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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 ...


6

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


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 ...


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 ...


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 ...


3

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


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 ...


2

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


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 ...


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 ...


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) ...


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 $$\underset{\|x\| = 1}{\sup} \left \| (A_1 + A_2)x \right \| = \sup_{\| x \| =1} \| A_1 x+ A_2x \| \leq \sup_{\|x \| = 1} \| A_1 x \| + \sup_{\| x \| = 1} \| A_2 x\|.$$


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 ...


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 ...


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 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. ...


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 ...


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

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

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.)



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