Tag Info

New answers tagged

0

(1)For any $x$ with $||x||\leq 1$ there exists $y$ with $||y||=1$ and $|T(x)|\leq |T(y)|$. PROOF.If $x=0$ let $y$ be any vector of norm $1$ because $T(0)=0$. If $ 0<||x||\leq 1 $ let y=x/||x|| (Because $||x||\leq 1 \implies |T(y)|=|T(x)|/||x|| \geq |T(x)|$.) .....................(2) Therefore $$||T||= \sup \{|T(x)| : ||x||\leq 1\} \leq \sup \{|T(y)| : ...


0

Suppose we have differentiable functions $F_1:R\to R^n$, $F_2:R^n\to R^m$, and $F_3:R^m\to$. Then the composition function $F_3\circ F_2\circ F_3:R\to R$ is also differentiable, and by the chain rule, it's derivative at any point $x\in R$ is given by: $$(F_3\circ F_2\circ F_1)^{'}(x) = D_{F_2\circ F_1(x)}F_3\circ D_{F_1(x)}F_2\circ F_1^{'}(x)$$ Where the ...


0

Hint: You're missing something here. It seems that what you wanted to do was define $$ x_0 = [1,1,1,\dots] $$ Then, define $$ x_1 = -[1,0,0,\dots]\\ x_2 = -[0,1,0,\dots] $$ and so forth. Note, however, that this won't work, since we always have $\|S_n\| = 1$. On the other hand, we could begin by defining $$ x_0 = [1,1/2,1/3,\dots] $$ why does this work ...


1

As the comment above indicates, there is no such inner product. One can deduce that this is the case by noting that the parallelogram identity fails to hold.


1

This proof is unnecessarily complicated. It's easier to start with $$ \|\mathbf v+s\mathbf w\|^2\ge0\;, $$ which yields $$ \mathbf v\cdot\mathbf v+2s\mathbf v\cdot\mathbf w+s^2\mathbf w\cdot\mathbf w\ge0\;, $$ and then differentiate with respect to $s$ to find the "worst case" of $s$, yielding $$ 2\mathbf v\cdot\mathbf w+2s\mathbf w\cdot\mathbf w=0\;. ...


0

Assuming you mean $A\in M(n,\,\mathbb{C})$=$n\times n$ complex element matrices, that $x,\,y\in\mathbb{C}^n$, proceed as follows. The quotient $\frac{\|A\,x\|}{\|A\,y\|}$ is invariant when you scale $x,\,y$ by any $\alpha\in\mathbb{C}$ (i.e $\frac{\|A\,\alpha x\|}{\|A\,\alpha y\|}=\frac{\|A\,x\|}{\|A\,y\|}$) so: ...


1

The spectral radius is the biggest eigenvalue of a matrix. There is a linked norm called the spectral norm, which is infact the square root of the biggest eigenvalue of the matrix $A^*A$. (So not linked to its own spectral radius, but to the spectral radius of $A^*A$) Norms are always convex. Due to triangle inequality and to the homogeneity. Edit: ...


1

You can think of the norm as the maximum amount the map $T$ can expand a vector. You begin with a unit vector $v$, i.e. $\Vert v\Vert =1$ and create a new vector $Tv$ with $\Vert Tv\Vert \leq \Vert T\Vert$. The norm of $T$ is the smallest value that you can safely put on the right hand side of the inequality and still be certain the inequality is true. In ...


1

The $L_p$ norm is more general, but you need to specify a measure space for it to make sense. You're integrating over $\mathbb X$ after all! The $l_p$ norm can be seen as a particular case of the above, as @Stephen Montgomery-Smith noted, with the counting measure on positive integers. So I don't think there really is any source of ambiguity: either I ...


1

Since $\rho(A)<1$, there is an induced norm $N(.)$ s.t. $N(A)<1$. Let $\mu=N(A)$; thus $N(A^k)\leq \mu^k$. Since the norms are equivalent, there is a fixed $C$ s.t. $||.||\leq CN(.)$. Finally $||A^k||\leq CN(A^k)\leq C\mu^k$. EDIT. If $E$ is a Banach space and $A\in L(E)$ is bounded, then one has the same result. According to Gelfand, there is ...


6

If a function $m$ satisfies your condition, then $m(x) = 1$ for all $x \ne 0$. Just observe the following inequality: $$2^{m(x)}||x|| = ||2x|| = ||x + x|| \le ||x|| + ||x|| = 2||x||$$


0

Let $x=(x_1,\dots,x_n)\in\mathbb{R}^n$ and $||x||_3^3=\sum_{i=1}^n|x_i|^3$. Using the fact that $\forall 1\leq i\leq n$, $|x_i|\leq||x||_{\infty}$, we can chose $m=n^{-\frac{1}{3}}$. On the other hand: $$||x||_{\infty}^3=\left(\max_{1\leq i\leq n}|x_i|\right)^3=\max_{1\leq i\leq n}|x_i|^3\leq\sum_{i=1}^n|x_i|^3=||x||_3^3$$ So you can chose $M=1$.


0

We have the following: Suppose $z:=(x,y)\in R^2$. First prove that $| x |^3 + | y |^3\leq(| x | + | y |)^3$. Therefore, $2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 \geq \Vert z\Vert_3$. On the other hand, $|x|^3+|y|^3\geq \Vert z \Vert_\infty^3$. Thus, $\Vert z \Vert_3 \geq \Vert z \Vert_\infty$. Summing up: $2 \Vert z\Vert_\infty \geq \Vert z\Vert_1 ...


2

It is enough to show that when one of the norms is 1, the other is bounded between two positive constants, $m$ and $M$. It is easiest to look at the case when $||(x,y)||_\infty=1$. What does that equation tell you about $x$ and $y$? You can use this to bound $||(x,y)||_3$. In terms of the shape of $||.||_3$, you really just want to sketch the curve ...


3

To elaborate on my comment on Michael's answer: The symbol $\left\Vert\mathbf{u}\right\Vert$ for a vetor $\mathbf{u}$ usually stands for the norm of that vector. A norm is "a function that assigns a strictly positive length or size to each vector in a vector space" (quoted from wikipedia). Having a normed vector space enables you to talk about e.g. the ...


1

The norm of a vector $(1,3,4,11,13)$ is $\sqrt{1^2+3^2+4^2+11^2+13^2}$. It is an extension of Pythagoras' Theorem.


2

No, and a counter-example can come from the simple cases. Take $X=\mathbb{R}^2$, $v_0=\hat{i}$ and $v=\hat{i}+\hat{j}$. In this case, $v=v-v_0+v_0=(1,1)$ and so $\|v-v_0+v_0\|=\sqrt{1+1}=\sqrt{2}$ while $\|v-v_0\|=\|\hat{j}\|=1$ and $\|v_0\|=1$. In fact I think the equal sign holds only when $v-v_0$ is a non-negative multiple of $v_0$. For the Pythagorean ...


3

Take $\mathbb R$ over $\mathbb R$ endowed with the absolute value norm, $v = 1$, and $v_0 = 2$. Then $$1 = |v| = |v - v_0 + v_0| \neq |v - v_0| + |v_0| = 1 + 2 = 3.$$ For inner product spaces in particular, $||x||^2 = \langle x, x \rangle.$ We have \begin{align*} ||x + y||^2 &= \langle x+y, x+y \rangle \\ &= \langle x,x \rangle + 2\langle x,y ...


0

This depends very much on convention. If the basis is indexed by lower indices, the tensor product should be lower indices too. One could write the coefficients with upper indices and omit the summation sign like so: $$K = k^{ij}e_i\otimes e_j.$$ That would be the Einstein summation convention which is quite popular in mathematical physics. You can also ...


2

This is a matter of convention that differs between literatures and depending on what you're doing. If you're going to use raised indices in a meaningful way, then I'd expect to see the basis written as $$ B = \{ u^i \}_{i=1}^{\infty} $$ with the index up on vectors. Then your second option for $K$ would be correct. Since you wrote the vectors in the ...


4

Hint: choose any constant function $g$ on the interval $[a,b]$. Then $g \in B$ but $g$ is not the zero function.


4

(a) Consider $g(t)=1$. (b) Consider $g(t)=t^2(1-t)^2$.


0

Balls in general have to be convex. I dont think its possible.


3

A necessary and sufficient condition for a subset $B$ of a vector space $V$ to the unit ball of a norm on $V$ is that $B$ is non-empty, convex and symmetric about the origin ($-B = B$). If $B = \{(x, y) : x^2-1 \le y \le 1-x^2\}\subseteq\mathbb{R}^2$, then $B$ is non-empty, convex and symmetric about the origin and so it is the unit ball of a norm on ...


2

Claim 1: The topology induced by $d_2$ is finer than the topology induced by $d$. Proof of Claim 1: It is enough to show that for every $\epsilon > 0$ we can find $\delta$ such that $$B_{d_2}(0,\delta) \subset B_{d}(0,\epsilon).$$ This turns out to be easy: let $\delta := \frac{\epsilon}{2}.$ Then if $x \in B_{d_2}(0,\delta)$, we have $$\sum_{n = ...


2

No, certainly not. Convergence in $\mathbb {R}^{\mathbb {N}}$ is equivalent to convergence is each slot. Thus $e_n$ converges to the $0$ sequence in $\mathbb {R}^{\mathbb {N}}$ (here $e_n$ is the sequence with $1$ in the $n$th slot, $0$'s elsewhere), while $e_n$ does not converge in $l^2(\mathbb N ).$


1

Notice that for $x \neq 0$,$$\frac{\delta}{2}\frac{x}{\|x\|_1} \in B^1_{\delta}(0) \subset B^2_1(0) \quad \Rightarrow \quad \Big\|\frac{\delta}{2}\frac{x}{\|x\|_1}\Big\|_2 \le 1 \quad \Rightarrow \quad\|x\|_2 \le \frac{2}{\delta}\|x\|_1.$$ The other inequality is obtained similarly, using the other known inclusion. Clearly if $x = 0$ the inequality is ...


2

Given the Banach spaces $\mathcal{c}_0$ and $\ell^\infty$. Consider the identity: $$T_0:\mathcal{c}_0\to\mathcal{c}_0:\quad T_0:=\mathbb{1}$$ It has No continuous extension: $$T:\ell^\infty\to\mathcal{c}_0:\quad T\restriction_{\mathcal{c}_0}=T_0$$ For the details see: Werner


1

No. Consider $F=\mathbb R^2$. And $T(e_1)=e_1$, $T(e_2)=2 e_2$. The subspace generated by $e_1$ could be your $E$. The norm of $T_0$ then is 1 and the norm of $T$ is 2. Even better: Let $F$ be some Banach space and $T:F\to F$ a bounded non-zero operator. Take $E=\{0\}$. Then we have norm 0 for $T_0$ and a positive norm for $T$ because $T$ is non-zero. ...


0

The limit as written does not exist. In particular, there is no one value that works as the limit. For example: note that if we take $x(t) = ta$, then we have $$ \lim_{t \to \infty} \frac{a^Tx(t)}{|x(t)|} = |a| $$ On the other hand, if we take $x(t) = tb$ where $b$ is perpendicular to $a$, then we have $$ \lim_{t \to \infty} \frac{a^Tx(t)}{|x(t)|} = 0 $$ ...


0

This is false, it does not hold disregarding wether the matrix is diagonalizable or not. For a non diagonalizable matrix, take $A=[-2,1;0, -2]$. For a diagonalizable matrix, take $A=[-1,0; 3, -0.2]$. Edit: the textbook I mencioned in the question was incomplete. The correct theorem is that there exist a basis for which this holds. For source, go to ...


2

The proximal operator for $\|CX\|_1$ does not admit an analytic solution. Therefore, to compute the proximal operator, you're going to have to solve a non-trivial convex optimization problem. So why do that? Why not apply a more general convex optimization approach to the overall problem. This problem is LP-representable, since $$\|CX\|_1 = \max_j \sum_i ...


1

The condition $$\tag{1} B=\sum_jp_jU_jAU_j^*,\ \ \ \sum p_j=1,\ \ p_j\geq0$$ implies that $B$ has the same trace as $A$. If $$ A=\begin{bmatrix}1&1\\1&-1\end{bmatrix}, $$ neither $A$ is a convex combination of unitary conjugates of $|A|$, nor $|A|$ is a convex combination of unitary conjugates of $A$. $$\ $$ It does work, as you say, for ...


1

$\delta(p,r)$ is a square with sides of length $2r$ and centre at $p$. For simplicity I'm going to elaborate taking $p$ to be the origin but this carries through for any $p$ by translation (which is an isometry). $$\delta(\vec{O},r) = \{ x \in \mathbb{R^2} : \| x \|_{\infty} < r \} = \{x \in \mathbb{R^2} : |\mbox{max}(x_i)| < r \}$$ where $x_i$s are ...


1

$\delta$ is just a symbol used to denote this type of set; the author could have used $B(p,r)$, $D(p,r)$, or really any symbol, as long as the notation is consistent. Don't let it bother you too much. For a geometric understanding, the unit ball centered at $0$, $\delta(0,1)$, is the square of side length $2$ centered at $0$. If you write out the definition ...


5

Thanks for the help from user zhw. I claim that $$\lim_{p\rightarrow \infty} \int_{\mathbb R^N}\Bigg\{\left(\frac{|\nabla u|}{\|\nabla u\|_p}\right)^{p-2}\frac{\nabla u}{\|\nabla u\|_p}\Bigg\} \cdot \nabla v dx = \int_K \frac{1}{m(K)}\frac{\nabla u}{\|\nabla u\|_\infty} \cdot \nabla v dx$$ where $K:= \{|\nabla u| = \|\nabla u \|_\infty\}$. We assume ...


0

For the case $p\neq 2$ the famous parallelogram law dos not hold so the space $l^p$ is not Hilbert. It is a theorem in many functional analysis text book such as Erwin Kreszig (Intoductory functional analysis with application)


3

What it means is not that hard: We have a sequence of real numbers $\langle x_n,y_n\rangle$ which converges to a real number $\langle x,y \rangle$. It doesn't matter what your interpretation of what an inner product "means" is. So it means that the absolute value of $\left|\langle x_n,y_n \rangle - \langle x,y \rangle \right|$ gets as small as we like by ...


1

The differentials of the Manhattan and Frobenius norms are $$\eqalign{ d\,\|X\|_1 &= {\rm sign}(X) : dX \cr d\,\|X\|_2 &= \frac{X}{\|X\|_2} : dX \cr }$$ where the sign function is applied element-wise, and the colon represents the Frobenius product, i.e. $\,\,A\!:\!B = {\rm tr}(A^TB)$ Letting $Y=(AX-B)$, the current function and its differential ...


4

The relaxed condition also implies $$ \left\|\frac1\alpha \alpha x\right\|\le\left|\frac1\alpha\right|\|\alpha x\|$$ and hence $$ \|\alpha x\|\le |\alpha|\|x\|\le |\alpha|\left|\frac1\alpha\right|\|\alpha x\|=\|\alpha x\|,$$ which implies equality throughout.


1

Pick bases such that one inner product is given by the identity matrix and the other is given by $M$. Since $M$ is symmetric it can be written $M = N^TN$. Now left-multiplication by $N$ is an isometry between the two inner product spaces.


1

Let us calculate the derivative of the map $x\mapsto\|x\|_1$. For all $x=\left(x_1,\ldots,x_N\right)\in\mathbb{R}^N$, one has $$\|x\|_1=|x_1|+\ldots+|x_N|$$ so that $$\frac{\partial}{\partial x_j}\left(x\mapsto\|x\|_1\right)=\frac{x_j}{|x_j|}\quad\quad\quad 1\leq j\leq N$$ provided that $x_j\neq 0$. Whence ...


2

In an inner product space $$ \langle v, w \rangle = \frac{1}{2}(\|v + w\|^2 -\|v\|^2 - \|w\|^2) $$ So the norm determines the inner product. The norm on a normed vector space is induced by an inner product iff the function $\langle \cdot, \cdot \rangle$ defined by the above formula satisfies the axioms for an inner product. The Jordan-von Neumann theorem ...


1

Well, this depends entirely on what you mean by a norm being induced by an inner product. The way this expression is usually used, the norm induced by the inner product $\langle \cdot, \cdot \rangle$ is by definition the norm $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$. If you were to use the expression more loosely, you might for instance ...


1

Your question is a corollary of the fact that all norms on finite dimensional vector spaces (hence all inner product norms) are equivalent. Therefore, as Pedro Tamaroff said, You can just use the proof of that result. See e.g.: http://www.math.colostate.edu/~yzhou/course/math561_spring2011/norm_equiv.pdf


1

Hint: For any $x$, consider $z = \frac{\delta x}{2 ||x||_1}$.What do you know about that $||z||_1$? Notice that this will imply $||z||_2 < 1$. Now, unravel that statement by substituting $z$ for $\frac{\delta x}{2 ||x||_1}$. This should yield you one side of the inequality. Other should be rather similar.


8

Remember the definition of equivalence is that there exists numbers $c,d$ such that $c\Vert x\Vert_2 \leq \Vert x\Vert_1 \leq d\Vert x\Vert_2$. What the c and d do in the picture is to stretch or shrink the shapes you've drawn. What the inequality represents is one shape fitting inside another. What equivalence means is that I could shrink the circle for ...


1

Hints: $y$ is a composition of two functions, $w \mapsto w w^T x-x$ and then $w \mapsto \|w\|_2^2$. Let me remark that $$ w w^T x - x = \langle w,x\rangle w - x, $$ where $\langle -,- \rangle$ denotes the $L^2$-inner product. To apply the chain rule, you should first compute $$ \frac{\partial}{\partial w} \left( \langle w,x\rangle w - x \right) = \langle -,x ...


4

If $f_n(x) = (n+1)x^n,$ then $\|f_n\|_1 = 1, \|f_n\|_2 =(n+1)/\sqrt {2n+1}, \|f_n\|_u = n+1.$


2

Jensen's inequality? Oh geeze, I'm a bad analyst, so I can never remember something like that. Here's a trick you might try. Suppose that there is a constant $c > 0$ so that $||f||_{L^1(\mathbb{R}^n)} \le c ||f||_{L^2(\mathbb{R}^n)}$. Let's make the change of variables $x \to \epsilon x $ for some $\epsilon > 0$. That is, let's let $y = \epsilon ...



Top 50 recent answers are included