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1

The only example is $\Bbb C$. Say $A$ is a Banach algebra with the given property and say the identity is $e$. First suppose that $a$ is invertible and $\lambda\in\sigma(a)$, the spectrum of $a$. Then $|\lambda|\le||a||$. And $\lambda^{-1}\in\sigma(a^{-1})$, so $|\lambda^{-1}|\le||a^{-1}||=||a||^{-1}$. So $|\lambda|=||a||$. Now if $a$ is any element of $A$,...

0

First note that in the SVD of $A$, the diagonal entries of $D$ are guaranteed to be nonnegative. So the absolute values are unnecessary in the first definition of the trace norm. The two definitions of the trace norm are equivalent. Note that $A = UDV^T \implies A^T A = V D^T D V^T \implies A^T A V = V D^T D$, which shows us that the columns of $V$ form a ...

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The $\mathcal{H}^{\infty}$ norm is specified from one vector input to one vector output. Right now, you have written things with two inputs. So you need to reorganize your $A,B,C,D$ matrices.

2

Let $\sigma_i, i=1,...,n$ the $\Bbb{Q}$-isomorphisms of $K$ where $n=[K;\Bbb{Q}]$ , let $x\in K$ , a primitive element of $K$ so by definition $N(x)=\prod_{i=1}^{i=n}\sigma_i(x)$ then $N(-x)=\prod_{i=1}^{i=n}\sigma_i(-x)=(-1)^n\prod_{i=1}^{i=n}\sigma_i(x)=(-1)^nN(x)$ so the equality $N(-x)=-N(x)$ , then give $n$ odd. Edit: In Number Field Sieve, we take $... 1 If$A$is symmetric and positive definite, then its eigenvalues are real and positive. The spectral norm of$A$is given by $$\|A\|_2 := \sigma_{\max} (A) = \sqrt{\lambda_{\max} (A^T A)} = \sqrt{\lambda_{\max} (A^2)} = \sqrt{\lambda_{\max}^2 (A)} = |\lambda_{\max} (A)| = \lambda_{\max} (A)$$ 1 You should have a look of the various definitions of convergence again. Convergence in the$||.||_\infty $norm is uniform convergence, and it does matter, because only uniform convergence allows you to conclude that the limit of a sequence of continuous functions is continuous. Having said that it's in fact almost trivial resp a tautological statement ... 1 We calculate $$\DeclareMathOperator{\tr}{trace} \|X - XWG^T\|_F^2 = \\ \tr[(X - XWG^T)^T(X - XWG^T)] =\\ \tr(X^TX) - 2\tr(X^TXWG^T) + \tr(GW^TX^TXWG^T)$$ Note, however, that order matters in that last term. 0 Here's a start: The gradient of$\|x_i - y\|_2$with respect to$y$(if$y \ne x_i$) is the unit vector in the direction of$y - x_i$. At a minimum that is not one of the$x_i$, those unit vectors will add to$0$. The minimum might also be one of the$x_i$: if that happens, then the sum of the unit vectors for the other$x_i$has norm$\le 1$. 0 Hint You have to find a vector$x$such that$||Ax||_1 = \max_\limits{j=1,...,d} \sum_\limits{i=1}^d |a_{ij}|$Let$j_0$such that:$\max_\limits{j=1,...,d} \sum_\limits{i=1}^d |a_{ij}| = \sum_\limits{i=1}^d |a_{ij_0}|$Let$e$the vector such that$e_j = 1$if$j=j_0$and$e_j = 0$otherwise. Now compute$||Ae||_1$. 1 A norm is a thing we sometimes define on a vector space. The quotient space you describe is not a vector space; it has no concept of addition. So one can't even begin talking about a norm on it. The quotient space is naturally identified with unit sphere in$V$, since each equivalence class contains exactly one element with unit norm. This leads to a ... 2 This is one of those cases where maybe it's better --and more instructive--to prove the general case first. Then, adapting the proof to your specific function is easy. So, suppose$v,w\in \mathbb F^{n}$. Then,$w=v+(w-v)$so$\Vert w\Vert \leq \Vert v\Vert +\Vert w-v\Vert \Rightarrow\Vert w\Vert-\Vert v\Vert\leq \Vert w-v\Vert$. Interchanging the roles of ... 0 This is not a norm. Consider$x=(1,0,0,0,\ldots)$and some$|\lambda| \neq 1$For e.g.$\lambda = 2$we get $$||\lambda x|| = \frac{|\lambda|}{2(1+|\lambda|)} = \frac{1}{3}$$ but on the other hand $$|\lambda|||x|| = 2 \frac{1}{2(1+1)} = \frac{1}{2}$$ 0 This is not a norm. In fact, it is only a F-norm. It is usually defined in a Fréchet space. And it is always defined by using a series of semi-norms. I am not sure what it is called formally. But I call it F-norm. 0 No. In$L^1(0,2)$, let$f=\chi_{[0,1]}$and$g=\chi_{[1,2]}$. Then $$||f+hg||=1+|h|.$$ 1 Divide by$n_k$in the inequality $$\tag{*}\|f(x_{n_k})\| > \varepsilon + n_k \|g(f(x_{n_k}))\|, \quad k \in \mathbb{N}$$ and letting$k$going to infinity, we get, by boundedness of$\left(f\left(x_{n_k}\right)\right)_{k\geqslant 1}$, that$\lim_{k\to +\infty}g\left(f\left(x_{n_k}\right)\right)=0$. By continuity of$g$we also have that$\lim_{k\to +\...

2

Let $V$ be any real or complex vector space, $S$ the corresponding projective space (i.e. the equivalence classes of $V \backslash \{0\}$, where two elements are equivalent if they are scalar multiples of each other), and $\widehat{p}$ any positive real-valued function on $S$ such that $\sup_{s \in S} \widehat{p}(s) \le 2 \inf_{s \in S} \widehat{p}(s)$. ...

0

$\|f(x_{n_k})\|\leq \|x_{n_k}\| \|f\|=\|f\|$. This implie that from your inequality, $\|f\|\geq \|f(x_{n_k})\|>\epsilon +n_k\|g(f(x_{n_k}))\|$ since $f(x_{n_k})$ converges towards $y$, $lim_n\|g(f(x_{n_k}))\|=\|g(y)\|$. We deduce: $\|f\|\geq \epsilon+n_k\|g(y)\|$. This implies that $g(y)=0$ (since $n_k\|g(f(x_{n_k}))\|$ has to be bounded) and $y=0$ ...

0

Inner products In $\mathbb C:\;\,\qquad \langle x, y\rangle=\bar yx$ In $\mathbb C^2:\qquad\langle x, y\rangle=\bar y^\top x$ Norms In $\mathbb C:\;\,\qquad ||x||_1^2=x\bar x$ In $\mathbb C^2:\qquad||x||_2^2=||x_1||_1^2+||x_2||_1^2$ Calculation $$||e^{\theta}x-e^{i\phi}y||_2^2=(e^{i\theta}x_1-e^{i\phi}y_1)\overline{(e^{i\theta}x_1-e^{i\phi}... 0 The proof is fine. But if you do not find it convincing yet this might mean you should add more details. For example, you could justify the last inequality a bit more by saying that for each j one has |x_j| \le \max\{|x_1|,...,|x_n|\} and |y_j| \le \max\{|y_1|,...,|y_n|\}  and thus |x_j|+|y_j| \le \max\{|x_1|,...,|x_n|\} +\max\{|y_1|,...,|y_n|\} . 3 The norm \|\cdot\|_\infty is quite special for the space C([a,b]), since this space is complete with respect to this norm. In fact, the convergence in norm \|\cdot\|_\infty implies the uniform convergence of functions in C([a,b]) and the uniform limit of continuous function is a continuous function. If one consider other norms, for example the norm "... 2 Looking at the definition of a norm on wikipedia (which matches the definitions I encountered in textbooks) (link here) if V is a vector space, then a norm is a function \rho: V\to \mathbb R that satisfies certain properties so it is quite explicitly stated that the norm of every vector is a real number, thus finite. 0 This paper describes six different measures, namely: Hausdorff-distance (HD), Trucco-distance (TD), Modified line segment Hausdorff-distance (MHD), Modified perpendicular line segment Hausdorff-distance (MPHD), Midpoint-distance (MD), Closest point-distance (CD) and our new Straight line-distance function (SD). WIRTZ, Stefan; PAULUS, Dietrich. ... 2 Normed spaces are sets along with norms. If you want to remove the norm, and just treat it as a space, you're free to do so. We say "normed" if we have a norm in mind. Some spaces are "normable" but we haven't chosen a specific norm. And different norms induce different topologies, in particular different ways of things converging. As you're possibly aware, ... 1 You know \|\delta\|\le 6. Can you find some f such that \|f\|_{\sup}=1 and \delta(f)=6? This should not be hard to find. 0 The question of continuity is one of "can we put the limit inside?" So what this question asks is: "if f_n\to f in the following norms, then is it true that \displaystyle\int_{-1}^1f_n\to\displaystyle\int_{-1}^1 f?" So for the first one, if f_n\to f uniformly, then is it true that \displaystyle\int_{-1}^1f_n\to\int_{-1}^1f? Well, yes, the ... 1 The notation S^\perp means \{x\in H\mid \langle x,y\rangle=0,\text{ for all }y\in S\}. Let x\in\ker T; you need to prove that, for every m with \lambda_m\ne0, you have \langle x,\varphi_m\rangle=0. You know that \sum_n\lambda_n\langle x,\varphi_n\rangle\varphi_n=0, so also$$ \Bigl<\sum_n\lambda_n\langle x,\varphi_n\rangle\varphi_n,\...

0

HINT: You need to use the fact that $\phi_n$ is orthonormal. That is, $\left\langle \phi_n,\phi_m \right\rangle = \left\lbrace \begin{array}{cc} 0 & n \neq m \\ 1 & n = m \end{array} \right.$. Start by supposing that $f \in \ker(T)$ so that $T(f) = \sum_{n=1}^\infty \lambda_n \left\langle f,\phi_n \right\rangle \phi_n=0$. What can you say if $f$ is ...

1

Note that each $\varphi_n$ is of norm one, and $F(\varphi_n) = \lambda_n\varphi_n$, so $|F|$ is at least $|\lambda_n|$ for each $n$; that is $|F|\geqslant \sup |\lambda_n|=M$. On the other hand by Cauchy Schwarz you get $\langle x,\varphi_n\rangle$ is of norm at most $|x|$ so that $|Fx|\leqslant M|x|$, and the other inequality holds.

1

For any $\lbrace u_n \rbrace_{n \in \mathbb{N}} \subset \mathbb{R}^n$ you can build an orthonormal system $S=\lbrace e_j \rbrace_{j \in \mathbb{N}}$ with Gram-Schmidt: Let $e_1= u_1 / \left \| u_1 \right \|_{n}$ and by recurrence define $e_j = v_j / \left \| v_j \right \|_{n}$ where $v_j = u_j - \sum_{k=1}^{j-1} (u_k, e_j)_n e_j$ for $j=2,3,...$ Note that ...

1

gram-schmidt concerns orthogonality. This means we must have an inner product around. This means you must also have an induced norm. You use that norm. The induced norm is $$||x||=\sqrt{\langle x,x\rangle}$$

1

For a real Hilbert space $H$, the square norm $N(h):=\langle h , h ⟩$ is (Fréchet) differentiable with derivative $dN(h)v = 2⟨ h, v ⟩$. In this case we have $H=L^2_x$ where the subscript indicates that its the space of spatially $L^2$ functions. We then have the chain rule for functions $f=f(t,x)∈ H^1_tL^2_x = H^1([0,T],L^2(Ω)) = H^1(0,T;L^2(Ω))$, $$\frac{... 0 With one such norm, the space is called a weighted \ell^1 space. I'm not aware of any term for the space you get using the collection of norms. The norms will remain norms after completion. Indeed, any Cauchy sequence with respect to these norms converges componentwise, which allows us to identify its limit with a sequence indexed by I. Clearly, for any ... 0 Assume that x = a+bi where i^2 = -1. Then, the derivative of \ell_p norm of x (i.e., \|x\|_p) is given by$$d/dx (|x|_p) = d/da (|x|_p) - d/db (|x|_p)i$$0 Hint for the C(X) question: Suppose X is a compact metric space with more than one point. (Note that if X is not compact, it's not clear that \| \,\|_\infty is a norm on C(X).) Consider f(x) = 1/(1+d(x,a)),g(x) = 1/(1+d(x,a))^2. 0 First, note that the map X \mapsto X^{-1} is a continuous map from GL back to itself. Moreover, the map \|\cdot\| : M_{m \times n} \to \Bbb R is continuous. It follows that the map X \mapsto \|X^{-1}\| is a continuous map from GL_{n \times n} to \Bbb R, since it is the composition of two continuous map. So, since (A_n) is a sequence in GL... 5 There are several good answers here, one accepted. Nevertheless I'm surprised not to see the L^2 norm described as the infinite dimensional analogue of Euclidean distance. In the plane, the length of the vector (x,y) - that is, the distance between (x,y) and the origin - is \sqrt{x^2 + y^2}. In n-space it's the square root of the sum of the ... 2 OK let's see if this helps you. Suppose you have two functions f,g:[a,b]\to \mathbb{R}. If someone asks you what is distance between f(x) and g(x) it is easy you would say |f(x)-g(x)|. But if I ask what is the distance between f and g, this question is kind of absurd. But I can ask what is the distance between f and g on average? Then it is$$...

3

I'm not sure how intuitive is this, but the norm $||f||_{L^2([a,b])}$ of an integrable function defined on $[a,b]$ is the square root of the "area under the graph" of $|f|^2$ on the interval $[a,b]$. For example, if $f \equiv C$ is a constant function, then the area under the graph of $f^2 = C^2$ is the area of a rectangle given by $(b - a)C^2$ and so $||f||... 3 If you have some physics background, then$L^2$norm can often be interpreted as the "energy" of the wave functions. Physical interpretation of L1 Norm and L2 Norm In quantum physics, the$L^2$norm represents the probability of detecting a particular pure state amount many mixed states. In statistic, minimizing the$L^2$norm of the difference between 2 ... 0 By the Chinese remainder theorem it is enough to consider ideals of the form$I = P^n$for prime ideals$P$. We show that$N(P^n)=N(P)^n$by induction. The case$n=1$is clear. Assume now that it holds for$n$. We want to show it for$n+1$. The map$A_P/P^{n+1}\rightarrow A_P/P^n$is surjective with kernel$P^n/P^{n+1}$, which is a$1$-dimensional$A_P/P$-... 0 As pointed out by PhoemueX, the proof is correct. However, we do not have$\lVert \varphi_n\rVert\leqslant 1/2$. For example, if we take$x$such that$x_k=\sqrt k$for$1\leqslant k\leqslant n$and$x_k=0$for the other$k$'s, then$\$\frac{\left|\varphi_n(x)\right|}{\lVert x\rVert}=\frac 1n\sum_{k=1}^n k\cdot \frac 1{\sqrt{\sum_{k=1}^n k}}=\frac 1n\sqrt{\...

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