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4

If $A$ is nonsingular, then $AA^{-1} = I$, so $$ 1 = ||I|| = ||AA^{-1}|| \leqslant ||A||\cdot||A^{-1}||. $$ In general, then $1 \leqslant ||A||\cdot||A^{-1}|| \implies ||A||^{-1} \leqslant ||A^{-1}||$. Equality is thus not necessarily guaranteed for arbitrary nonsingular $A$; however, the inequality above implies that equality may occur. Consider an ...


4

You want to prove the intersection of the closed balls of radius $3$ around $(1,1,1)$ and $(-1,-1-1)$ has infinite cardininality. It suffices to prove the intersection of the open balls of radius $3$ around $(1,1,1)$ and $(-1,-1-1)$ is non-empty (because non-empty open sets have infinite cardinality). To see this notice $(0,0,0)$ is in this intersection.


4

As Travis said in the comments, $$\| a b\| = \| a\| \| b\|$$, in this case we have $ab=1$ so we can write, $$ 1 = \|a\| \|b\|$$, and if we square both sides we can write, $$ 1 = (a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2).$$ All the $a_n$ and $b_n$ are integers. We then have that the product of two integers is equal to $1$. Under what ...


3

$\newcommand{\norm}[1]{\left\|{#1}\right\|}$ First of all you have to know or demonstrate that $$\norm{S \otimes T} \leq \norm{S}\norm{T}$$ Then notice that the matrix $M$ you address is $$M=J \otimes I$$ where $I$ is the identity matrix and $J$ the generalization of this matrix $$J = \left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 1 ...


3

Let $f$ have 3 indices corresponding to voxels: $f_{ijk}$, with maximal indices $i_\text{max}, j_\text{max}, k_\text{max}$. The gradient of $f$ has an additional Cartesian index $\alpha$: \begin{align} g^\alpha &=\left(\nabla f\right)^\alpha=\partial_\alpha f\\ g^\alpha_{ijk} & = \partial_\alpha f_{ijk}. \end{align} The TV norm is the sum of the ...


3

$$|x\cdot y| \le \sum_{i=1}^n |x_i\bar y_i|\le \sum_{i}\left((\sup_j|x_j||y_i|)\right)=\sup_j|x_j|\sum_i|y_i|=\|x\|_{\infty}\|y\|_1$$


3

It is meant to be 1-Lipschitz with respect to $\| \cdot \|_0$. Otherwise the claim would be false, consider for example the norm $\| \cdot \|_0 := 2 \| \cdot \|$ for $y = 0$ and an arbitrary $x \ne 0$.


3

This follows from Ptolemy's inequality, which states that for any quadrilateral $ABCD$, we have $$\overline{AB}\cdot \overline{CD}+\overline{BC}\cdot \overline{DA} \ge \overline{AC}\cdot \overline{BD}.$$ Identifying $A,B,C,D$ with $0,z,x,y$, this gives $$\left\|z-0\right\|\left\|y-x\right\|+\left\|x-z\right\|\left\|0-y\right\|\ge ...


2

Note that $$\dfrac{1}{\inf_{\|{\bf x}\|=1} \|A{\bf x}\|} = \sup_{\|{\bf x}\|=1} \dfrac{1}{\|A{\bf x}\|} = \sup_{\|{\bf x}\|=1} \dfrac{\|A^{-1}(A{\bf x})\|}{\|A{\bf x}\|} = \sup_{{\bf x} \in \mathbb{R}^n \setminus \{{\bf 0}\}} \dfrac{\|A^{-1}(A{\bf x})\|}{\|A{\bf x}\|}$$ Since $A$ is nonsingular, we have $A(\mathbb{R}^n \setminus \{{\bf 0}\}) = \mathbb{R}^n ...


2

Define $z:=x-\frac{\left\langle y,\,x \right\rangle}{\left\langle x,\,x \right\rangle}y$. Rearrange $\left\langle z,\,z \right\rangle \geq 0$, with equality iff $z = 0$, to $\left|\left\langle x,\,y \right\rangle\right|^2 \leq \left\langle x,\,x \right\rangle \left\langle y,\,y \right\rangle$. This inherits the above equality condition, which implies $x,\,y$ ...


2

I'll leave some hints. For a constant $c$, $$ \left\Vert cf\right\Vert ={\it \int_{a}^{b}\left|cf(x)\right|dx=\int_{a}^{b}\left|c\right|\left|f(x)\right|dx=\cdots} $$ The triangle inequality is an application of the triangle inequality on $|\cdot|$: $$ \left\Vert f+g\right\Vert ...


2

I am also new to analysis like you if almost having $\epsilon$ level of experience in "real analysis". Let $f \in \mathcal{C}[a,b]$ and suppose further that $f \neq 0$, hence there is some $c \in [a,b]$ with $f(c) \neq 0$. If $c = a$, then $f(a) \neq 0$ and since $f$ is countinuous at $x = a$, using definition of continuity at $a$ for $f$ we have: $\epsilon ...


2

I think there's no general inequality which holds: The sequences $$a_0 = 2, a_k = 0, k > 0$$ and $$b_0 = \frac{1}{2}, b_k = 0,k > 0$$ should show that (take i.e. $p = 1, q = 2$). To prove $l^p \subseteq l^q$, you have to take a sequence $(a_k) \in l^p$ and show that it is in $l^q$. First, observe that we can assume without loss of generality that ...


2

You know that $\lvert x_i\rvert \le \max_{j=1,\dotsc,n}\lvert x_j\rvert$ for each $i=1,\dotsc, n$. Hence $\lvert x_i\rvert\le \lVert x\rVert_\infty$ for each $i=1,\dotsc,n$. Now, \begin{align*} \lvert x\cdot y\rvert &= \left\lvert\sum_{i=1}^nx_iy_i\right\rvert \le \sum_{i=1}^n \lvert x_i\rvert\cdot \lvert y_i\rvert\\ &\le \sum_{i=1}^n\lVert ...


2

Your large matrix is also symmetric, so computing the 2-norm is a question of computing the eigenvalues. I recommend that you write your large matrix as a Kronecker product $E \otimes A^{-1}$ for a suitable symmetric matrix $E$. Known properties of the Kronecker product will then solve your problem.


2

As Carl Christen said you just need to find the eigenvalues of your $A^{-1}_n$ and get the max. The characteristic polynomial of $A^{-1}_n$ is: $$ p_0(\lambda)=1$$ $$ p_1(\lambda)=\lambda - 4 $$ $$ ... $$ $$ p_n(\lambda)=(\lambda - 4) p_{n-1} + p_{n-2} $$ Being the resolvant a tridiagonal matrix of the form $$A^{-1}_n- \lambda I = \left[\begin{array}{cccc} ...


2

Apply the triangle inequality $\|a+b\| \leq \|a\| + \|b\|$, with $a=x-z$ and $b=z-y$. $$ \|x-y\| = \| x -z + z - y \| = \| (x -z) + (z - y) \| \leq \| (x -z) \| + \|(z - y) \| < 2+3=5 \\ \implies \|x-y\| < 5.$$


2

The problem is that, rather, we have the identity $$\overline{\alpha\beta}=\bar\beta\bar\alpha$$ for any $\alpha,\beta\in\Bbb H.$ Hence, $$N(\alpha\beta)=\alpha\beta\bar\beta\bar\alpha=\alpha N(\beta)\bar\alpha.$$ All that remains is to justify that $N(\beta)$ commutes multiplicatively with every element of $\Bbb H,$ regardless of our choice of ...


2

As comments already pointed out, you can't have a physical space with more than 3 spatial dimensions. So the concept of length as something which can be measured with a ruler breaks down. You can simply continue to use the term “length” for the norm, since it helps intuition, it is compatible with everyday experience in lower dimensions and there is no ...


1

No. Don't satisfies Triangle inequality. For example d([0 0 0],[2 2 2])=12 > d([0 0 0], [1 1 1]) + d( [1 1 1], [2 2 2])=6.


1

We're looking for a vector norm $|\cdot|_T$ on $X$ such that $$\|A\|_T = \max_{x \in X} \frac{|Ax|_T}{|x|_T} \left(= \max_{x \in X, |x|_T = 1} |Ax|_T\right),$$ i.e. $\|\cdot\|_T$ is induced by $|\cdot|_T$. Let $|x|_T = |T^{-1}x|$. (Is this a norm?) Then we have: $$\max_{x\in X} \frac{|Ax|_T}{|x|_T} =\max_{y\in X} \frac{|ATy|_T}{|Ty|_T} =\max_{y\in X} ...


1

The best intuition I can get for your result comes from one dimension. Think of $u$ as being $1$ on an interval $(a,b)$ and then diminishing down to $0$ outside a bigger interval $(c,d)\supset (a,b)$. As you exponentiate $u$ to higher and higher powers, it becomes dominated by what happens on $(a,b)$ and goes to zero outside $(a,b)$. Also, no matter how ...


1

Apparently this is not true, according to link: If $\Omega = (0,1)$ and $f(x) = g(x) = x^{-1/3}$, we have $\|fg\|_2 = \infty$ but $\|f\|_2 = \|g\|_2 < \infty$.


1

Not necessarily. First of all note that if $d$ is a metric induced by a norm then it must satisfy the following two properties (1) $d(x,y)=d(x+z,y+z)$; (2) $d(ax,ay)= |a|d(x,y)$ for all scalars $a$ and vectors $x,y,z$. So, the discrete metric will be a trivial counter example to your question.


1

Your notation is a little idiosyncratic. More precisely, $\|x\| = \inf \{ r \ge 0 | x \in r B \}$. Suppose $x = 0$, then $x \in rB$ for all $r >0$, hence $\|x\| = 0$. If $\|x\| = 0$, then there are $r_k \ge 0$ such that $r_k \to 0$ such that $x \in r_k B$. Since $\cap_k r_k B = \{0\}$ (from 5.), we see $x = 0$. Suppose $\lambda = 0$, then $0 = \| ...


1

To generalize a little bit using the fact that the rationals are dense is the right idea: Let $X$ be a metric space with a countably dense subset $A$. We will show that it has a countable basis. Let $\mathbb{B}=\{b(a,1/n) \mid a \in A, n \in \mathbb{Z_{+}}\}$. Clearly, $\mathbb{B}$ is countable. Let $x \in X$. Let $U$ be a neighborhood of $X$. Then ...


1

Only for almost all $x$. This follows from the definition of the "norm". One possible (equivalent) definition is that $\|f\|_\infty$ is the smallest number $M>0$ such that $|f(x)|\le M$ for almost all $x$.


1

The "functions" in $L^p$ spaces, including $L^\infty$, are actually equivalence classes, where $f$ and $g$ are equivalent if they are equal almost everywhere. So any statement you make about $f(x)$ can only be interpreted in the "almost everywhere" sense. However, for $f \in L^\infty$ you can choose a representative of the equivalence class that is bounded ...


1

A good first step is to perform some experiments with small matrices! Here tools such as MATLAB can be valuable when disproving identities. You will find that the statement under investigation is false in general, but it is certainly true if the matrices $B^{-1}A$ and $A^{-1}B$ are symmetric! There is condition which occurs frequently in pure mathematics, ...


1

I'd say the main argument would be that this formulation is not really advantageous. The thing is that for $\dim X < \infty$, $B[0,1]$ and $S$ are compact and $\left\Vert Tx \right\Vert$ is continuous. Hence, there exists some $x_0 \in S$ such that $$\left\Vert T \right\Vert = \left\Vert Tx_0 \right\Vert$$ making the supremum a maximum. The supremum over ...



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