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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 ...

4

Norms are a measure of distance. One has different ways to define what is the distance between points in multiple dimensions, which collapse to the usual notion of the absolute value in 1D. in particular, Euclidean distance is defined by the 2-norm, $$\left\| \begin{pmatrix} x_1 \\ \vdots \\ x_k \end{pmatrix} \right\|_2 = \sqrt{\sum_{i=1}^k x_i^2}$$ ...

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$norm of the difference between 2 ... 3 The norm$\|\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$that satisfies certain properties so it is quite explicitly stated that the norm of every vector is a real number, thus finite. 2 Normed spaces are sets along with norms. If you want to remove the norm, and just treat it as a space, you're free to do so. We say "normed" if we have a norm in mind. Some spaces are "normable" but we haven't chosen a specific norm. And different norms induce different topologies, in particular different ways of things converging. As you're possibly aware, ... 2 OK let's see if this helps you. Suppose you have two functions$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 $$... 2 This is one of those cases where maybe it's better --and more instructive--to prove the general case first. Then, adapting the proof to your specific function is easy. So, suppose v,w\in \mathbb F^{n}. Then, w=v+(w-v) so \Vert w\Vert \leq \Vert v\Vert +\Vert w-v\Vert \Rightarrow\Vert w\Vert-\Vert v\Vert\leq \Vert w-v\Vert. Interchanging the roles of ... 2 Let V be any real or complex vector space, S the corresponding projective space (i.e. the equivalence classes of V \backslash \{0\}, where two elements are equivalent if they are scalar multiples of each other), and \widehat{p} any positive real-valued function on S such that \sup_{s \in S} \widehat{p}(s) \le 2 \inf_{s \in S} \widehat{p}(s). ... 2 Let \sigma_i, i=1,...,n the \Bbb{Q}-isomorphisms of K where n=[K;\Bbb{Q}] , let x\in K , a primitive element of K so by definition N(x)=\prod_{i=1}^{i=n}\sigma_i(x) then N(-x)=\prod_{i=1}^{i=n}\sigma_i(-x)=(-1)^n\prod_{i=1}^{i=n}\sigma_i(x)=(-1)^nN(x) so the equality N(-x)=-N(x) , then give n odd. Edit: In Number Field Sieve, we take ... 2 Let us consider the unit sphere under || \cdot ||_1: all polynomials p(x) such that \int_0^1|p(x)|dx \le 1. We have$$\int_0^1 |p(x)| dx \le \int_0^1 \sum |a_i||x^i| dx = \int_0^1 \sum |a_i|x^i dx = |a_0|+\frac{1}{2}|a_1| + \dots + \frac{1}{n+1}|a_n| $$by the triangle inequality and similar standard inequalities. Thus, for any polynomial on the L1 ... 1 Seems to me the only example is \Bbb C. Say \hat a is the Gelfand transform. Since ||\hat a||_\infty\le||a||, similarly for a^{-1}, and ||\hat a||_\infty||1/\hat a||_\infty\ge1 it follows that |\hat a| is constant whenever a is invertible. Hence, whether a is invertible or not, |\hat a-\lambda| is constant for every \lambda not in the ... 1 If A is symmetric and positive definite, then its eigenvalues are real and positive. The spectral norm of A is given by$$\|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) $$Note, however, that order matters in that last term. 1 A norm is a thing we sometimes define on a vector space. The quotient space you describe is not a vector space; it has no concept of addition. So one can't even begin talking about a norm on it. The quotient space is naturally identified with unit sphere in V, since each equivalence class contains exactly one element with unit norm. This leads to a ... 1 Divide by n_k in the inequality$$\tag{*}\|f(x_{n_k})\| > \varepsilon + n_k \|g(f(x_{n_k}))\|, \quad k \in \mathbb{N}$$and letting k going to infinity, we get, by boundedness of \left(f\left(x_{n_k}\right)\right)_{k\geqslant 1}, that \lim_{k\to +\infty}g\left(f\left(x_{n_k}\right)\right)=0. By continuity of g we also have that \lim_{k\to +\... 1 Suppose 0<\mu (A)<\infty and 0<\mu (B)<\infty, and A\cap B=\phi. (1).Let f(x)=1/\mu (A) when x\in A and f(x)=0 when x\not \in A. Let g(x)=1/\mu (B) when x \in B and g(x)=0 when x\not \in B. Then f, g are linearly independent, and \|(f+g)/2\|_1=\|f\|_1=\|g\|_1=1. (2). Let f(x)=1 when x\in A and f(x)=0 when x \... 1 You know \|\delta\|\le 6. Can you find some f such that \|f\|_{\sup}=1 and \delta(f)=6? This should not be hard to find. 1 The notation S^\perp means \{x\in H\mid \langle x,y\rangle=0,\text{ for all }y\in S\}. Let x\in\ker T; you need to prove that, for every m with \lambda_m\ne0, you have \langle x,\varphi_m\rangle=0. You know that \sum_n\lambda_n\langle x,\varphi_n\rangle\varphi_n=0, so also$$ \Bigl<\sum_n\lambda_n\langle x,\varphi_n\rangle\varphi_n,\... 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{...

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