# Tag Info

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No. Chose the norm $\|\cdot\|_\text{me}$ defined by $\|x\|_\text{me} :=\frac12 \|x\|_\infty$. You can easily check that this is in fact a norm, but $\|2\cdot e_1\|_\text{me} = 1$. There is however the following theorem (a special case of the fact that all norms on $\mathbb R^n$ are equivalent): For any norm $\|\cdot\|$ on $\mathbb R^n$ there are ...

4

It's clearly a norm: If $\|\nabla u\|_{L^2}=0$ then $u$ is constant, and the only constant in $L^2$ is $0$. On the other hand, if $H^1=W^{1,2}$ were complete under this norm we would have, by the bounded inverse theorem, that both norms are equivalent, which we know to be false (the scaling below should give you an idea of how to build explicit ...

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I have long suspected that this practice of calling the cardinality function the "$\ell_0$ norm" would cause problems, and this post is evidence of that. The so-called "$\ell_0$ norm" is not a norm, and it is not convex. If Wikipedia is to be believed, the term "$\ell_0$ norm" was coined by David Donoho, in his work on using the $\ell_1$ norm (a true ...

3

For the first question: Let a sequence be defined by letting $x_m=1$, and $x_k=0$ for $k \neq m$. Then what happens when $m$ goes to infinity? For the second question: $$| \phi ( \{ x_n \} ) | \leq \sum_{n=1} ^{\infty} |(1 - \frac{1}{n} )x_n| < \sum_{n=1} ^{\infty} |x_n| \leq 1.$$ The strict inequality is true for any sequence where $||\{x_n\}|| \neq ... 3 If you are talking about basic standard matrix norm which is also equivalent to spectral norm, then just compute$A_{sub}^TA_{sub}$and find the maximum absolute value eigenvalue of that, and take square root, which is equivalent to the basic standard matrix norm. You should find that the product matrix is all$1$on the diagonal except for one zero, and ... 3 We can get a quick bound on the operators by using some straighforward inequalities. Call our operator$T$. Then$\|T\| = \sup_{\|x_n\|_\infty = 1} \|Tx\|_2$. Take$\{ x_n \} \in l^\infty$and suppose that$\|x_n\|_\infty = 1$. We can see that $$\| Tx_n \| = \sqrt{ \sum_{n=1}^\infty \frac{|x_n|^2}{n^2} } \le \left(\sup_n |x_n|\right) \sqrt{\sum_{n=1}^\infty ... 3 Let's say you assumed the existence of x_0 such that$$ \Vert f(x_0) \Vert > \Vert f\Vert \Vert x_0\Vert. $$Then you know there exists an \epsilon \in\mathbb{R}_{>0} with$$ \Vert f(x_0) \Vert > (\Vert f\Vert +\epsilon) \Vert x_0\Vert $$because the inequality holds strictly (so we can make the right side a little bigger and the inequality ... 2 Multiplying by the transpose of the matrix gives$$ A^TA = \begin{bmatrix}1&3\\3&10\end{bmatrix}$$The eigenvalues of this matrix are found from the characteristic equation \det(\lambda I - A^TA) = (\lambda-1)(\lambda-10)-9 = 0. Solving this gives \lambda^2-11\lambda+1=0, or \lambda = \frac{11\pm\sqrt{117}}{2}. The norm is the square root of ... 2 \sum_{i}\sigma_i^2=Trace(\Lambda \Lambda^T) where M=U\Lambda V^T. Then,$$\|M\|_F^2=Trace(MM^T)=Trace(U\Lambda V^TV\Lambda^T U^T)=Trace(U\Lambda \Lambda^TU^T)=Trace(\Lambda\Lambda^T U^T U)=Trace(\Lambda\Lambda^T)$$2 You can use the triangle inequality in \Bbb C to prove this inequality. Suppose x = (x_1,\ldots, x_n) and y = (y_1,\ldots, y_n). For k = 1,2,\ldots, n,$$|x_k + y_k| \le |x_k| + |y_k| \le \|x\|_\infty + \|y\|_\infty.$$Therefore$$\max_{1 \le k \le n}|x_k + y_k| \le \|x\|_\infty + \|y\|_\infty.$$That is,$$\|x + y\|_\infty \le \|x\|_\infty + ... 2 Hint: In "intuitive" terms, the 1-norm is a max row-sum of$A$and the$\infty$-norm is the max column sum of$A$(in absolute value). What happens to$A$when you transpose it? 2 Let$\|\cdot\|$be any matrix norm induced by a vector norm. Then we have $$\|A\|_2^2= \rho(AA^H) \leq \|AA^H\| \leq \|A\|\|A^H\|.$$ Here the first inequality follows from a "famous theorem" (see e.g. Proposition 4.4) and the second inequality follows from the fact that$\|\cdot\|$is a matrix norm induced by a vector norm and thus is submultiplicative. ... 2 Recall that an infimum is always attained for continuous maps defined on compact sets. The map you consider can be shown quite directly to be continuous. The set$B$is closed by definition, yet it might not be compact if it is unbounded. However, it is "clear" that the infimum cannot be attained at a point far far away. (Yet give a formal argument for ... 2 You're close, but I think the notation somewhat confusing due to$y$showing up in two places where it is not the same. Given$y \in B_r^A(x)$, what you want to show is that there is some$y' \in B_r(0)$so that$y = x + A^{-1}y'$, hence showing$y \in \{x + A^{-1}y' : y' \in B_r(0)\}$and giving$\subset$. Working backwards by multiplying both sides of$y ...

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This is not a full answer, but maybe helpful anyway, as it clarifies, that an answer depends on the properties of $P$ itself. The induced matrix norm of the $L_2$-vector norm is the spectral norm, which is the maximal singular value of the matrix under consideration, so in order to find out something about $||APA^{-1}||_2$, we should look at ...

1

@ ziutek , a stochastic matrix does not necessarily admit a stationary distribution. cf. below. Note that $||APA^{-1}||_2=\sqrt{\rho (MM^T)}=\max_i \sigma_i$ where $M=APA^{-1}$ and $(\sigma_i)_i$ are the singular values of $M$ in decreasing order. Moreover $\rho(M)=\rho(P)=1$ and (it is true for every matrix) $\sigma_1\geq \rho(M)$. Thus your question is ...

1

This norm (according to conventional notations) is called the nuclear norm and is defined as $\|A\|_*=\sqrt{Tr(A^*A)}$ where $A^*$ is the Hermitian conjugate of $A$.

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If we let $q(x) = \sum_{i=0}^n a_i x^i$ then, if we define $a_{-1}=a_{n+1}=0$, we have $$q(x)(x-\alpha) = \sum_{i=0}^{n+1} (a_{i-1}-\alpha a_i) x^i$$ $$q(x)(\overline{\alpha}x-1) = \sum_{i=0}^{n+1} (\overline{\alpha}a_{i-1}-a_i) x^i$$ From the definition of the two norm being equal we get the equation \sum_{i=0}^{n+1} |a_{i-1}-\alpha a_i|^2 = ... 1 \begin{align} \left\| Ax \right\|_{2}^{2} :=\left<Ax, Ax\right> = \left|\left<v,x\right>\right|^{2}\left\|u\right\|_{2}^{2} \le \left\|u\right\|_{2}^{2}\left\|v\right\|_{2}^{2} \left\|x\right\|_{2}^{2} \end{align} So \left\|A \right\|_{2} \le \left\|u\right\|_{2}\left\|v\right\|_{2}. On the other hand, taking x = v/\left\|v\right\| gives ... 1 You have\|M(X)\cdot Y\|\le\left(\sum_{i=1}^{n} {\|A_i\|_\infty}^2\right)^{\frac{1}{2}}\|X\|$$Now observe that this inequality holds for all Y with \|Y\| \leq 1  and that the right side of the inequality is independent of Y, which makes it an upper bound for \|M(X)\cdot Y\|, therefore it must be greater or equal to the supremum of this ... 1 Let x \in \mathbb{R}\setminus\mathbb{Q} be fixed. For the construction of a counterexample, the following fact is useful: For every k \in \mathbb{N}, there exists \varepsilon > 0 : x_n \not\in (x-\varepsilon, x+\varepsilon) for n \le k. Now, construct a function f with the following properties: 0 \le f \le 1, f(x) = 1, f(x_n) = 0, n ... 1 Hint: Use that$$||{\bf X} \times {\bf Y}|| = ||{\bf X}||\,||{\bf Y}|| \sin \theta,$$where \theta is the angle between {\bf X} and {\bf Y}. 1 You just need to "matricize" your result a bit. Let X\in {\mathbb R}^{p\times n_1}, Y\in {\mathbb R}^{p\times n_2}, D\in {\mathbb R}^{n_1\times n_2}, v_1\in {\mathbb R}^{n_1}, and v_2\in {\mathbb R}^{n_2}; where the elements of the v-vectors are all ones. Then$$ D = {\rm diag}(X^TX)\,v_{2}^T + v_{1}\,({\rm diag}(Y^TY))^T - 2 X^TY $$... 1 Using \triangle-inequality & CS-inequality: LHS = ||\displaystyle \sum_{i=1}^n \lambda_iy_i||^2\leq \left(\displaystyle \sum_{i=1}^n |\lambda_i|\cdot ||y_i||\right)^2\leq RHS  1 Let denote (e_n)_{n\in \mathbb{N}} the canonical basis. Then \phi (e_n)= (1-1/n), so for every n\in \mathbb{N}, \| \phi \| \geq (1-1/n). For your second question, observe that |\phi (x) |\leq \sum |(1-1/n)x_n| < \sum |x_n | =1, if \|x \| = 1. 1 Hints: To show the norm is at least 1, try it on the "standard basis vectors" that have a single 1 and everything else 0. To show there is no member of \ell^1 where that is attained, note that if x_m \ne 0, |\phi(x)| \le \|x\| - |x_m|/m. 1 If x\in[0,1],$$|f(x)-f(0)|\leq x\sup_{y\in[0,x]}|f'(y)|\leq x\|f'\|_\infty$$by Lagrange's theorem. Hence$$|f(x)|\leq|f(0)|+x\|f'\|_\infty\leq |f(0)|+\|f'\|_\infty$$so, taking the supremum as x varies in [0,1], you get$$\|f\|_\infty\leq |f(0)|+\|f'\|_\infty$$so$$\|f\|_\infty+\|f'\|_\infty\leq |f(0)|+2\|f'\|_\infty\leq 2|f(0)|+2\|f'\|_\infty$$... 1 Hint:  |f(t)| = |f(0)+\int_{0}^{t}f'(x)dx| \le |f(0)|+\int_{0}^{t}|f'(x)|dx \le |f(0)|+\|f'\|_{\infty} for 0 \le t \le 1. 1 For real valued functions, both notions you write are the same, because if we set x_n := \int |f-f_n|^2 \, dx, then convergence in mean means x_n \to 0, while convergence in (L^2)-norm means \sqrt{x_n}\to 0. Since x_n\geq 0, these statements are equivalent (use continuity of the square root and of x \mapsto x^2). Finally, continuity of the norm ... 1 Let A=(a_{ij}) and A^H=(b_{ij}) where b_{ij}=\overline{a_{ji}} so$$||A^H||_{\infty}=\max_i\sum_{j=1}^n|b_{ij}|=\max_i\sum_{j=1}^n|\overline{a_{ji}}|=\max_\ell\sum_{k=1}^n|{a_{k\ell}}|=||A||_1

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