# Tag Info

1

As pointed out by Clement C in the comments: Equivalent norms induce the same topology. Also the other direction of implication is true: When two norms induce the same topology then they are equivalent. Take two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space and you ask yourself: When are the topologies from those norms the same? This is the case ...

1

Some application of this result and more generally that all norms are equivalent on finite dimensional spaces: The compact subspaces are the closed bounded spaces. All linear maps are continuous. More generally all multilinear maps are continuous. All linear maps are bounded on the unit ball.

0

A random variable $X$ is symmetric if for each Borel set $B$ we have $\mathsf{P}(X\in B) = \mathsf{P}(-X\in B)$. Regarding your first question, we are looking for a measurable simple function $$f=-\mathbf{1}_{A_1} + 0\cdot \mathbf{1}_{A_2}+ \mathbf{1}_{A_3}$$ (once we have one, we have a sequence of independent copies) such that $\mu(A_1)=\mu(A_3)$ ...

2

Other than $0$? No. Try $A = -I$.

0

Ok satisfactory enough: Given the Hilbert space $\ell^2(\mathbb{N})$. Extend to a Hamel basis: $$\mathcal{B}:=\{\delta_n:n\in\mathbb{N}\}\cup\{\beta,\beta',\ldots\}$$ Define an operator by: $$T(\delta_n):=n\quad(n\in\mathbb{N})$$ $$T(\beta):=0\quad(\beta,\beta',\ldots)$$ Then it is unbounded.

0

There is no separable commutative Banach algebra that contains isometric copies of all separable commutative Banach algebras. Indeed, if $p$ and $q$ are commuting projections in a Banach algebra then $\|p-q\|\geqslant 1$ so each set of commuting projections is discrete. Now, projections in a Banach algebra can have arbitrarily large norms. Consequently, a ...

0

PROBLEM Answer Explanation Demonstration Given the real line $\mathbb{R}$. Consider the metrics: $$d(x,y):=|y-x|$$ $$d'(x,y):=\arctan|y-x|$$ Then one obtains: $$\mathcal{N}=\mathcal{N}'\quad\mathcal{U}\neq\mathcal{U}$$ Concluding problem. PROOF Identification Given normed spaces $\Omega$ and $\Omega'$. Regard the category: ...

1

You can start with "the norms induce the same topology". Then use the fact that a linear transformation is continuous if and only if it is bounded. And this is one of your inequalities. For the other direction, use the inverse of that linear transformation.

1

Maybe it will be useful to consider an example of two norms $F$ and $G$ of a vector space $X$ not being equivalent to each other. What it means is that at least one of the quantities $\sup\limits_{x \in X}\frac{F(x)}{G(x)}$ or $\sup\limits_{x \in X}\frac{G(x)}{F(x)}$ is unbounded, i.e. there is a sequence $(x_n)_{n \geq 0}$ of vectors in the space such that ...

1

Suppose you have a sequence which converges in $G$. The lower bound implies it converges in $F$ to the same limit. Suppose you have a sequence which ddoes not converge in $G$. The upper bound implies it does not converge in $F$. That's all you need, since metric spaces are sequential spaces.

0

In what you wrote first, $v_n$ should be an element of the dual space and then it means convergence in the dual norm (norm of the functionals over $W_0^{1,p}$). The second thing, that you wrote, i.e $\langle x^*,v_n\rangle\rightarrow 0\quad \forall x^*\in (W_0^{1,p})^*$ means weak convergence of the sequence $\{v_n\}_{n=1}^\infty\subset W_0^{1,p}$ to $0\in ... 0 As Martin pointed out,$\langle f,v,\rangle$is just notation for the action of the functional$f\in X^*$on the element$v\in X$; it does not mean (in general) an inner product. To be less ambiguous, one can write the pairing as $$_{X^*}\langle f,v\rangle_X$$ to emphasize the fact that$f$and$v$live in two different spaces, so the action is not an ... 1 The symbols$\langle x^*,v_n\rangle$just express$x^*(v_n)$, the functional$x^*$evaluated at$v_n$. It is a common notation, inspired in the Hilbert space case, where the dual is the same original space. 3 Let$Af = \int_{0}^{x}f(t)dt$in$L^{2}[0,1]$. Then$A : L^{2}\rightarrow L^{2}$is bounded. Let$W$consist of all continuously differentiable$g \in L^{2}[0,1]$for which$g(0)=g(1)=0$.$W$is dense in$L^{2}[0,1]$because$\{ \sin(n\pi x) \}_{n=1}^{\infty}\subset W$is an orthogonal basis of$L^{2}[0,1]$. However,$A^{-1}W$is not dense because$f \in ...

1

For any $(a,a')\in Y$, we have that $(0,-a')\in Z$. So $(a,0)\in Z$. In other words, $Y+Z$ contains the subspace $W=\{(a,0):\ a\in C^1[0,1]\}$. So now we need a Cauchy sequence in $W$ that is not convergent in $W$. For instance $\{(a_n,0)\}$, where $a_n(t)=(t+1/n)^{1/2}$.

6

Yes, $$C([0,1]) = \bigcup_{n = 1}^\infty \underbrace{\{ f \in C([0,1]) : \lVert f\rVert_\infty \leqslant n\}}_{A_n},$$ and $A_n$ is closed for each $n$ - if $\lVert g\rVert_\infty > n$, then there is a $\delta > 0$ and a non-degenerate interval $[a,b] \subset [0,1]$ such that $\lvert g(x)\rvert \geqslant n+\delta$ for all $x\in [a,b]$, and hence ...

0

Given the Hilbert space $\ell^2(\mathbb{N}_0)$. Consider the left shift: $$L_0:\mathcal{D}(L_0)\subseteq\ell^2(\mathbb{N}_0)\to\ell^2(\mathbb{N}_0):\quad L_0:=L$$ For nondense domain: $$\mathcal{D}(L_0):=\ell^2(\mathbb{N}):=\{\varphi\in\ell^2(\mathbb{N}_0):x_0=0\}$$ But it has an inverse: $$0\not\in\sigma(L_0):\quad RL_0=1_0\quad L_0R=1$$ Concluding ...

1

This was answered in the negative in this post at MathOverflow. See the first answer there and Bill Johnson's comment to the second answer.

0

We're considering finite subsets of the dual space, which are less general than bounded subsets. The basis by the latter contain the former and hence this gives the weakest topology. Since we're interested in the weakest topology where the functionals are continuous (hence bounded), it makes sense to consider finite subsets of the dual space.

1

It is not true, and it is easy to construct counterexamples. For instance, take $X = \Bbb R^2$ with the usual norm $\Vert y \Vert = \langle y, y \rangle^{1/2}$, where $\langle \cdot, \cdot \rangle$ denotes the standard real inner product, viz. $\langle (w_1, w_2), (v_1, v_2) \rangle = w_1v_1 + w_2 v_2$; let $T$ be the matrix $T = \begin{bmatrix} a & 0 ... 3 Invertible and nonivertible operators may have arbitrary norms: just note that$\|cT\|=|c|\|T\|$and$T$is ivertible if and only if$cT$is invertible (provided$c \neq 0$) 0 The composition sign in the line marked (chain rule) should be a product sign. By "normal" differentiation the final answer would be$f''(g(a))g'(a)+f'(g(a)g''(a)$. 1 Let$(V, ||\cdot ||)$be a Banach space and$V_1 $,$V_2 $two closed subspaces of$V$such that$V_1 + V_2 $is not closed. Consider a Banach space$X=V\times V\times V$with norm$||(x,y, z) ||= \max\{||x|| ,||y||, ||z||\} $and subspaces$S=V_1 \times V_1 \times V_2 ,T=V_2\times V_1 \times V_2 ,U=V_2 \times V_2 \times V_1 . $Then the subspaces ... 0 This is possible when the Hamel dimensions of$V$and$W$are the same infinite cardinal. This is not necessary but only sufficient, since countable dimensional space can be dense in an infinite dimensional space: consider the Hamel span of a countable Hilbert basis. Let us assume this and denote by$B$and$C$the bases of$V$and$W$so that$C\subset B$. ... 0 You don't even need to consider a sequence. If$z$is a different fixed point you can take$y=z-x^*$to obtain the contradiction $$\|y\|=\|z+x^*\|=\|f(z)-f(x^*)\|\le \alpha \|y\|.$$ 0 You can't use the theorem, but you don't need to. Start at any$y_0$and take$y_{n+1} = f(y_{n})$. What can you say about$\|y_n - x^*\|_2$? 2 The uniform bounded principle does work for any collection of operators, see wiki or any book on functional analysis. Note that convergent nets may not be bounded (in contrary to convergent sequences), see the post to which David Mitra is referring. 2 To clarify notation that wasn't clear to me at first: Say$f:X\to\mathbb C$. We say $$\sum_{x\in X}f(x)=s$$if for every$\epsilon>0$there exists a finite set$F\subset X$such that if$E$is finite and$F\subset E\subset X$then $$\left|s-\sum_{x\in E}f(x)\right|<\epsilon.$$ Lemma: Suppose$f:X\to\mathbb C$and$\sum_{x\in X}f(x)$exists. Suppose ... 10 Generally, without completeness, you can't deduce that a weak$^\ast$convergent sequence is bounded. Let$X = c_{00}$be the space of sequences with only finitely many nonzero terms, endowed with the$\lVert\,\cdot\,\rVert_\infty$-norm. Its completion is$c_0$, the space of sequences converging to$0$, and its dual therefore isometrically isomorphic to ... 2 Your argument is right. To express it more rigorously, you can do something like this: for every$\|x\|<1$, $$\sum_{n=1}^{\infty} \frac{x_n}{2^n} < \sup_{\|x\|=1} \sum_{n=1}^{\infty} \frac{1}{2^n}=1$$ where the inequality is strict because$x_n\le 1$for all$n$and$x_n<1$for some$n$. On the other hand, the vector$x = (1,\dots,1,0,\dots)$... 2 Many problems can be reformulated to precisely ask for a fixed point of a function. A very simple example is the following. Suppose that$A$is an$n\times n$matrix, thought of as a function$A\colon \mathbb R^n\to \mathbb R^n$. Solving$Ax=b$is an extremely important problem (I hope this does not require elaboration). Now,$Ax=b$holds if, and only if, ... 1 Seems to me that the definition given for$H^s$can't be right for$s<0$; it "must" be that$H^s$is actually the space of tempered distributions$f$such that$\hat f\in L^2(\mu)$, where$d\mu(\xi)=(1+|\xi|^2)^{s/2}\,d\xi$. Assuming so, this is easy: Say$(f_n)$is Cauchy in$H^s$. Then$(\hat f_n)$is Cauchy in$L^2(\mu)$. So$\hat f_n\to g$in ... 1 The thing about$C(\Omega)$works because$\mathbb R$is complete: If$(f_n)$is a Cauchy sequence in$C(\Omega)$then the definition of the metric shows that for every$x\in\Omega$the sequence$(f_n(x))$is a Cauchy sequence of reals. Since$\mathbb R$is complete there exists a real number$y$such that$f_n(x)\to y$; we define$f(x)=y$and proceed. The ... 3 You know the$u_m$are converging to a function, because$\mathbb{R}$is complete. For every$x \in \Omega$, you have $$| u_n(x) - u_m(x) | \leq \| u_n - u_m\|_\infty \to 0$$ So$u_n(x)$is a cauchy sequence, and by completness of$\mathbb{R}$, converge. The function$u$is then defined by : $$u(x) = \lim_{n\to \infty} u_n (x)$$ For your second ... 0 @Yongyong's answer is very good indeed. Here I just want to point out the idea behind this remark is that$W^{1,1}$is not a reflexive space. (Note$W^{1,p}$is reflexive for any$1<p<\infty$). Hence, when we try to use weak compactness to extract a weak subsequence, the limiting function may not lie in$W^1{1,1}, and hence the theorem fails. If you ... 0 Consider the function \begin{align} &f(x)=-1,x\in[-1,0];\\ &f(x)=1,x\in(0,1]. \end{align} Now for sufficiently smallh>0$, and some compact set which is a subset of$[-\alpha,\alpha]$for some$\alpha\in(0,1)$, we have: for$x\in[-\alpha,-h]\cup(0,\alpha]$, $$\frac{1}{h}|u(x+h)-u(x)|=0;$$ for$x\in(-h,0)$, ... 2 Nets that converge do not need to be bounded. Even when you use the uniform boundedness principle, you see that the operators applied to individual elements need not be bounded. For an example of this, imagine a net that is indexed by two copies of$\mathbb{N}$cascaded. (So every element of one copy is greater than the other.) I believe this is typically ... 0 A function$\psi: V \to R$is differentiable at$p \in D$when $$\lim_{v \to 0} \psi(p+v) - \psi(p) = L(p)v + r(v)$$ and$\frac{r(v)}{\|v\|}\to 0$as$v \to 0$. (Here$V$and$R$are normed vector spaces) To check for the existence of a derivative you take a general$v$and try to guess$L(x): V \to R$. In our case take$ p = (u,x)$and$v= (d,h)$(as ... 1 A Banach space is a complete normed space. If it is not complete, you loose a lot of things. Just like you loose things if you don't work with$\mathbb{R}$, and work only with$\mathbb{Q}$(since not every Cauchy sequence in$\mathbb{Q}$converge to a point in$\mathbb{Q}$...but every Cauchy sequence in$\mathbb{R}$converge to a point in$\mathbb{R}$; ... 1 The norm induces a topology on the vector space in which addition and scalar multiplication are continuous operations. This is the jist of a topologica vector space. A locally convex vector space is one in which there is a basis of neighborhoods which are convex as subsets of the linear space. (They contain the line segment between any two points, where the ... 3 The norm naturally provides a topology. The triangle inequality implies that the unit ball is convex, for if$\|x\|,\|y\| < 1$then$\|tx + (1-t)y\| \le t\|x\| + (1-t) \|y\| < 1$. Thus a normed linear space is locally convex. 1 As the commenters said, your argument is flawed in the part where you conclude that (using notation$s_n=x_1+\dots+x_n$) $$s_n\to x \text{ and } Ts_n\to \infty \overset{?}{\implies} Tx=\infty$$ You did prove something, however: your argument shows that an unbounded operator must be discontinuous. (This isn't obvious; in fact, on an incomplete normed space ... 2 You also have a continuous projection$p \colon X \to E$, with$p\circ i = \operatorname{id}_E$. Looking at the dual map$p^\ast \colon E^\ast \to X^\ast$, we see that$p^\ast$is a continuous injection of$E^\ast$into$X^\ast$, which by the closed range theorem has closed range. Since$i^\ast \circ p^\ast = \operatorname{id}_{E^\ast}$, it follows that ... 1 HINT: Use the chain rule and induction on$k$to prove that$F,G\in C^k \implies G\circ F\in C^k$. (Remember that$F\in C^k \iff DF\in C^{k-1}$.) 1 This is immediate from the definitions, plus the ordinary scalar-valued Cauchy's Integral Formula. Suppose$\Lambda\in X^*$, and let$g=\Lambda\circ f$. So$g$is analytic (by definition or not, depending on which definition of "analytic" you took). CIF shows that $$\Lambda(n(\gamma;\lambda)f(\lambda))=n(\gamma;\lambda)g(\lambda)=\frac1{2\pi ... 1 Yes. For example, in \ell_1 let Y_1 be the subspace consisting of elements whose even coordinates are zero and Y_2 be the subspace consisting of elements whose odd coordinates are zero. Take Tx=(x_2-x_1, x_4-x_3, x_6-x_5,\ldots). T is an isometry on both Y_1 and Y_2 but T(1/2,1/2,1/2^2,1/2^2,1/2^3,1/2^3,\ldots)=0. 1 Thanks alot to T.A.E.!! Counterexample Given the Hilbert space \ell^2(\mathbb{N}_0). Consider the domain:$$\mathcal{D}_0:=\langle\{e_0+\frac{1}{n}e_n:n\in\mathbb{N}\}\rangle$$It is dense since:$$e_0=\lim_n(e_0+\frac{1}{n}e_n):\quad e_0\in\overline{\mathcal{D}_0}\Rightarrow e_n\in\overline{\mathcal{D}_0}$$Define the operator:* ... 1 By uniform boundedness principle, there is a c such that for all x\in E, \left\|\dfrac{a_x}{\|x\|}\right\|≤c. So ∥a_x∥≤c\|x\|_E. 6 The inclusion \mathscr{C}^1[a,b] \hookrightarrow \mathscr{C}^0[a,b] is continuous, so F is also closed in \mathscr{C}^1[a,b]. Thus F is a Banach space with respect to the \mathscr{C}^0-norm and with respect to the \mathscr{C}^1-norm. Hence these two norms are equivalent on F (since they are comparable). Say we have$$\lVert ... 2 I would say that it is important because of the Rosenthal's$\ell_1$-theorem which asserts that Given a bounded sequence in a Banach space, it either contains a basic subsequence equivalent to a basis for$\ell_1$or a weakly Cauchy subsequence. Thus, if a space$X$does not contain copies of$\ell_1$all bounded sequences in$X\$ have weakly Cauchy ...

Top 50 recent answers are included