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Here is a try. The matrix entries are Bernoulli random variables, that get a value of $\pm 1$. A Bernoulli random variable is zero mean subgaussian variable with finite moments. If we define $\lambda = m/n$, then for large values of $m,n$: $\Vert A \Vert_2 \rightarrow \sqrt{n}\left(1+\sqrt\lambda \right)$ and therefore, $\Vert B \Vert_2 \rightarrow ... 4 Note that$ a $and$ b $must be nonnegative scalars, not vectors, for your calculation to work. 3 First, in order that$P(f)$be defined for$x\in[0,1]$, I prefer to define$P$as follows: $$P(f)(x)=\cases{1&if \quad x=0\cr \displaystyle1+kxf(x)\int_0^1\dfrac{f(s)}{x+s}ds&if \quad x\in(0,1]}$$ In this case$P(f)\in C([0,1])$for every$f\in C([0,1])$. Indeed, we only need to prove the continuity of$x\mapsto P(f)(x)$at$x=0$, and this ... 3 No, nothing interesting can be said, and almost anything is possible. 3 Your reasoning is valid. You are uncertain about the use of the same letter$x$on both sides... it's not wrong, since on the right$x$is a dummy variable. But if you want to avoid this repetition, write $$\frac{\|Tx\|}{\|x\|} \leq \sup_{y\ne0} \frac{\|Ty\|}{\|y\|} = \|T\|$$ You may also want to add that for$x=0$the inequality holds because$Tx=0$. 2 If you take the definition of$\|T\|$to be$\|T\| = \sup\left\{\dfrac{\|Tx\|}{\|x\|} : x \in X \right\}$, then it is a triviality that$\|Tx\| \le \|T\|\|x\|$for every$x\in X$. But sometimes the definition is taken to be$\|T\| = \sup\left\{\dfrac{\|Tx\|}{\|x\|} : \|x\|\le 1 \right\}$or$\|T\| = \sup\left\{\dfrac{\|Tx\|}{\|x\|} : \|x\|\ = 1 \right\}$. ... 2 Not in general. Here is a counterexample. Let$X = C([-1,1])$. For$t \in [-1,1]$, let$\delta_t$denote the point mass / evaluation functional$\delta_t(x) = x(t)$. Let$D = \{ \delta_q : q \in [-1,1] \cap \mathbb{Q}\}$. Then$D$is countable and we have$\|x\| = \sup_{f \in D} |f(x)|$for every$x \in X$. Let $$y(t) = \begin{cases} 4t, & -1 \le t ... 2 The notation$$\langle\cdot,\cdot\rangle$$is used to denote the inner product, which in Euclidean space is essentially the dot product. Using the dot product definition in the case of the inner product of a vector \mathbf{v} and itself,$$\mathbf{v}\cdot\mathbf{v}=v_x^2+v_y^2+\cdots$$Notice that this is simply the magnitude of \mathbf{v} squared, ... 2 Your're almost there... Recall the definition of convex conjugates. Recall that the convex conjugate of a norm \|.\| is the indicator function of the unit ball of the dual norm \|.\|_*. Now, \begin{split} g(\nu) &:= \underset{x}{\inf }L(x,\nu) = \underset{x}{\inf }\|x\| + \nu^T Ax - \nu^T b = \nu^Tb -\underset{x}{\sup }x^T(-A^T\nu) ... 2 they are not equivalent : you should consider$$a_{n,p} = \left\{1 \; if \;n < p\;,\; 1/n \;if \;n \ge p \right\}$$For any p (a_{n,p})_n is l_2 but there is a p to have a contradiction 2 I'll assume that \|A\| denotes the supremum norm, that is, for A \in \mathrm{Mat}_{n \times n}(\mathbb R) (or \mathbb C),$$ \|A\| \overset{def}= \sup_{x \neq 0} \frac{\|Ax\|}{\|x\|}, $$where \|x\| is the Euclidean norm on \mathbb R^n or \mathbb C^n. Your inequality implies$$ \|BA^{-1} - I\| \le \|A^{-1}\|\|B-A\| < 1 $$where I is the ... 2 You've shown that for a convex function f:[0,a+b]\to\Bbb R (where a,b\ge 0) we must have f(a+b)+f(0)\ge f(a)+f(b). In particular, if f(x)=|x| then equality must hold. But this simply says that (a+b)+0= a+b, and there is no contradiction. 2 I get the same answer as @Galc27 We have$$ \|L(x,y)\| = \sqrt{(x+3y)^2+(y-x)^2} = \sqrt2\sqrt{x^2+2xy+5y^2}. \tag 1 $$Now, \|(x,y)\|=1 implies x^2+y^2=1, so (1) simplifies to$$\sqrt{1+2xy+4y^2}. $$Substituting x = \sqrt{1-y^2} yields$$\sqrt2\sqrt{1+2\sqrt{1-y^2}y+4y^2}. $$When y=\sqrt{\frac12+\frac1{\sqrt5}}, we have$$\left\|L\left(1 - ... 2 The matrix of$ L $with respect to standard basis is \begin{bmatrix} 1 & -1\\ 3 & 1\\ \end{bmatrix}. Hence$L^*L$is \begin{bmatrix} 10 &2\\ 2 &2 \end{bmatrix} The eigen values of the above matrix are$6+\sqrt 5,6-\sqrt 5$Hence the required norm is$(6+\sqrt5 )^{1/2}=1+\sqrt5$1 In addition to the norms suggested in comments, here is one more (inspired by B.S.Thomson): for$t\in [0,1]$, let $$\|f\|_t = |f(t)| + \int_0^1 |f(x)|\,dx$$ The integral term is only needed to make this a norm rather than a seminorm. The fact that these are mutually nonequivalent follows by considering$f_{a,n}(x)=\max(0,1-n|x-a|)$which satisfies $$... 1 For linear operators the norm reduces to being the eigenvalue with the largest modulus. Given L can be written as a 2x2 matrix acting on (x, y) you shouldn't have any problems using such an approach. 1 For any u with \|u\|_\infty\le 1 we have \forall j\in\mathbb{Z}$$|(Au)_j|=|\sum_k\rho_{j-k}u_k|\le\sum_k|\rho_{j-k}|\underbrace{|u_k|}_{\le 1}\le\sum_k|\rho_{j-k}|=\|\rho\|_1. $$Hence, \|A\|\le \|\rho\|_1. For a fixed j, i.e. j=0, and u\colon u_k=\text{sign}\rho_{j-k} we get$$|(Au)_j|=|\sum_k\rho_{j-k}u_k|=\sum_k|\rho_{j-k}|=\|\rho\|_1. $$... 1 The solutions of the heat equation are smooth (even up to the boundary, because one can reflect across the boundary). So, u_x\in L^2([0,\pi]) is assured. Reflection also helps with the Poincaré inequality. Extend u_0 to [0,2\pi] so that u_0(2\pi -x)=u_0(x). Consider the function v that solves the heat equation on [0,2\pi]\times [0,\infty) with ... 1 If \Vert Ax \Vert=\Vert Bx \Vert for every x, then in particular: \sup \Vert Ax \Vert = \sup \Vert Bx \Vert for \Vert x \Vert=1. Then by the definition of the induced norm, \Vert A \Vert_2 = \Vert B \Vert_2 (the largest singular value of A is equal to the largest singular value of B). 1 Your arguments are fine. The only thing that might feel a little awkward is your notation of the inner product \langle \cdot,\cdot \rangle: Obviously it is not the usual dot product of two column vectors since you have the row vector z^tL and the column vector L^t z in there together. What you mean in the 3^\text{rd} line is the usual matrix ... 1 In the symmetric case, the answers are given by the unit eigenvectors with the smallest and largest eigenvalue (respectively). This occurs in your case. 1 This problem uses a basic fact about quadratic forms, symmetric matrices and their eigenvalues and eigenvectors, but you don't need to know anything about the latter to solve it. Let v=(\cos\theta, \sin\theta)^T. Then the problem becomes that of finding the maxima and minima of$$(\cos\theta, \sin\theta)\pmatrix{6&-2\\-2&6}\pmatrix{\cos\theta\\ ... 1 We claim that the line segment joining$x$and$y$is in the unit sphere: Set$x_t=tx_1+ (1-t)x_0, t\in [0,1]$and$x_1=x,x_0=y$. Now since the norm is convex, we have that: $$\|x_t\| = \|tx_1+(1-t)x_0\| \leq t\|x_1\| + (1-t)\|x_0\|$$ $$\leq t+(1-t)=1$$ So we have that the line segment is in the closed unit ball, we need to show that it isn't in the open ... 1 Hint:$ \|u+v\|^2=\|u\|^2+\langle u,v\rangle+\langle v,u\rangle+\|v\|^2.Edit: We have: \begin{align} \|u+v\|^2&=\|u\|^2+\langle u,v\rangle+\langle v,u\rangle+\|v\|^2 \\ &= \|u\|^2+\langle u,v\rangle+\overline{\langle u,v\rangle}+\|v\|^2 \\ &=\|u\|^2+2\Re\langle u,v\rangle+\|v\|^2 \\ &\le \|u\|^2+2\left|\langle u,v\rangle\right|+\|v\|^2 ... 1 As you have not defined the norm on\mathbb{R}^n$and$\mathbb{R}^m$and all norms are equivalent, I will go with the supremum norm, thus$\| \left( \begin{matrix} b_1 \\ \cdots \\ b_m \end{matrix} \right) \| = \max_{1 \le j \le m} | b_j |$. Then for$x = e_i \in \mathbb{R}^n\$ we have $$\| Ae_i \| =\| \left( \begin{matrix} a_{1,i} \\ \vdots \\ ... 1 Assuming that the norm in \mathbb R^n is the Euclidean norm we have (using Cauchy-Schwarz), for any i,j,$$ |a_{ij}|=|\langle Ae_j,e_i\rangle|\leq \|Ae_j\|\,\|e_i\|=\|Ae_j\|\leq\|A\|_{\rm op}\,\|e_j\|=\|A\|_{\rm op}. $$1 Hint. For x \in V, what is a collinear vector y to x such that \Vert y \Vert <a? What happens then to \Vert y \Vert^\prime? And vice versa? 1 This is not an answer to your question, but merely an attempt to encourage you to formulate the question in a way that will encourage more attention and answers. You write: By definition of a dual space Lip_0(X)^*, every element is an evaluation function \mu : Lip_0(X) \rightarrow \mathbb{F}. This is not true. The norm-closure in Lip_0(X)^{*} ... 1 Differentiate with respect to t both sides of the equation$$ A(t) [A(t)]^{-1} = I, $$getting (via the product rule)$$ A(t) \frac{d}{dt} [A(t)]^{-1} + \Bigl(\frac{d}{dt} A(t)\Bigr) [A(t)]^{-1} = 0. $$Now multiply both sides of the equation above by [A(t)]^{-1} on the left, getting$$ \frac{d}{dt} [A(t)]^{-1} = - [A(t)]^{-1}\Bigl(\frac{d}{dt} ...