# Tag Info

## New answers tagged normed-spaces

0

The equation simply says that $\sup \{\frac {\{|\sum c_ig_i(x)|} {|\sqrt {\sum |c_i|^{2}}}\}=\sup \{|\sum c_ig_i(x)|: \sum |c_i|^{2}=1\}$ where the supremum on the left is taken over all choices of $c_1,c_2....,c_N$. [I have used the orthonormlaity of $g_i$'s to get $\|f\|_1^{2}=\sum |c_i|^{2}$]. Now this identity has noting to do with $L^{2}$ functions. ...

0

Show that the left-hand side can be rewritten as $$\sup_{f \in A_N, \|f\|_2=1} |f(x)|.$$ The supremum over functions $f \in A_N$ can be instead over functions of the form $\sum_{i=1}^N c_i g_i(x)$. Show that then the $\|f\|_2=1$ constraint then becomes $\|\sum_{i=1}^n c_i g_i\|_2^2 = \sum_{i=1}^N c_i^2 = 1$.

1

Yes. Let's say that $B(a,r)$ denotes the ball of center $a$ and radius $r$, included in your subspace $U$. Then for every $x \in V \setminus \lbrace 0 \rbrace$, you have $$x= \frac{2||x||}{r}\left[\left( \frac{rx}{2||x||}+a\right)-a \right]$$ which belongs to $U$, since both $\left( \frac{rx}{2||x||}+a\right)$ and $a$ belong to $B(a,r) \subset U$, and $U$ is ...

0

Yes, it is. Intuitively, since a subspace is closed under scalar multiplication you can extend this ball as much as you want. For example, assume $U$ contains a ball with radius $\epsilon$ around $0$. Then for every $0\ne v\in V$ we have $\frac{\epsilon}{2}\frac{v}{||v||}\in U$, and so: $v=\frac{2}{\epsilon}||v||(\frac{\epsilon}{2}\frac{v}{||v||})\in U$ The ...

2

Reducing the problem to vectors of norm $1$ is the first step. The second step is to notice that your bilinear form has a positive minimum on the unit sphere $\mathbb{S}^{n-1}=\{x\in\mathbb{R}^n:\|x\|=1\}$, because it is continuous and the unit sphere is compact.

2

Let $\displaystyle{F(x)=\sum_{n=1}^\infty x_n}$. $|F(x)|\leq \|x\|_1<\|x\|_1+\|x\|_p$ so that $\|F\|^*\leq 1$. Let $u_k$ be defined by $u_{k,n}=1/k$ if $n\leq k$ and $u_{k,n}=0$ if $n> k$. Since $F(u_k)=1$ and $\|u_k\|_1+\|u_k\|_p= 1+k^{1/p-1}\rightarrow 1$ as $k\rightarrow\infty$, $\|F\|^*= 1$.

0

It seems that there is no flaw. (1) Fact : A function $g(y)=\| x-y\|$ is continuous in a metric space $(X,\|\ \|)$ i.e. When $\|y_n -y\|\rightarrow 0$, then $g(y)=\lim_n\ g(y_n)$ (OP use this fact) (2) Define $f(x)=\| x- M\|$ (For convenience, we use this notation) Hence $f(x)= \inf_{y\in M}\ \{ g(x,y) \}$ where $g(x,y)=\|x-y\|$. Then prove that $f$ is 1-...

1

A positive answer to this question would help me prove the following assertion: If $\mathbb{R}^n$ is a unitial division algebra, then the sphere $S^{n-1}$ is an $H$-space. You don't need the fact you're trying to prove to prove this. $S^{n-1}$ can be defined without any reference to the Euclidean norm, as the quotient of $\mathbb{R}^n \setminus \{ 0 \}$ by ...

2

(Norms on $\mathbb{R}^n$ are necessarily continuous wrt the Euclidean topology.) A norm on $\mathbb{R}^n$ is completely specified by specifying its unit ball $\{ x \in \mathbb{R}^n : \| x \| \le 1 \}$, which must be symmetric (meaning closed under $x \mapsto -x$), convex, and compact (equivalently, closed and bounded (wrt the Euclidean norm)). Conversely, ...

-1

First observe that the space $R[0,1]$ of all Riemann integrable functions is a subspace of the complete space $B[0,1]$ formed by all bounded functions. So $R[0,1]$ is complete iff it is closed in $B[0,1]$. Theorem. $R[0,1]$ is closed in $B[0,1]$, and hence complete. Proof. Given a sequence $\{f_n\}_n$ in $R[0,1]$ converging to some $f$ in $B[0,1]$, let $D_n$...

2


1

Based on the comments above, it's easy to see that $$||\phi||_{op} \le \int^1_0 |a(t)|dt$$ and for equality you can consider the sign function $f = sgn(a)$, then for $a(t) \neq 0$ $$||\phi||_{op} \ge |\phi(f)|= \left|\int^1_0 sign(a) \cdot a(t)dt\right|$$ but the product, $sign(a) \cdot a(t) = |a(t)|$. Finally $$||\phi||_{op} \ge \int^1_0 |a(t)|dt$$ ...

1

A proof can be found here (page 399)

2

Let $$r'=\frac{1+r}{2}$$ so that $1 < r' < r$, and let $0 < \varepsilon < r'-1$. Let $V = \mathrm{Vect}(x_1,..., x_n)$, and suppose that $X \neq V$. Then there exists $y \in X \setminus V$. Let $d=d(y,V)$ the distance from $y$ to the subspace $V$ : because $V$ is finite-dimensional, hence it is closed, hence $d>0$. By definition of $d$, there ...

2

For $(a)$ take $z=\frac{1}{d}\cdot (x-y)$ where $d=d(x,M)$. Then, $||z||=1$. And, for $y'\in M$ we have \begin{align} ||z-y'||&=\bigl|\bigl|\frac{1}{d}(x-y)-y'\bigl|\bigl|\\ &=\frac{1}{d}\bigl|\bigl|x-y-dy'\bigr|\bigr|\\ &\geq \frac{d}{d}=1 \end{align} Hence, $d(z,M)\geq 1$. For the other inequality, $$d(z,M)\leq ||z-0||=||z||=1$$ Therefore, d(z,... 1 You have\begin{align}\|y\|&=\|y-x+x\|\\&\leqslant\|y-x\|+\|x\|\\&<\frac65\|x\|\end{align}and\begin{align}\|y\|&=\|y-x+x\|\\&\geqslant\bigl|\|y-x\|-\|x\|\bigr|\\&=\|x\|-\|y-x\|,\end{align}since\|x\|\geqslant\|y-x\|$. But$\|x\|-\|x-y\|>\frac45\|x\|$. 1 It is pretty straightforward, just using the same remark you have for$g$More precisely, for any$N>0we have: \| \sum_{i \ge 1} f_n(i)-f(i) \| \le \| \sum_{i \ge 1}^N f_n(i)-f(i) \| + \| \sum_{i \ge N+1} f_n(i)-f(i) \| \le \underbrace{\| \sum_{i \ge 1}^N f_n(i)-f(i) \|}_{ \longrightarrow 0 \text{ when } n \rightarrow +\infty}+2 \sum_{i \ge N+1} \| ... 3 Indeed the key point is that p-norm is a continuous function as stated in the comments. Although, a small argument is needed: Fix x=(x_1,...,x_n)\in \mathbb{R^n} (i assume that your p is >1). Then, by Hölder's Inequality used in the vectors x,y\in \mathbb{R^n} where y=(1,...,1) we have \begin{align} \sum_{k=1}^{n}|x_k|&=\sum_{k=1}^{n}|x_k|\... 0 If u,v\in H^2(\Omega)\cap H_0^1(\Omega), then \int_{\Omega}u\Delta v\,dx=-\int_{\Omega}\nabla u\cdot\nabla v\,dx. $$Hence$$ \int_{\Omega}u\Delta u\,dx=-\int_{\Omega}|\nabla u|^2\,dx $$and thus$$ \|u\|_{L^2}\|u\|_{H^2} \ge \int_{\Omega}u(u-\Delta u)\,dx=\int_{\Omega}(u^2+|\nabla u|^2)\,dx=\|u\|_{H^1}^2 $$which implies tha 2 We wish to investigate the statement$$ \|y-x\|\le\|y\|\land\|z-x\|\le\|z\|\implies\|y+z-x\|\le\|y+z\|\tag0 $$Inner Product Spaces Suppose that we are in an inner product space: |u|^2=\langle u,u\rangle. Proposition 1:$$ |y-x|\le|y-o|\land|z-x|\le|z-o|\implies\left|\frac{y+z}2-x\,\right|\le\left|\frac{y+z}2-o\,\right|\tag1 $$Proof:$$ \begin{align} &... 0 Consider the special case ofX=\mathbb R$. Then,$d(x,y)=\frac{|x-y|}{|x|+|y|}$is not continuous at$(x,y)=(0,0)$. Along$x=y,\ (x,y)\to (0,0)\Rightarrow d(x,y)\to 0$whereas along$x=0$we have$d(x,y)\to 1.$So even if we define$d(0,0)=0$the triangle inequality can not be satisfied. 1 Notice that your condition implies that$(x_n)_n$is a Cauchy sequence. Namely, for any$\varepsilon > 0$by definition of$\lim_{j\to\infty}$there exists$j_0 \in \Bbb{N}$such that $$j \ge j_0 \implies \lim_{k\to\infty} \|x_j-x_k\| \le \frac\varepsilon2.$$ Now for any$j \ge j_0$by definition of$\lim_{k\to\infty}$there exists$k_0 \in \Bbb{N}$such ... 2 The proof you have written works. Denoting the inclusion by$i:V\to H$you have for$f\in H$and$v\in V$that: $$|f(i(v))| ≤\|f\|_{H'} \cdot \|i(v)\|_H ≤ \|f\|_{H'}\cdot \|i\|_{L(V,H)}\cdot \|v\|_{V}$$ where you applied boundedness of$f$and$i$as linear maps to get that$v\mapsto f(i(v))$is bounded as a linear map (with bound$\|f\|_{H'} \cdot \|i\|_{L(...

1

It's false if you assume $D$ to be a proper dense subspace of infinite codimension, let alone more general dense sets. As an example, if we take $X = C[0, 1]$ (with the $\infty$-norm), and $D$ to be the subspace of polynomials on $[0, 1]$, then Stone-Weierstrass implies $D$ is a dense subspace of $X$. As $D$ has countable dimension, it must have (uncountably)...

1

Update: Total rewrite! … And then partial retraction. Hopefully, some part of this is still helpful and I will continue to think about how to get a possibly complete answer. This is a great question. I've spent quite a while thinking about it now, as well as perusing Dan Amir's Characterizations of Inner Product Spaces for possibly related conditions. I ...

2

I can give you a partial answer, because I couldn't check all the details, especially for $\Phi_2$. Let's start with $\Phi_1$, First note $Q$ is an orthogonal projector on $L^2(\Omega)$ then $Q^2v=Qv$ then  \begin{align} |\Phi_1(v)| = \|Qv\|_{L^2(\Omega)}\leq\|v\|_{L^2(\Omega)} \leq \|v\|_{H^1(\Omega,\mathcal{T} )} \quad \forall v \in H^1(\Omega) \end{...

0

Here is a more formal argument presenting your idea. This is taken from one of my old questions. Let $\varepsilon > 0$. Since $\gamma$ is continuous, $\gamma^{-1}(\langle 0, +\infty\rangle)$ and $\gamma^{-1}(\langle -\infty,0\rangle)$ are open sets so they can be written as countable (or finite, that case is easier) disjoint unions of intervals: $$\gamma^{... 2 I'm giving a more basic proof (considering the first two answers). The technique I'm using is rather standard and tries to infer global convergence of a Cauchy family of sequences by analyzing some sort of vertical convergence given componentwise. Let's first prove that (V, \| \cdot \|) is a normed space. This means that we have to prove first that \| \... 4 Note that |B_n(f)(x)|\leq \|f\|_{\infty} \sum_{k=0}^n {n \choose k} x^k (1-x)^{n-k}. Now,$$ 1=1^n=(x+(1-x))^n=\sum_{k=0}^n {n\choose k} x^k(1-x)^{n-k}, $$by the binomial identity. 0 One possible answer is that out of the L_p spaces, the p-norm \|\cdot\|_p satisfies the Parallelogram Law:$$2\|u\|^2+2\|v\|^2=\|u+v\|^2+\|u-v\|^2$$only for p=2. It is a well-known result that by satisfying the above, one can define an inner product via:$$(u,v) := \frac{1}{2}\left(\|u\|^2+\|v\|^2-\|u-v\|^2\right).$$Conversely, you can show that the ... 0 We can say \lvert \int_0^t f(s)\,ds \rvert^2\leq \|1\|_{L^2([0,t])}^2\,\|f\|_{L^2([0,t])}^2=t \|f\|_{L^2([0,t])}^2 \leq t \|f\|_{L^2([0,1])}^2. Hence, \|Tf\|^2=\int_0^1 \lvert \int_0^t f(s)\,ds \rvert^2\,dt \leq \|f\|_{L^2([0,1])}^2 \int_0^1 t\,dt =\frac 12 \|f\|_{L^2([0,1])}^2. Taking square roots shows$$ \|Tf\|\leq \frac{\sqrt{2}}{2}\|f|. $$What am I ... 1 d(x,U)=0 if and only if x \in \overline U. So a dense porper subspace gives a counter-example to the first part. For a specific example take U=\ell^{0} in X=\ell^{2}. Since finite dimensional subspaces are closed we have d(x,U) >0 when U is finite dimensional. 1 Hints: take X to be a space of convergent sequences and U to be the subspace of X comprising sequences that are finitely non-zero. This shows that claim (1) is false. choose a basis \langle u_1, \ldots u_k \rangle for U and extend it to the basis \langle u_1, \ldots u_k, x \rangle for \mathrm{span}(U, x). If d(x, U) = 0, then x \in U. (... 1 Hint: Let g(t)=1 for t \geq 0 and g(t)=0 for t <1. Then Tf=f*g. Show that T^{n}f=f*g^{(n)} where g^{(n)} stands for g*g*...*g (n- fold convolution). Show by direct computation that g^{(n)}(t)=\frac {t^{n}} {n!} for t >0 and 0 for t <1. This gives \|T^{n}\| \leq \frac 1 {n!}. So \|T^{n}\|<\|T\|^{n} for n >1. 3 The inequality can fail for X=\Bbb R^2 endowed with \|\cdot\|_\infty norm. Indeed, let x=(1,0), y=(-1,2), and z=(-1,-2). Then \|x-y\|=\|y\|=\|x-z\|=\|z\|=2, but \|x-(y+z)\|=3>2=\|y+z\|. 2 Define a measure space as (\mathbb N,\mu), where for any subset E\subset \mathbb N,$$\mu(E)=\sum_{n\in E} 2^n.$$Then L^1(\mathbb N,\mu) is precisely the set V you describe, with$$\|a\|=\int_{\mathbb N} |a|\,d\mu =\|a\|_{L^1(\mathbb N,\mu)}.$$Since every L^1 space is a complete normed linear space, we are done. Of course, if you haven't studied ... 2 You can prove that (C([0,1]),||.||) is not complete by comparing the norm ||f||=\sup_{0\leq t\leq 1}|tf(t)| with the usual norm ||f||_{\infty} but first let me answer to your questions: No, you dont need to show that (f_n) is Cauchy in neither of two norms in order to show that there is no constant c>0 such that$$\tag{1}||f||_{\infty}\leq c||f|...

2

What you did is fine and it proves indeed that there is no constant $c$ so that we always have$$\|f\|_\infty\leqslant c\|f\|.\tag1$$Note that $(1)$ is equivalent to $\frac{\|f\|_\infty}{\|f\|}\leqslant c$. So, proving that $(1)$ doesn't hold for some $c$ and for every $f\in C\bigl([0,1]\bigr)$ is equivalent to asserting that for every $c$ there is some $f_c\... 3 Let$l_1=\{(\alpha_1,\alpha_2,...): \sum_{n=1}^{\infty}|\alpha_n|<\infty\}$and let$T:l_1\to V$with $$T(\alpha_1,\alpha_2,...,\alpha_n,...)=\bigl(\frac{1}{2}\alpha_1,\frac{1}{2^2}\alpha_2,...,\frac{1}{2^n}\alpha_n,...\bigr)$$ By the identity $$||T(\alpha)||_V=\sum_{n=1}^{\infty}2^n|\frac{\alpha_n}{2^n}|=\sum_{n=1}^{\infty}|\alpha_n|=||\alpha||_{l_1}$$$...

1

($\alpha$) It doesn't matter. All you need is an inequality, and $\frac{1}{m}(\sqrt n -\sqrt m)$ is an upper bound. ($\beta$) It's the same. Being Cauchy is $\lim_{m,n\to\infty}\|f_n-f_m\|=0$, which is the same as the common phrasing of "Cauchy" that you use. ($\gamma)$ Because it doesn't fit the expression $1/\sqrt{t_0}$, and it doesn't hurt the ...

2

We start out with the assumption that $K$ is Hilbert, which includes completeness in its definition. Therefore $K$ is already a closed convex subset of $H$ and step 1 is not needed. Just compose projection of $H$ onto $K$ with $T$ to get the desired map. $$H \stackrel{P_K}{\to} K\stackrel{T}{\to} X$$

1

You correctly identified the mistake, we have to first prove that $(l_n)_n$ converges and then that $(x_k)_k$ converges to $\lim_{n\to\infty} l_n$. To prove this, we can show that $(l_n)_n$ is a Cauchy sequence. Pick $\varepsilon > 0$. Sequence $(x_n)_n$ is converges in $\|\cdot\|_\infty$ so it is in particular Cauchy. Therefore we can pick $n_0 \in \Bbb{... 5 Indeed it follows that$\dim (\overline{V}/\overline{W}) \leqslant 1$. Let$x \in V \setminus W$. Then, since$\dim (V/W) = 1$, we have$V = W + \mathbb{K}\cdot x$(where$\mathbb{K}$is the scalar field, be that$\mathbb{R}$or$\mathbb{C}$). Note that this is just the vector space sum, it need not be a topological sum. Now, the sum of a closed subspace and ... 1 Suppose$f$is an interior point of$L^{2}$in$L^{1}$. Then there exists an open ball$B(f,r)$contained in$L^{2}$. Now suppose we can find a sequence$(f_n)$in$L^{1}\setminus L^{2}$converging to$f$. Then$f_n \in B(f,r)$for$n$sufficiently large. But this is a contradiction because$f_n \in B(f,r) \subseteq L^{2}$and$f_n \notin L^{2}$. 0 It's not just sine. The Weierstrass approximation theorem makes clear that every function in$C[0, 1]$can be uniformly approximated by a sequence of polynomials, even something like$f(x) = |x - 0.5|$, a blancmange curve, Cantor's devil's staircase, or the Minkowski question mark function. If we want to treat the sine function as a polynomial because it can ... 5 Even when a vector space is infinite-dimensional, each vector is written as a finite linear combination of basis vectors. Similarly, the definition of$\textrm{span}(S)$is "the set of all finite linear combinations of elements from$S$". Thus, even though the sine function can be written as an infinite series, we still define the set of all ... 0 After a rumble and a tumble, I think I got it figured out. (I followed a construction I found on Wikipedia, but the details still took a bit of work and I'm spelling them out here for completeness.) Fix some vectors$x_i$as well as$x$, all of norm$\geq 1$, let$x_A$be as above and let${\cal F}$be the family of$A\subseteq S = [n]$such that$\lVert x_A ...

2

This statement is not true. As an example, consider the matrix $$A = \pmatrix{0 & 1\\0 & 0},$$ and let $\|\cdot\|_k$ denote the matrix norm defined by $$\|A\|_k = \|S_kAS_k^{-1}\|, \quad S = \pmatrix{k & 0\\ 0 & 1},$$ where $\|A\|$ denotes the induced Euclidean norm (maximal singular value, AKA spectral norm) of $A$. We find that for ...

1

It is not closed in $l^{2}$. Consider the elements $(\frac 1 n,\frac 1 n,...,\frac 1 n,0,0,...)$ where $\frac 1 n$ is repeated $n$ times. These are elements of $S$ and they converge in $\ell^{2}$ norm to $(0,0,...)$ which is not in $S$.

0

First prove an estimate on a fixed cell $K$ of the triangulation: $$\|v_h\|_{L^2(K)}^2 \le c \cdot |K| \cdot (v_{K,1}^2 + \dots + v_{K,d+1}^2),$$ where $v_{K,1}... v_{K,d+1}$ are the nodes values of $v_h$ in $K\subset\mathbb R^d$. Use element mass matrix to prove this. Constant $c$ is independent of $K$. Then use the fact that each node is element of at ...

Top 50 recent answers are included