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1

Perhaps an example will help. Let $A = \{7, 13\}$. Then $E$ is the vector space of complex-valued functions on $\{7, 13\}$. Let $a = 7$; then $N_7$ is a seminorm on $E$. Letting $a=13$, it's also true that $N_{13}$ is a seminorm on $E$. For instance, letting $f(7) = i, f(13) = 0$, we have $N_7(f) = |f(7)| = |i| = 1$, whereas $N_{13}(f) = |f(13)| = 0$. Since ...


0

$N_a(f)=0$ does not imply $f(x)=0$ for all $x\in A$. It only implies $f(a)=0$.


0

If $V,W$ are vector spaces and $|\cdot|$ is a norm (ort seminorm) on $W$, and $T\colon V\to W$ is linear, then $v\mapsto |Tv|$ is a seminorm on $V$


2

Let's work with $A=M_n(\mathbb{C})$. Consider $x\in A$, then $\|x\|<1$ for some subordinate matrix norm iff $\;\mathrm{Spec}(x)\subset D^{\circ}$ where $D^{\circ}=\lbrace z\in\mathbb{C}\text{ s.t. }|z|<1\rbrace$ is the open unit disc. Indeed, if $\|\cdot\|$ is subordinate to some norm $|\cdot|$ on $\mathbb{C}^n$, then for any nonzero ...


0

By the Mazur-Ulam theorem every bijective isometry between normed spaces, in particular between identical normed spaces, is affine. The linked paper also contains an example of injective non-affine isometry. See also Should isometries be linear?


1

First, your definition of $||f||_s$ is wrong: the square root on the right-hand side is missing. Now, if we use the correct definition $$ ||f||_s=\sqrt{\sum_m(1+m^2)^s|\hat f(m)|^2}, $$ then everything is pretty straightforward: $$ \sup_x|f(x)|=\sup_x\left|\sum_m\hat f(m)e^{imx}\right|\le\sum_m|\hat f(m)|=\sum_m |\hat f(m)|(1+m^2)^{s/2}(1+m^2)^{-s/2} $$ (by ...


1

The first three properties of a norm should be very easy to verify. The one that is giving you grief is triangle inequality, but it is no more difficult than you think it is since $$\underset{\|x\| = 1}{\sup} \left \| (A_1 + A_2)x \right \| = \sup_{\| x \| =1} \| A_1 x+ A_2x \| \leq \sup_{\|x \| = 1} \| A_1 x \| + \sup_{\| x \| = 1} \| A_2 x\|.$$


0

The FOIL rule with column vectors $u, v \in \mathbb R^n$ tells us that \begin{align} \|u - v\|^2 &= (u - v)^T (u - v) \\ &= u^T u - u^T v - v^T u - v^T v\\ &= \|u\|^2 - 2 u^T v + \|v\|^2. \end{align} Applying this to our particular problem, we find that \begin{align} \|Y - X \beta \|^2 &= \|Y\|^2 - 2 Y^T X \beta + \| X \beta \|^2. ...


1

We have that $\def\norm#1{\left\|#1\right\|}$ \begin{align*} \norm{u(t)-v(t)}^2_{L^2} &= 2\pi\norm{\widehat{u(t)} - \widehat{v(t)}}^2_{\ell^2(\def\Z{\mathbb Z}\Z)}\\ &= 2\pi\sum_{m\in\Z}\def\abs#1{\left|#1\right|}\abs{e^{tP(im)}\bigl(\hat u(0,m)-\hat v(0,m)\bigr)}^2 \end{align*} Now note that if $P$ maps the imaginary axis into itself, that ...


0

Try checking linearity of the norm with $-3$ as the scalar instead, or by checking the additive property - you should see that it fails. The dot product is linear in the sense that the map $x\mapsto x\cdot y$ for some fixed $y$ (independent of $x$) is linear; the map $x\mapsto x\cdot x$ is not linear.


6

Does $$\|x+y\|=\|x\|+\|y\|\quad?$$ Find a counterexample and recall the triangle inequality.


1

They aren't equal only equivalent, which metrics are if they induce the same topology. This means that they have the same open sets, and since a field is also a one-dimensional vector space, this just means they need to have the same unit balls. But then $$\lVert \cdot \rVert_1\le 1 \iff \lVert \cdot\rVert_2\le 1$$ so choose any $y$ such that $0\ne \lVert ...


2

Observe that the $2$-norm of a matrix is $$\|A\|_2=\sup_{\|x\|=1}\sqrt{\langle x,A^*Ax\rangle}=\sup_{\|x\|=1}\sqrt{\|Ax\|^2}=\|A\|,$$ i.e. the 2-norm and the operator norm coincide. Knowing only $\|B\|\leq\|A\|$, I would not expect to be able to control $\|CBC\|$ in terms of $|\|CAC\|$. The following example shows that you need some type of extra condition ...


0

Having duly noted Martin Argerami's answer, I thought it worth mentioning that for normal $A,B$ in a $C^*$-algebra, we have $$ \|(A+B)^m\|^{1/m} \leq \|A+B\| \leq \|A\|+\|B\|=\|A^m\|^{1/m}+\|B^m\|^{1/m}. $$ Observe also that for commuting normal operators, we have the following Cauchy-Schwartz type inequality $$ \|AB x\|^2=\langle A^*Ax,B^*Bx\rangle\leq ...


1

The concept of this "norm" does not come from the desire to generalize $L^p$ norm. It comes from the desire to compress information. When you save a picture as JPEG file, information is stored in the form of a sequence of Fourier coefficients. To reduce the size of the file, we want to store as few coefficients as possible, and still have a decent image. ...


0

The following adds nothing to the above proofs other than a slightly geometric flavour. It is too long for a comment, so I am including it as an answer. I am taking the space to be $\mathbb{C}^n$. It is not hard to show (see http://math.stackexchange.com/a/424335/27978 for example) that $p \mapsto \|x\|_p$ is non-increasing, $\lim_{p \to \infty} \|x\|_p = ...


3

The $p$ norm of a vector is defined as such: $\|x\|_p = (\sum_{i=1}^{n}|x_i|^p)^\frac{1}{p}$. Notice that when $p=2$ this is the simple euclidean norm. You asked about the infinity norm. When $p$ tends to infinity, we can see that: $$\lim_ {p \to \infty} \|x\|_p = \lim_ {p \to \infty} (\sum_{i=1}^{n}|x_i|^p)^\frac{1}{p}$$ Convince yourself that if ...


9

General Approach As shown in this answer, $$ \lim_{p\to\infty}\left(\int_A|f(x)|^p\,\mathrm{d}x\right)^{1/p}=\sup_{x\in A}|f(x)|\tag{1} $$ Using a discete measure, $(1)$ shows that $$ \lim_{p\to\infty}\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}=\max_{1\le k\le n}|x_k|\tag{2} $$ Since $$ \|x_k\|_p=\left(\sum_{k=1}^n|x_k|^p\right)^{1/p}\tag{3} $$ if we take the ...


4

Here's the way I like to think about it (which is not too rigorous but can be made rigorous). We have, more generally, the $L^p$ norms ($1\le p < \infty$): $$(\|x\|_p)^p:=\sum_{i=1}^n |x_i|^p.$$ The Euclidean norm is a special case of this (take $p = 2$); the taxicab norm is also a special case (take $p=1$). Suppose $|x_i|\ge |x_j|$ for all $1\le j\le ...


1

Let $v=\frac{5}{9}$ and $w=-\frac{5}{9}$. Then $(v,w)$ is clearly not in the first set, but it is in the second since 1) $|v+1w|=0\le1$ 2) $|v+(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)w|=\frac{5}{9}|\frac{3}{2}-\frac{\sqrt{3}}{2}i|=\frac{5}{9}\sqrt{3}\le1$ 3) ...


2

By homogeneity and the triangle inequality $$\left\|\frac{1}{n}\sum_{k=1}^nx_k \right\|=\frac{1}{n}\left\|\sum_{k=1}^nx_k \right\| \leq \frac{1}{n}\sum_{k=1}^n\|x_k \| \leq\frac{1}{n}\sum_{k=1}^n\|x_m \| =\|x_m\| $$


4

Works with $C=4$. Can be improved a bit if someone wants to. Let $S=\sum_{n \geq 1} a_n$. Case 1: There exists $N$ such that $a_N\ge S/4$. Then the right hand side of the above inequality is at least $$\Big(2^N (S/4)^2\Big)^{1/4}\Big(2^{-N} (S/4)^2\Big)^{1/4} = \frac{S}{4}$$ Case 2: There is no $N$ as above. Then let $N$ be the smallest integer such ...


2

The quantity $\frac{x^T A x}{||x||_2^2}$ is the Rayleigh quotient of $A$ and its maximum value is the largest eigenvalue of $A$, $\lambda_{max}$. Noting that $||A||_F = \sqrt{tr(A A^T)} = \sqrt{tr(A A)} = \sqrt{tr(A^2)} = \sqrt{\sum_i \lambda_i^2} \geq \sqrt{\lambda_{max}^2} = |\lambda_{max}| \geq \lambda_{max}$, we get the inequality, where the steps in ...


3

$$\|\frac{1}{2}(1,0)+\frac{1}{2}(0,1)\|_0=2 > 1=\frac{1}{2}\|(1,0)\|_0+\frac{1}{2}\|(0,1)\|_0$$


1

The formula you wrote combines the two halves of the answer. It shows that a norm is Euclidean iff it's restriction to each 2-d linear subspace is a Euclidean norm. Now, see if you can prove that a norm on a 2-plane is Euclidean iff the unit ball is an ellipse. (Use the same formula.)


1

Not in general, for example the infinity norm is a norm on any vector space over a totally ordered field that does not come from an inner product. E.g. for a vector $\vec{v} = (v_1, \dots, v_n) \in \mathbb{R}^n$, defined $||\vec{v}||_\infty = \max\{v_i\}$. You can check that this is a norm, and does not come from an inner product.


1

Note that, in your case with the $2$-norm, you have $\|Ax-b\|_2^2 = \langle Ax-b,Ax-b \rangle $, where $\langle \cdot , \cdot \rangle$ is the euclidian scalar product. But this is not always possible, for example the $p$-norm with $1 \leq p < \infty$ defined by $$\|x\|_p := \left(\sum_{k=1}^n |x_k|^p\right)^{1/p},$$ is such that there is no scalar product ...


1

Note that for the characteristic function of an interval, it is simple to prove that $$ \lim_{h\to0}\int_{\mathbb{R}}|\chi_I(x+h)-\chi_I(x)|\,\mathrm{d}x=0\tag{1} $$ Approximating simple functions by a finite linear combination of such characteristic functions, then Lebesgue integrable functions by simple functions, we can show that $$ ...


2

This is not possible, because Claim: For a $A$ being an $n\times n$ matrix $A$ we have $$1/n\cdot \|A\|_1\le \|A\|_\infty \le n\cdot \|A\|_1$$ Proof: $$ \begin{align} \|A\|_\infty &= \max_i \sum_j |a_{ij}| \le \sum_j\sum_{i} |a_{ij}|\le \sum_j \max_{k} \sum_i|a_{ik}|=n\|A\|_1 \end{align} $$ This shows the first inequality. The second follows ...


2

Good observation skills! Let's chase definitions and then be happy :) Let me write $n = 10, m = 3$ and let your matrix be denoted $A$. Then we calculate! $$ \begin{align} f &= \|A\|_F \;\; \text{ your defn} \\ & = \sqrt{\sum_{i=1}^n\sum_{j=1}^m |A_{ij}|^2}\;\; \text{ defn of Frobenius norm} \\ & = \sqrt{\sum_{i=1}^n\sqrt{\sum_{j=1}^m ...


1

Consider whether, for a given $i$, $z_{i}$ and $w_{i}$ have the same or different sign. If they're the same, then it is trivial that $$\max\left\{\lambda z_{i}+(1-\lambda)w_{i},0\right\}=\lambda\max\left\{z_{i},0\right\}+(1-\lambda)\max\left\{w_{i},0\right\}$$ for all $\lambda\in[0,1]$. If they're different, then consider the inequality between $\lambda ...


4

To add to Qiaochu's answer, if $(A,\|\cdot \|)$ is a normed algebra such that multiplication is continuous, then there is an equivalent norm $\|\cdot\|'$ on $A$ such that $$ \|xy\|' \leq \|x\|'\|y\|' $$ In fact, by the uniform boundedness principle, one has that multiplication is jointly continuous. ie. $\exists M \geq 1$ such that $$ \|xy\| \leq M ...


5

You can think of $\| xy \| \le \| x \| \| y \|$ as a multiplicative version of the triangle inequality $\| x + y \| \le \| x \| + \| y \|$; in that sense, it's as natural to use as the triangle inequality. It's one of the axioms of a normed algebra for several reasons, including but not limited to: It guarantees that multiplication is continuous in the ...


3

This follows from $$ \|f\| = \sup_{\|x\| = 1} |f(x)|. $$ If you like, take a sequence $x_k$ such that $\|x_k\|=1$ and $|f(x_k)| \to \|f\|$. Then set $y_k = x_k / |f(x_k)|$, so $|f(y_k)| = 1$, and $\|y_k\| \to \frac{1}{\|f\|}$.


0

Because it's not differentiable when any row of $W$ is zero. $$\|W\|_{2,1} = \sum_i \left( \sum_j w_{ij}^2 \right)^{1/2}$$ When it is defined, the partial derivative with respect to an individual element of $W$ is $$\frac{\partial}{\partial w_{ij}} \|W\|_{2,1} = w_{ij} \left( \sum_k w_{ik}^2 \right)^{-1/2}$$


0

To clarify your question: By definition $\mathscr D'$ is the dual of the $\mathscr D$ (with its locally convex inductive limit topology) and the topology of uniform convergence on all bounded sets is the locally convex topology on $\mathscr D'$ defined by the semi-norms $p_B(u)=\sup\lbrace |u(\varphi)|: \varphi\in B\rbrace$ for all bounded sets $B$ of ...


0

Thank you, I understood your proofs. But, why mine is not correct? I used Cauchy–Schwarz inequality $||Ax||^2_{n}=(\sum\limits_{i,j=1}^n a_{i,j} x_j)^2 = \sum\limits_{i=1}^n (\sum\limits_{j=1}^n a_{i,j} x_j)^2 \le \sum\limits_{i=1}^n (\sum\limits_{j=1}^n a^2_{i,j})(\sum\limits_{j=1}^n x^2_j)= M ||x||^2_n $ now it's better?


0

I propose you here a way to show it regardless of the norm used: Note that for every $x \neq 0$, holds $$\|A\frac{x}{\|x\|} \| = \frac{\|A\|}{\|x\|}$$ and thus $$\|A\|:= \sup \left\{\frac{\|Ax\|}{\|x\|}: x \in \mathbb{R}^n\right\} = \sup \left\{\|A\frac{x}{\|x\|} \|: x \in \mathbb{R}^n\right\}= \sup \left\{\|Ax\|: \|x\|\leq 1, x \in \mathbb{R}^n\right\}. ...


4

We have $$ \left\{ \frac{\|\mathcal Ax\|_{n}}{\|x\|_n} : x \in \mathbb R^n \right\} = \left\{ \|\mathcal Ax\|_{n} : x \in \mathbb R^n, \|x\|_n=1\right\} $$ The set on the right is compact because $\mathcal A$ and norm are continuous and the unit sphere is compact. Hence, the set on the left is bounded.


2

Since you are using double sums, I highly advise you to use two sumation signs. For example, the $i$-th component of $Ax$ equals $(Ax)_i = \displaystyle \sum_{j=0}^n a_{ij}x_j$, so the value of $||Ax||^2$ is $\displaystyle\sum_{i=0}^n(Ax)_i^2$ which equals $$\sum_{i=0}^n \left(\sum_{j=0}^n a_{ij}x_j\right)^2$$ To prove what you need, I advise you use the ...


0

Assuming you are asking why it is finite, it is a finite sum of finite elements and therefore finite.


1

In the below, $\{e_1,\dots,e_n\}$ denote the standard basis vectors. It helps to write out the norm using dot products. In this case, we have $$ \|Av\|^2 = (Av)^*(Av) = v^*A^*A v \equiv v^*v = \|v\|^2 $$ In fact, we can rewrite this as $v^*(A^*A - I)v = 0$, as I will leave to you to verify. So the question is now if $B = A^*A - I$ satisfies $v^*Bv = 0$ ...



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