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Let $\phi:V\to\mathbb{K}$ (where $\mathbb{K}$ is the base field) be a discontinuous linear functional$^{(1)}$. Let $x\in W\setminus 0$. Define $T:V\to W$ by $Tv=\phi(v)x$. Let's show that $T$ is discontinuous: Since $\phi$ is discontinuous, it is unbounded, so there exists a bounded sequence of vectors $\{v_n\}$ in $V$ such that $\{\phi(v_n)\}$ is ...

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The function $\phi .f(x)$ is a continuous function, as a product of continuous functions. Since it is defined on a compact set, it takes on a maximum $M$, then $T_{\phi} \leq \int_0^1 Mdx =M(1-0)=M$ Then $T_{\phi}$ is a bounded linear operator, and so it is continuous. EDIT: As pointed out in the comments, to make this more rigorous ( I took shortcuts ) we ...

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Hard to beat the simple solution of @Keith: We'll give a proof for $X$ infinite dimensional Banach space (extra condition). First, show that there exists a sequence $x_n$ in $X$ such that $||x_n|| =1$ and $d(x_n, \langle x_1, \ldots x_{n-1}\rangle ) \ge \frac{1}{2}$. One constructs the sequence inductively. Once $x_1$, $\ldots x_{n-1}$ are obtained, take ...

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Another way, first we show linearity: $K(f+g)=\int_0^1 k(x,y)f(x,y)dy+ \int_0^1 k(x,y)g(x,y)dy=\int_0^1 k(x,y)(f(x,y)+g(x,y))dy$ , by linearity of the integral. Then we use 1st countability of $C[0,1]$ (since it is a metric space), and show sequential continuity, which is equivalent to continuity: Assume $f_n \rightarrow f$ in $C[0,1]$ , so that $Sup ... 3 let$f\in C[0,1]$and$M=\max_{[0,1]\times [0,1]} k$. Then $$\left|\int_0^1 k(x,y) f(y) dy\right| \le M \int_0^1 |f|\le M |f|_{C[0,1]}$$ Now take$\sup$over$x\in [0,1]$. (Note that the map in question is obviously linear) 0 First, since$V$is finite-dimensional, the image of all transformations in$L(V, W)$have finite rank, and are therefore bounded, so we do have the operator norm here. Otherwise I'd insist that we use$B(V, W)$instead. Additionally, we may assume without loss of generality here that$V$is$\mathbb{R}^n$with the$1$-norm, for some$n$, since every two ... 6 Every infinite-dimensional normed space has a non-closed subspace. Let$X$be an infinite-dimensional normed space, let$a$be a nonzero vector. Assume by induction that we have found vectors$x_1, x_2, \dots, x_{n-1}$for which$|x_i - a| < 1/i$and$a \not\in V_{n-1} = \Sigma_{i=1}^{n-1} \mathbf{R}x_i$. We will extend this sequence by finding an$n$th ... 0 Classically, the approach is to show that$T(B_V)$, where$B_V$is the unit ball of$V$, is bounded in$W$. That is,$T(B_V) \subseteq MB_W$for some$M$. Then, if we have any bounded set$A$, that is$A \subseteq NB_V$, then, $$T(A) \subseteq T(NB_V) = NT(B_V) \subseteq NM B_W,$$ i.e.$T(A)$is bounded, which is what we want to prove. So, by continuity at ... 1 Let$r>$be such that:$ A \subset B_r(0) $. With$B_r(0)$the ball of radius$r$centered in$0$. Suppose$T(A)$is not bounded. Then for each$n \in N$there exists a$x_n \in A$such that:$||Tx_n|| > n+1 $. The sequence$\frac{x_n}{n+1}$converges to$0$because$A$is bounded. But clearly$|| T \left( \frac{x_n}{n+1} \right)|| >1$. This ... 0 It seems the following. For each point$x=(x_n)\in\ell^\infty$put$x_*=\underline{\lim}_{n\to\infty} x_n$and$x^*=\overline{\lim}_{n\to\infty} x_n$. Then both values of$x_*$and$x^*$are finite,$x_*\le x^*$, and$x_*= x^*$iff$x\in c$. The set$c$is closed in$\ell^\infty$. Indeed, let$x=(x_n)\in\ell^\infty\setminus c$. Then$\varepsilon= ...

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a topological vector space that is not a metric space: take $V=C(\Bbb{R})$ where the topology is given by convergence on compact sets. A basis for this topology is given by sets of the form $$U_{K,f,\varepsilon} = \{ g : \sup_K |g-f| < \varepsilon \}$$ where $f \in V$ is continuous, $\varepsilon >0$ is a positive real number, $K \subset \Bbb{R}$ is a ...

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Let $\lambda_a \colon b \mapsto a\cdot b$. Then it's easy to see that the operator norm of the matrix is at most $$\max_i \sum_j \lVert \lambda_{a_{ij}}\rVert_{\operatorname{op}}.$$ For a general Banach algebra, it is possible that $\lVert \lambda_a \rVert_{\operatorname{op}} < \lVert a\rVert$ for some $a$. Still, even if we take the operatornorm of the ...

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An operator norm on $\operatorname{Hom}(V,W)$ is determined by its value on rank-one operators once we know it's an operator norm. Without this additional information, it is not so determined. First, I claim that the operators of rank $\le 1$ form a manifold $R_1(V,W)$ of dimension $$\dim R_1(V,W) = \dim V+\dim W-1$$ Indeed, fix a norm on $V$ and $W$. ...

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If $X$ is a Banach space then a subspace is norm-closed if and only if it is weakly closed. (This is a major difference between weak and weak*). Say $S$ is a weakly dense countable set. Let $V$ be the norm-closed span of $S$. Then $V$ is weakly closed, hence $V=X$. So the linear combinations of elements of $S$ with rational coefficients are dense in $X$, ...

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I think that does not exist. Indeed, if $X$ is a normed space, then $X$ is weak separable if and only if $X$ is norm separable. Proof. "$\Leftarrow$": Let $X$ be norm separable. Let $A$ be countable and norm dense in $X$. Then, we use the fact that the weak topology is indeed weaker than the norm topology to see that $\overline{A}^w\supset ... 0 For a Cauchy sequence$\{x^{(n)}\}_{n\in\mathbb{N}}$, we must show that$x^{(n)}\to x$in$(l^2,||\cdot||_2)$(the space of square-summable sequences under$2-$norm), which means showing$||x^{(n)}-x||_2<\epsilon$. The Cauchy sequence$\{ x^{(n)}\}_{n\in\mathbb{N}}$is bounded, hence you can bound it under$l^2$norm. Use this to conclude that the ... 3 Let$\|\cdot\|$be some C$^*$-norm on$C_c(X)$. Write$A=(C_0 (X),\|\cdot\|_\infty) $,$B=\overline {C_c(X),\|\cdot\|) }$. Fix$f_0\in C_c(X)$. Put$A_0=C^*(f_0)\subset A$, and$\pi:A_0\to B$the identity map (this works because$A_0\subset C_c (X)\subset B $; since we use the supremum norm to generate it, every element has support inside that of$f_0$). ... 2 I am just adding some detailes to the solution of 1999: Summary of the key points: (1)$\|T_{\varphi,w}\|$detrmines$\| \cdot \|_W, \| \cdot \|_{V*}$up to scalar multiple. (2) By a well known corollary of the Hahn-Banach theorem,$\| \cdot \|_{V*}$determines$\| \cdot \|_V$Observation:$\|T_{\varphi,w}\| = \|\varphi\|_{V^*}\|w\|_W$Proof: ... 2 Yes, you can reconstruct original norms up to common scalar multiple. It suffices to consider rank-one operators$T_{\varphi,w}x = \varphi(x)w$where$\varphi$is a linear functional on$V$. The key point is that the operator norm of$T_{\varphi,w}$is$\|\varphi\|_{V^*}\|w\|_W$. Fix one nonzero functional$\varphi$, and define$\| w\|_\# = ...

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Well, the classic way to do this is to use the Hahn-Banach separation theorem, which works in infinite dimensional normed linear spaces, and in fact, locally convex topological vector spaces. But, in the case of finite-dimensions, you can prove it by alternate means. Let $X = \mathbb{R}^n$ under the Euclidean norm, without loss of generality (by which I ...

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It holds true that $q>p\implies \ell^p\subseteq \ell^q$ Indeed, let $(a_n)_{n\in\mathbb N}$ such that $\sum_n |a_n|^p<+\infty$ Since $q>p$, as soon as $|a_n|<1$, which holds from a certain $N$ onwards, $|a_n|^q<|a_n|^p$. So $\sum_{n\in\mathbb ... 0 In general,$\|nx\| = n \|x\|$can't work. In the case of a field of nonzero characteristic, there are only finitely many different values for$n x$. 1 If$\xi\in \ell^p$with$\|\xi\|_p = 1$, then $$\|T\xi\|_p^p = \sum_{i=0}^\infty |\xi_{2i+1}|^p\leqslant\sum_{i=0}^\infty |\xi_i|^p=1,$$ so$T$is bounded and$\|T\|\leqslant 1$. Let$\xi_0=(1,0,0,\ldots)$, then$\|\xi_0\|_p =1$and$T\xi_0=\xi_0$, so$\|T\xi_0\|_p=1$and$\|T\|\geqslant 1$. 1 Fix a norm$||\cdot||$on a vector space$V$. In your previous question, you found that the isometry group$G$of$||\cdot||$also preserves an inner product$<\cdot,\cdot>$. Fix that inner product, then there is a dual norm$||\cdot||_*$given by $$||v||_* = \sup_{||u|| = 1}<v,u>$$ This is really just identifying$V$with its dual via the ... 2 I think the answer is yes and the corresponding inner product is proportionnal to$\langle \cdot , \cdot \rangle$. Let's denote$E$the vector space and$S$the sphere associated to the inner product$\langle \cdot , \cdot \rangle$, i.e. the set$S=\lbrace x \in E \; | \; \langle x , x \rangle =1 \rbrace$. The key for the demonstation is the fact that ... 2 The answer is yes, assuming that the dimension is$\geq2$. Define$$f:V\to\mathbb{R},\quad y\mapsto\|x+y\|-\|x-y\|.$$The function$f$is clearly continuous. Pick any$0\neq y_0\in V$. If$f(y_0)=0$, we're good. Otherwise, take some path$\gamma:[0,1]\to V$connecting$y_0$with$-y_0$, without passing through$0$. Since$f(\gamma(0))$and$f(\gamma(1))$have ... 1 If$i$is an embedding, we can consider$y=i(x)$as$x$. This just means "if we relabel$y$and call it$x$instead, we actually do have$X\subset Y$". Of course, it still has the usual properties of$X$as well$y_1=i(x_1),y_2=i(x_2)\implies y_1+y_2=i(x_1+x_2)$and similarly for scalar multiplication. The reason why this is valid is because$i(x)=i(x')$... 0 I suppose here that$\frac{1}{p} + \frac{1}{q} = 1$. Somehow, the$\frac{1}{p}$and$\frac{1}{q}$terms are annoying, so you want to get rid of them knowing that the sum of the inverse is$1$. 4 Note that$T$being continuous at$x_0$means, by definition: for any$\epsilon > 0$, there exists a$\delta > 0$such that $$\|y\|< \delta \implies \|T(x_0 + y) - T(x_0)\| < \epsilon$$ By linearity, we can rewrite this as $$\|y\|< \delta \implies \|T(0 + y) - T(0)\| <\epsilon$$ So,$T$is continuous at$x_0$(an arbitrary point) if ... 0 Let$x \in V$be arbitrary and define a sequence$\{p_n\}$such that$p_n \to x$. Now define for each$n \in \mathbb{N}$the sequence$\{a_n\}$where$a_n = p_n + x_0 - x$, certainly we have that$a_n \to x_0$. Now by continuity$T(a_n) \to T(x_0)$but observe that we have $$T(a_n) \;\; =\;\; T(p_n) + T(x_0) - T(x) \;\; \to \;\; T(x_0).$$ 0 Prove that the operator$T$is continuous if and only if $$T':x\mapsto T(x-x_0)$$ is continuous. Assume$T$is continuous. Notice the translation operator is continuous. So$T'$is the composition of continuous operators. The same works to prove the other direction. 1 Every vector space$M$has a so called Hamel basis. Namely, a set$ H = \{ v_{\lambda} \in M : \lambda \in \Lambda \}$such that every$v \in M$has a unique expression as finite linear combination of elements of$M$: $$v = \sum_{j=1}^{n(v)} x_j \, v_{\lambda_j} \, .$$ If$M$has infinite dimension then the index set$\Lambda$has infinite elements so ... 2 You are right.$L$is (obviously) linear and well-defined. To compute its norm, let's denote the natural projection$N \to N/K$by$\pi$. Then we have$T = L\pi$, hence $$\def\n#1{\left\|#1\right\|}\n{T} = \n{L\pi} \le \n L \n\pi \le \n L$$ as$\n\pi \le 1$. On the other hand, for$x \in N$and$\epsilon > 0$choose$k \in \ker Twith \n{x+k} \le ... 1 So far you are correct, but you need to justify \lim(x_n + M) = M! That is what you want to prove, you cannot use it. To prove the norm-bound (and hereby prove continuity), do as always. First recall the definition of the norm in N/M: \def\norm#1{\left\|#1\right\|}\norm{x+M} := \inf_{m\in M} \norm{x+m} As 0 \in M, we have \begin{align*} ... 1 \left \| x \right \|\leq \left \| x-y \right \|+\left \| y \right \| \left \| y \right \|\leq \left \| x-y \right \|+\left \| x \right \| so \left \| x \right \|-\left \| y \right \|\leq \left \| x-y \right \| and \left \| y \right \|-\left \| x \right \|\leq \left \| x-y \right \| which is what you want. 3 By triangle inequality:\|x\| = \|x - y + y\| \leq \|x - y\| + \|y\|$$implies \|x\| - \|y\| \leq \|x - y\|. And$$\|y\| = \|y - x + x\| \leq \|y - x\| + \|x\|$$implies \|y\| - \|x\| \leq \|y - x\| = \|x - y\|. Therefore$$\left|\|x\| - \|y\|\right| \leq \|x - y\|.$$1 Hint: Prove that if K, Y are metric spaces, with K compact space, then every f:K\to Y continuous is uniformly continuous. 2 Because continuity on a compact is the same as uniform continuity. Note that sequential compactness is the same as compactness. Here's a link to a proof. 0 Just use the definition: If S_{n}=\sum _{k=0}^{n}x^{n} converges, then it is Cauchy so for all \epsilon >0 there is an integer N such that whenever n,m\geq N, \vert S_{n}-S_{m}\vert <\epsilon. For this N, take n=m+1 and then \vert S_{m+1}-S_{m}\vert=\vert x_{m+1}\vert and this is <\epsilon . 0 Let the sum be s; then, for every \varepsilon>0, there exists m such that, for every n>m,$$ \biggl\|\sum_{i=1}^n x_i-s\biggr\|<\varepsilon $$By the triangle inequality, for n>m,$$ \|x_{n+1}\|= \biggl\| \biggl(\sum_{i=1}^{n+1} x_i-s\biggr)+ \biggl(s-\sum_{i=1}^{n} x_i\biggr) \biggr\| \le \biggl\|\sum_{i=1}^{n+1} x_i-s\biggr\|+ ... 0 Following your notation, lets_n = \sum_{i=1}^n x_i$. Assuming the series converges,$s_n$is Cauchy. Rewriting this, for any$\varepsilon > 0$, there exists$N \in \mathbb{N}$such that for all$m \geq n > N$, we have $$\sum_{i=n}^m ||x_n|| < \varepsilon.$$ In particular, setting$m = n$shows that$x_n < \varepsilon$for all$n \geq N$, as ... 3 Hint Try letting$p=1$. ........... 0 Hint Pointwise limit of differentiable functions needn't be differentiable. 1 Converting comments to answers: This is true in any normed space. (In fact, it is true in every metric space that spheres are closed.) A variant of the triangle inequality is the fact that $$\bigg| \|x\| - \|y\| \bigg| \le \|x-y\|. \tag{1}$$ To prove this, use the triangle inequality to see that $$\|x\| = \|x - y + y\| \le \|x-y\| + \|y\|$$ which shows ... 1 As the commenters said, this is true because the norm is a continuous function. Forget Hilbert space and prove that the level set of any continuous function on a topological space is a closed set. (Think in terms of preimage). 1 Yes: this is the same as proving that for a real sequence$(a_n)_k$there is a subsequence$(a_{n_k})_k$such that $$\lim_{k} a_{n_k} = \liminf_n a_n = l,$$ say. To do this, we basically imitate one of the proofs of Bolzano–Weierstrass (the "high points" one) (or see here): pick a sequence$\varepsilon_k \to 0$. Then for each$\varepsilon_k$, there are, ... 2 Yes, and this holds more generally for real sequences. In fact, the limit inferior of a sequence can be defined as the least limit point of any subsequence of the sequence. To prove that such a subsequence exists, recall that, $$\liminf\limits_{n\rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \inf\limits_{k \ge n} a_k$$ Let the limit inferior be$L$... 2 Your question can be greatly reduced. The normed space and its norm can be "forgotten" because both the limit and the$\liminf$are applied to sequences of real numbers ($||x_n||$is a real number!), and the question simply becomes: For a given sequence of positive real numbers$(a_n)_{n\in \mathbb N}\$, does there exist a subsequence ...

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Every Banach space is a metric space. However, there are metrics that aren't induced by norms. If this were the case, then Banach spaces and complete metric spaces would the same thing... if metric spaces had operations and an underlying field! The difference is more than just the metric/norm dichotomy. For completeness of the answer (no pun intended), a ...

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A complete metric space need not be a complete normed space. A complete normed space is also called a Banach space! Since every norm induces a metric, these Banach spaces reside in the collection of all complete metric spaces. Thus, complete metric space and complete normed space - two different notions but related indeed.

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