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0

Yes, it is exactly like that. For linear operators between normed spaces, continuity and boundedness are equivalent.


1

Since \begin{align} \left\|Af \right\|_{\infty} := \sup_{x\in [0,1]} \left| \int_{0}^{x} f(t) \, \mathrm{d}t \right| \le \sup_{x\in [0,1]} \int_{0}^{x} |f(t)| \, \mathrm{d}t \le \int_{0}^{1}|f(t)|\, \mathrm{d}t =: \left\|f\right\|_{L_1} \le \left\|f\right\|_{\infty}, \end{align} then \begin{align} \left\| A \right\| &:= \sup_{f \in C^{1}[0,1]} ...


1

As there is not a lot to prove here, I will instead answer the following questions: What does isomorphism/isometry mean? Often times we want to consider mathematical objects as being pretty much the same. We call such two objects isomorphic and the corresponding map between the inner bits and pieces the isomorphism. If you're told two things are pretty ...


2

As StevenTaschuk proposed, we can take $$\|x\| = \inf \{ \sum_i \lambda_i :x=\sum_i \lambda_is_i, s_i\in S, \textrm{ and }\lambda_i\geq 0\}$$ for $x\in \mathbb R^n$. Each $x=(x_1,\dots,x_n)\in \mathbb R^n$ can be obtained as $\sum_{i=1}^n \tilde \lambda_iz_i$ for $\tilde \lambda_i\geq 0$ and $z_i\in \mathbb Z^n$. Since $S$ generates $\mathbb Z^n$, we can ...


0

It is enough to show that: $$\left\|x-y\right\|-\left\|w-z\right\|\leq \|x-w\| + \|y-z\| $$ $$\left\|w-z\right\|-\left\|x-y\right\|\leq \|x-w\| + \|y-z\| $$ $$\left\|x-y\right\|-\left\|w-z\right\|\leq \|x-z\| + \|y-w\| $$ $$\left\|w-z\right\|-\left\|x-y\right\|\leq \|x-z\| + \|y-w\|$$ and so on. But this follows directly from triangle inequality. For ...


1

It usually goes like this; The 1-norm of a vector with components $x_n$ is $\sum_n |x_n|$ The 2-norm is the euclidean norm given by $\sqrt{\sum_n x_n^2}$ The p-norm is given by $\sqrt[p]{\sum |x_n|^p}$ The infinity norm is the limit as the powers of these things go to infinity which happens to have the nice form $max(|x_n|)$. Here I'm using $|a|$ to mean the ...


2

Let $v=(x,y,z)$. Then: $$||v||_1=|x|+|y|+|z|$$ $$||v||_2=\sqrt{x^2+y^2+z^2}$$ $$||v||_{\infty}=\max{(|x|,|y|,|z|)}.$$


0

Using the identity, it is easy to show that $\|A\|_2=\|A^*\|_2$ by using the fact that $y^*Ax=\overline{x^*A^*y}$ and hence $|y^*Ax|=|x^*A^*y|$. Next, using the identity again with the Cauchy-Schwarz inequality, $$ \|A^*A\|_2=\max_{\|x\|_2=\|y\|_2=1}|y^*A^*Ax|\leq\max_{\|x\|_2=\|y\|_2=1}\|Ay\|_2\|Ax\|_2=\|A\|_2^2. $$ We get the equality by choosing $x$ and ...


1

You cancelled by $\|v\|$ which is a nonnegative real number, moreover, as $v$ intends to be an eigenvector, $v\ne 0$, so $\|v\|\ne 0$, you can cancel it out.


0

To show $\lim_{n\to \infty} ||x_n||=||x_0||$, we need to show $\forall \epsilon >0,\exists N\in \mathbb{N}, s.t. \forall n\geq N$,$|(||x_n||-||x_0||)| <\epsilon$. Thus we try to find an upper bound for $|(||x_n||-||x_0||)|$. How can we do that? We try to use the assumption, which is $||x_n-x_0||\to 0$.Then how can we connect these two things? We try to ...


0

HINT: $| \, ||x|| - ||x_n|| \,| \leq || x - x_n ||$


2

As Daniel Fischer noted, condition (P) is equivalent to the norm being strictly convex. Indeed, if the norm is not strictly convex, then the unit sphere contains a line segment, and removing the midpoint of this line segment from the closed unit ball creates a nonconvex set. Conversely, if the norm is strictly convex, then any $X$ as you described is ...


1

If $\|v\| = 1$, you should have $$v = \sum_{i=1}^{n-1} a_i v_i$$ with $\sum_{i=1}^{n-1}a_i^2 = 1$. Thus $$T^*Tv = \sum_{i=0}^{n-1}a_iT^*Tv_i = \sum_{i=0}^{n-1} a_i \lambda_i v_i$$ and $$\|T^*Tv\|^2 = \sum_{i=1}^{n-1}a_i^2\lambda_i^2 \leq (\sum_{i=1}^{n-1}a_i^2)\max\lambda_i^2 = \max \lambda_i^2$$ And take $i_0 = \arg\max \lambda_i^2$ and $v = v_{i_0}$, ...


1

From your first expression $$\|x\|^{2}=\sup_{\|\xi\|=1}\overline{\xi}^{t}x^{\star}x\xi.$$ If $\lambda$ is an eigenvalue of $x^{\star}x$ then $\lambda$ is real and non-negative, and there exist a unit eigenvector $\xi$ of $x^{\star}x$ with eigenvalue $\lambda$, from which it follows that $\|x\|^{2} \ge ...


1

$z=e^{i\theta}=\cos\theta+i\sin\theta$. $\cos z=\frac12(e^{ie^{i\theta}}+e^{-ie^{i\theta}})=\ldots$


0

Write out $$\cos{z} = \frac{e^{iz} + e^{-iz}}{2} $$ and apply the triangle inequality. There will be some minor details to work out, but they should be fairly straightforward :)


2

Let $f:\mathbb{R}^n\to\mathbb{R}$ satisfy the desired condition. It follows that whenever $\|x_1-y_1\|=\|x_2-y_2\|, $ we have $|f(x_1)-f(y_1)|=|f(x_2)-f(y_2)|$, thus equilateral triangles are mapped by $f$ to equilateral triangles. Since all equilateral triangles on the line are trivial, and every $x,y\in\mathbb{R}^n$ are two verices of an equilaterla ...


1

As Daniel already stated in a comment, for $f$ constant this is trivially true. So let's assume $f$ is not constant. Then there exist at least two distinct points in $\mathbb R^n$ which map to two distinct values. Let's call these points $P$ and $Q$. Consider two distinct paths from $P$ to $Q$ which have no other points in common. I believe that you can ...


0

If $||.||_1 \sim ||.||_2$ (are equivalent) then there exists two constants $c, C > 0$ such that for all $ x \in E$ we have: $$c||x||_1 \leq ||x||_2 \leq C||x||_1$$ Now if $(x_n)$ is cauchy with respect to $||.||_1$, it means that given $\varepsilon > 0$, we have $$||x_n - x_m||_1 < \dfrac{\varepsilon}{C}$$ if $n,m$ are bigh enough. But this ...


2

Let $X$ be a metric space. Take a convergent sequence, say $(x_n) \subset X $. For convenience, I will write the metric of $X$ as: $d( x, y ) = |x-y| $. However, you must keep in mind that we are in an arbitrary metric space, and that I am using the standard metric just for convenience. Now, since $(x_n)$ is convergent (Say $x_n \to x$) we know that for a ...


0

Let d be the metric and the sequence $a_1, a_2, ...$ converges to $a$, which means For given $\epsilon>0$ exist N such that $n\geq N$ implies $d(a_n,a)<\epsilon/2$ For that N, if $n\geq N$ and $m\geq N$, we have $d(a_n,a_m)\leq d(a_n,a)+d(a,a_m)<2\cdot\epsilon/2=\epsilon$ therefore the sequence is Cauchy.


1

Suppose $\;x_n\xrightarrow[n\to\infty]{}x\;$ , and let $$\;\epsilon>0\implies \;\exists\,N_\epsilon\in\Bbb N\;\;\text{such that}\;\;n>N_\epsilon\implies d(x_n,x)<\frac\epsilon2$$ thus, for $\;n,m>N_\epsilon\;$ we get $$d(x_n,x_m)\le d(x_n,x)+d(x,x_m)<\frac\epsilon2+\frac\epsilon2=\epsilon\implies$$ $\;\{x_n\}\;$ is Cauchy.


2

The concept of operator norms becomes more clear when we consider normed vector spaces $(X, \|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$. The space of linear maps from $X$ to $Y$, $L(X,Y)$ is equipped with the operator norm $$\| f \|_L = \sup_{x\in X} \frac{\|f(x)\|_Y}{\|x\|_X} = \sup_{\|x\|_X = 1} \|f(x)\|_Y$$ Now the special case is $(X,\|\cdot\|_Y) = ...


1

The author wants to define an operator norm on $\mathbb M_n$, that is a map $\mathbb M_n\to \mathbb R$. As often when defining a function $f\colon A\to B$ one writes down $f(x)=\text{some expression}$, except that in the case of a norm one conventionally writes $\|x\|$ instead of $f(x)$. Thus the equation defines the operator norm considered by specifying ...


1

$x:\mathbb{C}^n\to \mathbb{C}^n$ is a complex linear map. The norm $\|x\|$ can be written as: $$\|x\|=\sup_{|\xi|=1} |x\xi|.$$ That is, the norm of the given linear is the maximum on the unit sphere (identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$). (We can say maximum because of continuity and compactness). In other words, $\|x\|$ it is the maximum of the ...


2

Here $x\in \Bbb M_n$. So operator norm of $M_n$ is calculated.


1

Let's write $\|z\|_r = \left(\sum_{k=1}^K |z_k|^r \right)^{1/r}$; this is called the $r$-norm of $z$. In particular $|z|=\|z\|_2$. We are now in the situation of Relations between p norms. Two cases: If $r>2$, then Hölder's inequality gives $\|z\|_2\le K^{1/2-1/r}\|z\|_r$, as shown in the aforementioned Q&A. This is attained by ...


1

$L$ is not compact even in the weak topology because it is not bounded: $$a_k(n)=\frac{1}{\sqrt{n}}\quad \text{if}\: n\leq k$$ $$a_k(n)=0\quad \text{if}\: n> k$$ this sequence does not admin a convergent subsequence because it escapes to infinity. $K$ is closed and convex in the strong topology, and as such it is closed in the weak topology. In the weak ...


0

Using $2ab\le a^2+ b^2$ and monotonicity of $x\mapsto \sqrt x$ we obtain the first $$ \|x\|_1 = \sqrt{ (|x_1|+|x_2|)^2} = \sqrt{ |x_1|^2+2|x_1|\cdot|x_2|+ |x_2|^2} \le \sqrt{ 2|x_1|^2+2|x_2|^2} = \sqrt2\|x\|_2. $$ Again by monotonicity $$ \|x\|_2 = \sqrt{ |x_1|^2+|x_2|^2} \le \sqrt{ \|x\|_{max}^2+\|x\|_{max}^2} = \sqrt2\|x\|_{max}. $$ Since $|x_i|\le ...


1

First let's do it in special case $\vec{x}_{1}=\vec{0}$ $d\left(\vec{0},H\right)=\left\Vert \vec{p}\right\Vert $ where $\vec{p}$ denotes the projection of $\vec{0}$ on $H$. Then $\vec{a}.\vec{p}=b$ (i.e. $\vec{p}\in H$) and secondly $\vec{p}=\lambda\vec{a}$ for some scalar $\lambda$. This leads to $\lambda=b\left\Vert \vec{a}\right\Vert ^{-2}$ and ...


1

Even for the special case of $\|\cdot\|_A$ it is not true. Let $N$ be the orthogonal complement of $K$. The minimum of $\|Ax-b\|_2$ occurs when $\nabla(\|Ax-b\|_2^2)\in N$, that is, $A^2x-Ab\in N$. Similarly the minimum of $\|Ax-b\|_A$ occurs when $A^3x-A^2b\in N$. In general, these are not the same. For example, let ...


1

If $A$ is an isometry you have $\|Ax\| = \|x\|$, and so $\|A\|=1$. If $A$ is an isometry, then so it $A^{-1}$. To see this, note that since $\|Ax\| = \|x\|$, $A$ is invertible. Letting $x=A^{-1} y$ gives $\|y\| = \| A^{-1} y \|$. Hence $\|A^{-1}\| =1$.


2

When we use the bound $$\tag{*}\int_0^t|x(\tau)|\mathrm{d}\tau\leqslant \lVert x\rVert_\infty,$$ there is no loss if and only if $x$ is constant so the norm is greater than $1$. Inequality (*) for general $x$ shows that the norm is actually $1$.


2

The minimum depends on the norm you are using. Take $b=(1,1)^T$, $A=I_2$, $\mathcal K=span\{(1,0)^T\}$. Then any point on the line between $(0,0)$ and $(2,0)$ is a solution of $$ \arg\min_{x\in\mathcal K} \|x-b\|_\infty, $$ while it holds $$ (1,0)^T=\arg\min_{x\in\mathcal K} \|x-b\|_p $$ for $p\in [1,\infty)$.


0

Assuming $A\in\mathbb{C}^{n\times n}$ with columns $a_1,\ldots,a_n$ and that you already know the equivalence constants for vector $p$-norms: For $\|A\|_{\infty}\leq\sqrt{n}\|A\|_2$: We have for all $x\in\mathbb{C}^n$, $\|x\|_{\infty}\leq\|x\|_2\leq\sqrt{n}\|x\|_{\infty}$, so $$ \|A\|_{\infty}=\max_{x\neq 0}\frac{\|Ax\|_{\infty}}{\|x\|_{\infty}} ...


2

HINT: Use the equality: $$||\frac{x_n - x_m}{2}||^2 + ||\frac{x_n + x_m}{2}||^2= 1/2\,(\,||x_n||^2 + ||x_m||^2)$$ $M$ is convex so $\frac{x_n + x_m}{2}$ also have norm approaching $d$.


1

It is easy. For any matrix $||B||_{\infty}=\max_i\sum_j|b_{i,j}|\geq \max_i|b_{i,i}|$. Here $B=A^{-1}$ and $b_{i,i}=1/a_{i,i}$; then $||A^{-1}||_{\infty}\geq 1/\min(|a_{i,i}|)$.


0

Let $C = A + B$. Then $|C|^2 = CC^\dagger = AA^\dagger + BB^\dagger + AB^\dagger + BA^\dagger$, right? The key isn't to eliminate those cross terms but to bound them. The classic statement of the triangle inequality is $$|C|^2 \leq (|A| + |B|)^2$$ Expand the right to get $$|C|^2 \leq |A|^2 + |B|^2 + 2 |A ||B|$$ The classic proof argues that ...


1

My intuition says that you are limited to +1 or -1 on the diagonal if it should hold for all x and all norms. General rotations won't survive taxi distance or max in $R^2$ already.


1

First I want make a remark on the fact of not specifying the space on which we work(your matrices are square or not?), it is also important to give us your norms definitions, so your question will be self-understood, especially if you consider the differences found in the notations used in literature. In case you are not restricted to search for a ...


4

This is a little more general perspective that I hope will be helpful with your intuition about norms. From this perspective the answer to your $1$D question will be obvious. For every norm $\lVert\cdot\rVert$, there is a "unit ball" consisting of all the points that have norm less than or equal to $1$: $$B = \{x : \lVert x\rVert \leq 1\}$$ In fact, this ...


0

Let $V$ be a Vector Space over the field $F$ where $F$ is a subfield of $\Bbb C$. One of the axioms that specify a Norm on a space is If $a \in F$ then $||a \cdot \underline v|| = |a| \cdot ||\underline v||$ for every $\underline v \in V$ Now consider the cases when $a = 1$ and $a = -1$. See how this related to the definition of the Absolute Value ...


1

Since you originally asked about $L^1$ spaces I dare add this comment. If one wants to preserve the integral in (finite-dimensional and with finite measure ) $L^1$ spaces rather than the norm of $\ell^p$, the matrices $M$ that do this have two components, namely $M= S * G$, where $*$ is the Hadamard product. The $S$ component is effectively a stochastic ...


2

The gradient of a function is a linear functional that gives an approximation to the function: $$ f(x+h) = f(x) + \color{blue}{\nabla f(x) h} + o(\|h\|), \quad \|h\|\to0 $$ (I use Fréchet gradient definition here.) In scalar case, $\nabla f(x) h$ is just $f'(x)h$, multiplication of two numbers. In $\mathbb R^n$, it is $\nabla f(x) \cdot h$, inner ...


3

No, there are not. Let $\|\cdot\|$ be a norm subordinate to an arbitrary pair of vector norms $\|\cdot\|_*$, $\|\cdot\|_{**}$: $$\|A\| = \max_{x\ne 0}\frac{\|Ax\|_*}{\|x\|_{**}}.$$ Then for rank-one matrices $$\|uv^H\| = \max_{x\ne 0}\frac{\|uv^Hx\|_*}{\|x\|_{**}} = \max_{x\ne 0}\frac{\|u\|_*\cdot\langle x, v\rangle}{\|x\|_{**}} = \|u\|_* \cdot ...


1

Presumably, you have the entry-wise definition, $$ \|A\|_F^2 = \sum_{i,j} |a_{ij}|^2 $$ In order to get to this from the trace of the product $AA^*$, use the summation definition of matrix multiplication: $$ [AB](i,j) = \sum_{k=1}^n a_{ik}b_{kj} $$ and of the trace: $$ \text{trace}(A) = \sum_{\ell = 1}^n a_{\ell \ell} $$


1

Its not defined a norm on $C^1[0,1]$ since, for example $\|1\|=0.$


1

It's actually easier if we work with function notation for awhile and not norms. So let's define $f_1(X)=\|X\|_*$ and $f_2(Y)=\|Y\|_1$ and write it as $$\begin{array}{ll} \text{minimize}_{X,Y} & f_1(X) + \lambda f_2(Y) \\ & X + Y = C \end{array}$$ The Lagrangian is $$\begin{aligned} L(X,Y,Z) &= f_1(X) + \lambda f_2(Y) - ...


3

Of course, if $n=1$ or $m=1$, the linear maps are essentially vectors and the $\ell^1$ norms on $\mathbb C^n$ and $\mathbb C^m$ make the norm $\|\cdot\|$ into an operator norm. I have the following partial answers: The answer is "no" if $\min\{n,m\}>1$, if the norm on the domain is the Euclidean one. Also, it is impossible to give a norm $|\cdot|_k$ to ...


0

Yes this is a norm. Operator norms. Let $(X,\|\cdot\|_X)$, $(Y,\|\cdot\|_Y)$ be two normed spaces and $T:X\to Y$ a linear operator. $T$ is bounded, if there exists a $c>0$, such that $$ \|Tx\|_Y\le c\|x\|_X, \quad\text{for all $x\in X$}.\qquad (\star) $$ The infimum of all $c$ satisfying the above, also satisfies the above and it is the operator norm ...



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