# Tag Info

0

If one of the members of the product was NOT invertible,it would map a sequence of vectors, that doesn't converge to zero, to a sequence converging to zero.But then the product would also do that, making the product non-invertible.

1

No. It is not true. Let $A$ be the unit ball centered in the origin, let $D$ be the points with rational coordinates. Let $U$ be the complement of $A$.

0

It is only true that $U\cap D$ and $A\cap D$ are non-empty. $A\cap U$ can happen to be the empty set. Why $U\cap D$ is nonempty: Because $U$ is open, then for each $u\in U$ there is a ball $B(u,r)$ with center $u$ and some positive radius $r>0$, such that $B(u,r)\subset U$. Now, because $D$ is dense in $\mathbb R^n$ it follows that for each ...

1

Take $V = \mathbb R^2$, $A$ a line segment, $D = V \backslash A$.

0

Take $V=C[0,1]$ with the uniform norm, $A$ is the subspace of all polynomials (convex), $D=C[0,1]\setminus A$ (dense). But $A\cap D=\emptyset$.

0

First, $U\cap D\cap V=U\cap D$ because $U\subset V$. Second, any non-empty open subset of $V$ intersects $D$, precisely because $D$ is dense in $V$ (this is a definition of dense subset). So yes, $U\cap V\cap D$ is non-empty. Note that we have only used the topology of $V$. There's no need of linearity, convexity, norm, etc.

0

Yes. It suffices to consider the case $n=2$ and here to show that there exists a path from $(0,0)$ to $(1,0)$ (assuming wlog. that these points are in the set). The obvious attempt is the curve $\gamma\colon [0,1]\to\mathbb R^2$, $t\mapsto (t,0)$, but for each disk it intersects we need to take a detour along the boundary of the disk. So try this $t\mapsto ... 4 Let$Q=S-T$. By assumption,$\langle Qv,v \rangle =0$for all$v=\alpha x + \beta y$. Now, $$0=\langle Q(\alpha x+y),\alpha x+y \rangle = |\alpha|^2 \langle Qx,x \rangle + \langle Qy,y \rangle + \alpha \langle Qx,y \rangle + \bar{\alpha} \langle Qy,x \rangle \\ = \alpha \langle Qx,y \rangle + \bar{\alpha} \langle Qy,x \rangle.$$ Choosing first$\alpha =1$... 1 By considering$S-T$, it suffices to show that if$(Sz,z)=0$for all$z$, then$S=0$. To show that$S=0$, it suffices to show that$(Sx,y)=0$for all$x,y\in X$. Now think about what$(Sz,z)=0$tells you when$z=ax+by$is a linear combination of$x$and$y$. By varying$a$and$b$, can you show that$(Sx,y)$must vanish? 0 The vector space of all polynomials of degree at most$n$is isomorphic with$\mathbb{R^{n+1}}$and is complete for every natural number. 1 As pointed out in a comment, the question doesn't quite make sense because$f\in BMO$is only defined almost everywhere. But it seems to me that in fact there does exist a continuous$f\in BMO(\mathbb R)$with small norm mapping$\mathbb R$onto$S$. I'm going to write$||f||_*$for the BMO seminorm. First a reduction to something a little simpler: Lemma: ... 0 The first axiom for a norm space is N1:$\left\Vert \mathbf{x}\right\Vert =0\implies \mathbf{x}=\mathbf{0}$. Now, if we take this away and have the other two viz. homogenity and the triangle inequality, then we're left with what is called a seminorm. Any vector space over$\mathbb{R}$or$\mathbb{C}can be converted into a seminorm space as follows: take ... 0 We have by definition $$\mathbf{x} = \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots\\ x_n \end{array} \right]$$ and $$\mathbf{x}^T = \big[ x_1, x_2, \cdots, x_n \big]$$ so we get \mathbf{x}^T \mathbf{x} = \big[ x_1, x_2, \cdots, x_n \big] \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots\\ x_n \end{array} \right] = x_1^2 + x_2^2 + \cdots + ... 0 The question does not make completely sense as it is now, but I think this should help. \begin{align} \sum\limits_{i=1}^n (x_i-c_i)^2 &= \sum\limits_{i=1}^n (x_i^2 + c_i^2 -2x_i c_i)\\ &=\sum\limits_{i=1}^n x_i^2 +\sum\limits_{i=1}^n c_i^2 - 2\sum\limits_{i=1}^n x_i c_i \\ &=\mathbf{x}^T \mathbf{x} + \mathbf{c}^T\mathbf{c} - ... 0 The left hand side equals to\rho^2\|x\|_2^2 + (1 - \rho)^2\|y\|_2^2 + 2\rho(1 - \rho)x \cdot y.While the right hand side equals to \begin{align*} & \rho\|x\|_2^2 + (1 - \rho)\|y\|_2^2 - \rho(1 - \rho)(\|y\|_2^2 + \|x\|_2^2 - 2y\cdot x) \\ = &[\rho - \rho(1 - \rho)]\|x\|_2^2 + [(1 - \rho) - \rho(1 - \rho)]\|y\|_2^2 + 2\rho(1 - \rho)x \cdot y ... 1 If an inverse of any kind exist, T is a bijection. As a consequence of the open mapping theorem, a bijective operator is bounded from below, meaning that there is c>0 such that \|Tx\|\ge c\|x\| for all x. This and the property TS=I imply that S is bounded. 0 All norms on a finite dimensional vector space are equivalent so the norm on (\Bbb{R}^n)^{\ast} is BOUNDED BY (thanks Ian) a constant times any other norm on (\Bbb{R}^n)^{\ast}. Practically there's little need to know exactly what the constant is, but the dual norm of any l^p norm is the q norm where \frac1p+\frac1q=1. Practically though, it's ... 1 I think the Theorem works better if we write L=\{Tx:\ \|x\|<1\}. Note that B_x(\delta)=x+B_0(\delta)=x+\delta\,B_0(1). So TB_x(\delta)=Tx+\delta\,TB_0(1)= y+\delta L. $$By the Theorem, there exists r>0 with B_0(r)\subset L. Then$$B_y(\delta r)=y+\delta B_0(r)\subset y+\delta L=TB_x(\delta).$$So V is open. 2 Actually, for any locally compact Hausdorff X, and Radon measure \mu on X, the set of all continuous compactly supported function C_c(X) is dense in L^p(X) for all 1\leq p<\infty (theorem 3.14 of "Real and complex analysis" by Walter Rudin). Hence C_c(X)\subset L^1(X) \cap L^p(X), and so L^1(X) \cap L^p(X) is dense in L^p(X), for all ... 2 It's not. Consider f_{n}(x)=\frac{1}{x^{1+1/n}}\in L^1([1,\infty))\cap L^2([1,\infty))\subset L^2([1,\infty)). Then:$$\|f_n(x)-\frac1x\|_2=\int_1^{\infty}\left(\frac{1}{x^{1+1/n}}-\frac{1}{x}\right)^2dx=\frac{1}{2/n+1}-\frac{2}{1/n+1}+1\to 0$$as n\to\infty. However \frac1x\not\in L^1([1,\infty)). 3 First, notice that H is linear, so you only need to prove linearity at 0. And you have this inequality :$$| H(f) | = |f(1) - f(0) | = \left| \int_0^1 f'(t) dt \right| \leq \int_0^1 | f'(t) | dt \leq \| f\|$$So clearly, H(f) \to 0 when f \to 0 : H is continuous at 0, and by linearity, everywhere 1 Thr natural thing is to prove the contrapositive. If T is not bounded, there exists a sequence \{x_n\}_X with \|x_n\|=1 and \|Tx_n\|>n^2. Then x_n/n is a sequence that converges to zero with its image through T unbounded. Conversely, if x_n\to0 with \{Tx_n\} unbounded, then T is unbounded. 2 The separable case still requires some form of the Axiom of Choice, although it's less clear why. Say (x_n) is a dense sequence of elements of X. Say Z_n is the span of Z and x_1,\dots,x_n. Now you simply extend your functional to Z_1, then to Z_2, etc. You find you've extended it to the union of the Z_n, which is dense in X, and now ... 1 We have = if and only if x and each row of A are linearly dependent. This follows directly from Cauchy Schwarz inequality. Proof: Let a_i^T denote the i-th row of A. Then, we have$$ \| Ax \|^2 = \sum_{i=1}^m (a_i^T x)^2 \le \sum_{i=1}^m \|a_i\|^2 \|x\|^2 = \|A\|_{\mathrm F}^2 \| x \|^2, $$where the inequality follows from Cauchy Schwarz. Now, ... 1 No it doesn't hold in L^1. Take f(x)=g(x)=\frac{1}{\sqrt{x}} for x \in (0,1) and f(x)=g(x)=0 elsewhere. \Vert f \Vert_1=\Vert g \Vert_1=2 but \int fg =+\infty. 2 Let me show you another way of proving that (\ell^{\infty}, d_{\infty}) is not separable. The proof is specific to \ell^{\infty} but it is a cute trick, nonetheless. Assume, to the contrary, that there is a countable dense set S in \ell^{\infty}. Enumerate S into a list, i.e. S=\{x_1, x_2, x_3, …\}. Since x_i\in\ell^{\infty}, we can write ... 2 If X is separable, then all uncountable subsets of X contain a point of accumulation. As a result, A contains a point of accumulation, say x_0. Consider B_d[x_0, \epsilon]. Then, \left( B_d[x_0, \epsilon] \setminus \{x_0\}\right) \cap A \neq \varnothing (since x_0 is a point of accumulation). Hence, there's y_0 \in A, y_0 \neq x_0, such ... 1 Hint: if D is dense in X, given x \in A, we must have D \cap B(x, \epsilon/2) \neq \varnothing. Take a point there and call it f(x). We have a map f: A \to D, then. Prove that f is injective, and conclude that D is not countable. Since D was an arbitrary dense set, X is not separable. 0 Yes, your reasoning is correct. 0 If T=0, then the adjoint T^\times is also equal to zero and the inequality is trivial. If T \neq 0, there exists x_0 with Tx_0 \neq 0. Theorem 4.3.3 is then applied to Tx_0. 1 For the first part, see this question: Assume T is compact operator and S(I- T) = I .Is this true that (I- T)S =I? For the second, let A = I - (I-T)^{-1}, then A(I-T) = -T and so A = -(I-T)^{-1}T. Since T is compact, so is this product. 0 Note that:$$I-(I-T)^{-1} = I-(I-T+T)(I-T)^{-1} = I-I-T(I-T)^{-1} = -T (I-T)^{-1}$$A compact operator composed (on the right or on the left) with a bounded operator is still compact. But -T is compact, and (I-T)^{-1} is bounded... 1 Here are a few things that fail: Distributive law: In general, (\alpha + \beta)A \ne \alpha A + \beta A. Examples: A=\mathbb N, \alpha=1, \beta=1: (\alpha + \beta)A = 2\mathbb N = \{0,2,4,6,8,\ldots\} \alpha A + \beta A = \mathbb N + \mathbb N = \mathbb N A=\mathbb N, \alpha=1, \beta=-1 (\alpha + \beta)A = 0\mathbb N = \{0\} ... 2 First, you can forget S,T,U, and V. Say A is that 2\times 2 matrix with operator entries. Suppose you could prove$$||\alpha F||\le||\alpha||\,||F||\quad(i)$$for all F\in L^p\oplus L^p. Then it would follow that$$||(\alpha A)F||=||\alpha(AF)||\le ||\alpha||\,||AF||\le||\alpha||\,||A||\,||F||,$$which is exactly what you want. So you only need to ... 0 If you use the operatornorm with respect to the Euclidean norm on \mathbb{C}^m, you will end up with isoemtrically isomorphic spaces: Both M_{kn}(\mathbb{C}) and M_n(M_k(\mathbb{C})) are obviously complete in this norm and the norms satisfy the C^\ast-property: \|x^\ast x\|=\|x\|^2 in both cases. But such a norm is uniquely determined. 0 Let f(x)=y where y(t) = x\left( \dfrac{t-a}{b-a} \right)\cdot\dfrac 1 {\sqrt{b-a}} for a\le t\le b. Then$$ \|y\|_2^2=\int_a^b (y(t))^2\,dt=\int_a^b \left(x\left( \frac{t-a}{b-a} \right)\right)^2 \frac{dt}{b-a} = \int_0^1 (x(u))^2\,du = \| x \|_2^2 $$Let g(x)=z where z(t)= x\left( \dfrac{t-a}{b-a} \right) Then$$ \|z\|_\infty = \sup_{a\le t\le ... 3 Consider a functiong\in C[0,1]$. Let$f[g]\in C[a,b]$be defined by$f[g](x)=\frac{1}{\sqrt{b-a}}g(\frac{x-a}{b-a})$. You should be able to prove that's an isometry just by u-substitution! The basic idea is, you translate/stretch the function so that it covers the new interval, then renormalize it to give it the same norm as before. You can apply the same ... 1 You have that $$d(x,z)^2 = \langle x-z, x-z \rangle = \langle x-y+y-z, x-y+y-z \rangle$$ $$= \langle x-y, x-y \rangle + 2 \langle x-y, y-z \rangle + \langle y-z, y-z \rangle$$ $$= d(x,y)^2 + 2 \langle x-y, y-z \rangle + d(y,z)^2$$ Then by Cauchy-Schwarz, $$\leq d(x,y)^2 + 2 d(x,y)d(y,z) + d(y,z)^2$$ $$\leq \left( d(x,y) + d(y,z) \right)^2$$ 3 Hint. For a) Develop$P(x_1,\dots, x_n)$for the first variable knowing that $$x_1=\sum_{i_1=1}^{n_1} x_1^{i_1} e_{i_1}$$ and using linearity of$P$for the first variable. You get $$\vert P(x_1,\dots, x_n) \vert= \left\vert P(\sum_{i_1=1}^{n_1} x_1^{i_1} e_{i_1},x_2, \dots,x_n) \right\vert =\left\vert \sum_{i_1=1}^{n_1} x_1^{i_1} P(e_{i_1}, x_2, \dots, ... 2 \def\norm#1{\left\|#1\right\|_1}\def\abs#1{\left|#1\right|}As you write correctly, we have$$ \norm{Ax} = \norm{\sum_{i=1}^n x^i Ae_i} \le \sum_{i=1}^n \abs{x^i} \norm{Ae_i} $$Now, note that for every i, we have$$ \norm{Ae_i} \le \sup_{1\le j \le n} \norm{Ae_j} $$Let's call the supremum S, then \norm{Ae_i} \le S for all i, giving above$$ ... 1 For example, consider the norm $$\|(x,y)\| = x ^2- xy + y^2$$ We note that $$\|(2,0)\| > \|(2,1)\|$$ A class of norms that act the way you might expect is the set of "symmetric gauge functions", as referenced here. 0 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 ... 0 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 ... 0 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 ... 1 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 ...

Top 50 recent answers are included