuming that $$|(Tx,x)|\le \|x\|^2, \quad \text{for all}\,\, x\in H.\tag{2}$$ Note that your inequality holds even for$T=-2I$, and thus we NEED to assume these two additional things:$(1)$and$(2)$. So ... 0 Let$(V,\mid\mid\cdot\mid\mid)$be a normed vector space over$\mathbb{R}$; then the closed unit ball$\overline{B}(0 , 1)$is convex. Proof: Clearly$\overline{B}(0 , 1)$is non empty, so let$x,y\in\overline{B}(0 , 1)$and let$\lambda\in[0,1]$. Consider$z=\lambda x+ (1-\lambda)y$Now; $$\mid\mid z\mid \mid= \|\lambda x+ (1-\lambda)y \| \leq ... 0 We simply use the triangular inequality repeatedly:$$ \|u\|=\|(u-v)+v\|\le \|u-v\|+\|v\|, $$and thus$$ \|u\|-\|v\|\le \|u-v\|\le \|u\|+\|-v\|=\|u\|+\|v\|. \tag{1} $$Similarly$$ \|v\|=\|(v-u)+u\|\le \|v-u\|+\|u\|, $$and thus$$ \|v\|-\|u\|\le \|u-v\|. \tag{2} $$Now (1) and (2) imply that$$ \big|\|u\|-\|v\|\big|\le \|u-v\|\le \|u\|+\|v\|. $$5 The first$$||u-v||=||u+(-v)||\le ||u||+||-v||=||u||+||v||$$and the second$$||u||=||u-v+v||\le ||u-v||+||v||\Rightarrow ||u||-||v||\le ||u-v||$$and by symmetry we have the other inequality so we conclude. 0 Hint: \left\|u\right\| = \left\|(u-v)+v\right\| \left\|v\right\| = \left\|(u-v)+u\right\| \left\|u-v\right\| = \left\|u+(-v)\right\| 0 For any f \in C(\mathbb{R}), the L^{\infty} norm of f is the same as the C(\mathbb{R}) norm of f. So the closure of C(\mathbb{R}) in L^{\infty}(\mathbb{R}) consists of all functions which are equal a.e. to continuous functions. Choose any function in L^{\infty} which is not equal a.e. to a continuous function, and that function cannot be the ... 5 The simplest counterexample is the non-zero constant function$$ f(x)=1. $$If g\in C_0^1(\mathbb R), then \lim_{|x|\to\infty}g(x)=0, and hence$$ \lim_{x\to\infty}|f(x)-g(x)|=\lim_{x\to\infty}|1-g(x)|=1. $$Thus$$ \|f-g\|_\infty=\sup_{x\in\mathbb R} |f(x)-g(x)|=1,$$and therefore$f$can not be approximated by$C_0^1\$ functions.

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The inequality is easy. For the equality, use Hahn-Banach.

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