New answers tagged

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

$\newcommand{\dist}{\text{dist}} \newcommand{\R}{\mathbb R} \newcommand{\diam}{\text{diam}} \newcommand{\and}{\quad \text{and} \quad} $The conjecture proposed by the OP is true: every non-reflexive space may be embedded into a larger space $Y$ in such a way that the minimization problem $$ \|y-x_0\| = \inf\{\|y-x\| : x \in X\} $$ admits no solution for ...


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 $N>0$ we 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 of $X=\mathbb R$. 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 ...


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