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4

Yes. This is a consequence of the geometric series and $\left\|AB \right\|_{\infty} \le \left\|A \right\|_{\infty}\left\|B \right\|_{\infty}$. To wit, since $\left\|M\right\|_{\infty} < 1$, then \begin{align} (I-M)^{-1} = \sum_{k=0}^{\infty} M^{k} \end{align} and \begin{align} \left\|M(I-M)^{-1}\right\|_{\infty} &\le ...


3

No, in general you only have $\le$ and not $<$. The triangle inequality gives $$ \|A\|=\|A+B-B\|\leq \|A+B\|+\|B\|. $$ I leave it to you to find an example where equality holds.


3

Let $x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$. $cx=(cx_1,cx_2,\dots,cx_n).$ The definition for the euclidean norm is the following: $||x||=\sqrt{\sum_{i=1}^{n}x_i^2}$ For 2, $||cx||=\sqrt{\sum_{i=1}^{n}(cx_i)^2}=\sqrt{\sum_{i=1}^{n}(c^2x_i^2)}=\sqrt{c^2\sum_{i=1}^{n}x_i^2}=|c|\sqrt{\sum_{i=1}^{n}x_i^2}=c||x||$. You can consult, for example, this article: ...


3

Since $\|-x\|=\|x\|$, this cetainly fails for odd-gons. -- Hagen von Eitzen On the other hand, it holds for even number of sides. See Minkowski functional.


3

$$ \| (x_1,y_1)+(x_2,y_2)\|=\|(x_1+x_2,y_1+y_2)\| =(\| x_1+x_2\|_{X}^{p}+\| y_1+y_2\|_{Y}^{p})^{\frac{1}{p}}\leq$$ $$ \leq ((\| x_1\|_X+\|x_2\|_{X})^{p}+(\| y_1\|_Y+\|y_2\|_{Y})^{p})^{\frac{1}{p}}\leq$$ (we use Minkowski inequality) $$\leq (\| x_1\|_{X}^{p}+\|y_1\|_{Y})^{p})^{\frac{1}{p}}+(\| ...


2

An equivalent definition is $\|x\|_C = \inf \{ t>0 | {x \over t } \in C \}$. This is a little more convenient here. Let $C = \bar{B}(0,1)$, with the $\|\cdot\|_p$ norm. If $\|x\|_p \le t$, then ${x \over t} \in C$, and so $\|x\|_C \le t$. If $\|x\|_p > t$, then for some $\epsilon>0$ we have $\|x\|_p > t+\epsilon$. Then $\|{x \over s} \|_p ...


2

In the first line of the proof, $\| x^*\|$ means the regular norm. We want to see that $\| m^*\|=\| \sigma(m^*)\|$. Till the use of Theorem 3.3 we have seen that $$\| m^*\|\leq \|\sigma(m^*)\|\quad \text{and} \quad \|\sigma(m^*)\|\leq \| x^*\|, \tag1 $$ where $x^*$ is any extension of $m^*$. Let $p:X \to [0,\infty)$ be defined by $p(x)=\| m^*\| \| x\|$. ...


2

The first inequality is the triangle (Minkowski) inequality for the $L^p$ norm, the second inequality is Minkowski inequality for the counting measure, http://en.wikipedia.org/wiki/Minkowski_inequality.


2

Let $\lambda$ be an eigenvalue of $A$ and $v$ an associated eigenvector. We may suppose without loss of generalities that $\|v\|_2 =1$. Thus we have $|v_i|\leq 1$ for every $i$ and there exists $j \in \{1,\ldots,n\}$ such that $|v_j|\geq n^{-1/2}$ since $$ 1 = \left(\sum_{k=1}^n |v_k|^2\right)^{1/2} \leq \left(n\max_{k=1,\ldots,n}|v_k|^2 \right)^{1/2}$$ ...


2

Ad 1): This is right, and proves that at the origin the directional derivative of $N(\cdot)$ is $=1$ in all directions. Here the directional derivative of a function $f$ at a point $p$ in direction $u\in S^{n-1}$ is defined as $$D_uf(p):=\lim_{t\to 0+}{f(p+tu)-f(p)\over t}\ .$$ The fact that $D_uN(0)=1$ for all $u\in S^{n-1}$ already proves that the function ...


2

Let $\alpha = -(x,y)/(y,y)$, assuming $y \ne 0$. Then, by assumption, $$ \|x\|^{2} \le \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2} $$ Using the Pythagorean Theorem: $$ \left\|\left(x-\frac{(x,y)}{(y,y)}y\right)+\frac{(x,y)}{(y,y)}y\right\|^{2} \le \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2} \\ ...


1

Let ${\mathbb M}_n$ be equipped with $\| \cdot \|_1$. Denote by $B_1=\{ X\in {\mathbb M}_n;\quad \| X\|_1\leq 1\}$, the closed unit ball in ${\mathbb M}_n$. For $A\in {\mathbb M}_n$, let $\rho_A:{\mathbb M}_n \to {\mathbb M}_n$ be defined by $\rho_A(X)=XA$. Then the problem is to describe the set $$ \rho_{A}^{-1}(B_1)=\{ X\in {\mathbb M}_n;\quad \rho_A(X)\in ...


1

$||A||_1 = \sup^{}_{||v||_1=1} ||Av||_1 \geq ||A(\mathbf1 *\frac{1}{n})||_1 = \frac{1}{n} ||A\mathbf1||_1 = \frac{1}{n} \sum_{i} \sum_{j} |a_{ij}| \geq \frac{1}{n} \sum_{i} \max_{j} |a_{ij}|$ $ \Leftrightarrow n ||A||_1 \geq \sum_{i} \max_{j} |a_{ij}|$ Recall $||A||_1 := \sup^{}_{||v||_1=1} ||Av||_1$ and that $||\cdots||_1$ satisfies all the ...


1

$$||x+\alpha y||^2=\langle x+\alpha y,x+\alpha y \rangle= ||x||^2+|\alpha|^2||y||^2+\bar{\alpha} \langle x, y \rangle+\alpha \langle y, x \rangle.$$ Show that if $\langle x, y \rangle\neq 0$ there is $\alpha $ such that this expresion is smaller than $||x||$. Clearly, if $\langle x, y \rangle= 0$ then $$||x+\alpha y||^2=||x||^2+|\alpha|^2||y||^2\geq ...


1

In the context of norms, the "largest" is going to be the norm of greatest numerical value. If you're using the Euclidean norm, this "largest value" is going to correspond to length. For $p$-norm vectors, this isn't the case (you can still call the magnitude "length", but it doesn't correspond to the same notion of length we think about in the real world). ...


1

Using Robert Israel's hint, let $y = \dfrac{x}{\|x\|}$. Note that $\|y\| = 1$ and $x = \|x\|y$, so we have $$z^Tx = z^T(\|x\|y) = \|x\|(z^Ty) \leq \|x\|\sup_y\{z^Ty \mid \|y\| \leq 1\} = \|x\|\|z\|_*.$$


1

I think you've already basically worked out the logic. The case for $l_1$ can be easily generalised. Suppose $\{ x_n \}_{n \in \mathbb{N}} \in l_p$, then $(\sum^{\infty}_{n=1}|x_n|^p)^{1/n}<\infty $ and so $\sum^{\infty}_{n=1}|x_n|^p<\infty $. Therefore there exists an $N$ such that for arbitrary $n>N$, $|x_n|^p<1$. i.e $|x_n|<1$. And so for ...


1

Cauchy-Schwartz inequality states: $|y^t b| \le \Vert y \Vert \Vert b \Vert$ for all vectors $y,b$ That you have equality in previous inequality if and only if the vectors are colinear. Are you able to conclude with this?


1

In your example, $$\langle x, y \rangle = \sum_{v=1}^n x_v y_v$$ So $$|\langle x,y \rangle| = \left| \sum_{v=1}^n x_v y_v \right|$$ Therefore, $$|\langle x,y \rangle| = \left| \sum_{v=1}^n x_v y_v \right| \leq \sum_{v=1}^n |x_v y_v| \leq \left(\sum_{v=1}^n |x_v|^p\right)^{1/p} \left(\sum_{v=1}^n |x_v|^q\right)^{1/q}$$ where the first inequality is the ...


1

Hint: Suppose that $x$ with entries $x_j$ satisfies $\|x\|_\infty = 1$. Then the $j$th entry of $Mx$ is given by $$ \sum_{k=1}^m M_{jk}x_k $$ We note that $$ \left|\sum_{k=1}^m M_{jk}x_k\right| \leq \sum_{k=1}^m \left|M_{jk}x_k\right| \leq \sum_{k=1}^m \left|M_{jk}\right| \|x\|_{\infty} $$ Then, for any $j$, we can consider the vector $x$ whose entries are ...


1

No. Consider $x^T=(1,0,0)$ and $A=\operatorname{diag}(\varepsilon,1,1)$, where $\varepsilon>0$ is small.


1

The inequality is not true. Let $p(t)=t^n$. Then $$ \int_0^1t\,|p'(t)|\,dt=n\int_0^1t^n\,dt=\frac{n}{n+1}, $$ while $$ \int_0^1|p(t)|\,dt=\int_0^1t^n\,dt=\frac{1}{n+1}. $$


1

In your definition of the functional $\phi$, what is $x$? Is it a given number in $[0,1]$? In any case. This functional is not going to be bounded. Take $p(x)=x^n$ for example. We have $||p||=1$ but $p'(x)=nx^{n-1}$, which take the maximum value $p'(1)=n$ (assuming the $x$ in your definition of $\phi$ is $1$). Examples like this can be constructed no ...


1

You need to use Hoelder's inequality. Here is a start $$ (f*g)(x) = \int_{-\infty}^{\infty} f(x-t) g(t) dt \implies |(f*g)(x)| \leq \int_{-\infty}^{\infty} |f(x-t)| |g(t)| dt \leq ||f||_p||g||_q , $$ which proves boundedness. It is an easy matter now to prove $f*g\in L^{\infty}$.


1

Note that $A^*A$ is necessarily positive semidefinite. To show this, note that for any $x \in \Bbb C^n$, $$ x^*(A^*A)x = (x^*A^*)Ax = \|Ax\|^2 \geq 0 $$ The result you showed regarding $(x,Mx)$ is called Rayleigh's theorem. See also the min-max theorem (AKA the Courant-Fischer principle).


1

Hint. The Frobenius norm of a matrix is just the $2$-norm of its vectorization.



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