Tag Info

Hot answers tagged

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 ...


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 $\|\cdot\|_1$.


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.


3

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 ...


3

So I'm just elaborating on the comment by cobber.hat. Take $x = (1,0,0,\ldots)$ and $y = (0,1,0,\ldots)$. then: $$\lVert x+y\lVert^2_\infty =1 \qquad \lVert x-y\lVert^2_\infty =1 \qquad \lVert x\lVert^2_\infty =1 \qquad \lVert y\lVert^2_\infty =1$$ Thus we have a counter example to your formula.


3

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 ...


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 ...


2

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 ...


2

Suppose that $\{x_k\}$ is a sequence in $B_1(0)$ that converges to a point $x \in V$. Then $$\|x\| \le \|x - x_k\| \le \|x_k\| < 1 + \|x - x_k\|$$ for all $k$. Take the limit as $k \to \infty$ to conclude that $\|x\| \le 1$. Thus $$ \overline{B_1(0)} = \{x \in V : \|x\| \le 1\}.$$ Can you show that $B_1(0)$ is open? Once you have established that, use ...


2

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 ...


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 ...


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 ...


2

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 ...


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 ...


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 ...


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 $f$ ...


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 $t$, ...


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

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 ...


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

Hint. It suffices to prove the following: $$ \|x_n-x\|\to 0 \quad\text{iff}\quad \|x_n-x\|_1\to 0. $$


1

For $x,y,z \geq 0$, note that, $\frac{x}{1+x+y}\leq \frac{x}{1+x}$ and similarly $\frac{y}{1+x+y}\leq \frac{y}{1+y}$. Adding them you will get $\frac{x+y}{1+x+y}\leq \frac{x}{1+x} +\frac{y}{1+y}$. Now using triangle inequality, we have $|a_n - b_n|\leq |a_n-c_n| +|c_n-b_n|$. Now using the fact that $\frac{t}{1+t}$ is an increasing function in $t$,we have ...


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.


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

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

The space $M(Z)$ consists of all random variables $Y$ such that there exists a Borel-measurable function $\phi\colon\mathbf R\to\mathbf R$ such that the equality $Y=\phi(Z)$ holds, and $Y$ belongs to $\mathbb L^2$. Let $(Y_n)_{n\geqslant 1}$ is a convergence sequence in $L^2$ of elements of $M(Z)$ to some $Y$. For each $n$, we may write $Y_n=\phi_n(Z)$ ...


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} ...


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 $||\cdot ||_{\infty}$


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

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible