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Write $\|x\|_\infty = \max |x_i|$. Clearly $\|x\|_\infty \le \|x\|_p$ for any $p$. Then for $p < q$ you have $$\|x\|_q^q = \sum_{k=1}^n |x_i|^q \le \|x\|_\infty^{q-p} \sum_{k=1}^n |x_i|^p \le \|x\|_\infty^{q-p} \|x||_p^p \le \|x\|_p^q.$$ Thus $\|x\|_q \le \|x\|_p$.

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Yes, the dual-norm is that one! Furthermore, you should consider following: First: The embedding $$H^1(\Omega)\hookrightarrow L^2(\Omega).$$ This is because $$\| \cdot \|_{_{H^1(\Omega)}}^{^2} := \| D^1\cdot \|_{_{L^2(\Omega)}}^{^2} +\| \cdot \|_{_{L^2(\Omega)}}^{^2}$$ is a "stronger" norm than $\|\cdot\|_{_{L²(\Omega)}}$. Second: it follows from the ...

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As stated, this is false, since even $A$ itself could be singular under the given assumptions. To conclude that $A-B$ is nonsingular when $B$ is "small", one needs: $A$ to be nonsingular $\|B\| < \|A^{-1}\|^{-1}$ The reason this works is that $\|A^{-1}\|^{-1} = \min_{\|x\|=1}\|Ax\|$. Hence, assumption 2 implies $\|Bx\|<\|Ax\|$ for all $x$, making ...

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The first inequality is clear by the definition of the matrix norm. The second is clear by the definition and $\frac{|ABx|}{|x|}=\frac{|ABx|}{|Bx|}\frac{|Bx|}{|x|}$.

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The first inequality is immediate, since $$\|A\| = \sup_{x \in \mathbb{R}^n \setminus \{0\}} \frac{|Ax|}{|x|}\ge \frac{|Ax|}{|x|}.$$ For the second note that for $x\ne0$ with $Bx\ne0$, we have $$\|AB\| = \sup_{x\ne0} \frac{|ABx|}{|x|}=\sup_{Bx\ne0}\left( ... 2 I assume you want this \forall K>0, not all t. Since f(t)^Tf(t)\geq0, you can take \beta=(\int_0^{\infty}f(t)^Tf(t)\,dt)^{1/2}. 1 Just note that \|f\|_K^2=\int_0^K f(t)^Tf(t) \, dt\leq \int_0^\infty f(t)^Tf(t) \, dt for all K, since f(t)^Tf(t) is always nonnegative. 1 Note that$$ \left(\frac{x}{e^x}\right)'=\frac{e^x(1-x)}{e^{2x}} $$vanishes at x=1 (which is a maximum of the function). Hence,$$ p(f)=\frac1e\sup_{\|v\|=1}f(v)=\frac1e\|f\|^*. $$0 It is not. In general, for a_i \in \mathbb{C}, we have$$ |a_1+a_1+\cdots+a_n|^2\neq|a_1|^2+|a_2|^2+\cdots+|a_n|^2. $$6 Suppose A and B are positive operators on a finite-dimensional inner product space V and \|A + B\| = \|A\| + \|B\|. Because A and B are positive, A+B is also a positive operator. Thus there exists x \in V such that$$\|x\| = 1 \quad \text{and} \quad\|A+B\| = \langle (A+B)x, x \rangle. Now \begin{align*} \|A+B\| &= \langle (A+B)x, x ... 1 Let v be an eigenvector of A corresponding to the eigenvalue \lambda, i.e. Av=\lambda v. Then (\lambda I-B)^{-1}(A-B)v=(\lambda I-B)^{-1}(\lambda I -B)v=v . Hence by definition of the operator norm, \left\|(\lambda I-B)^{-1}(A-B)\right\|\geq 1. Question: Where did we use that \lambda is not an eigenvalue for B? 1 If x is a unit eigenvector of A associated with \lambda, we have (\lambda I-B)^{-1}(A-B)x=(\lambda I-B)^{-1}(\lambda I-B)x=x $$and hence$$ 1=\|x\|=\|(\lambda I-B)^{-1}(A-B)x\|\leq\|(\lambda I-B)^{-1}(A-B)\|. $$0 Let A=\left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right). \left\| A^TA \right\|_F=\sqrt{trace(A^TAA^TA)}=\sqrt{17} However, trace(A^TA)=5. Hence the equality doesn't hold. 4 Notice that for each vector x, one has$$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$Thus$$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$1 Diagonalization will help here. (When in doubt and working with normal matrices, try utilizing diagonalization!) Write T = UDU^*, then T^* = UD^* U^*, giving that T^*T^2 = UD^*D^2U^*. However T^3 = UD^3 U^*. Since unitary conjugation does not change the operator norm, this boils down to considering D^*D^2 and D^3. Here Dg(x) = f(x)g(x) for ... 2 In the paragraph above Lemma 2.6, the authors explicitly state "[we write] \mathcal L^\infty for the space of measurable bounded functions" 0 Theorem 2.1 in the paper A Singular Value Thresholding Algorithm for Matrix Completion 1 Assuming continuity and differentiability of f: This comes from the Mean Value Theorem: If f is continuous on [a,b] \in \mathbb{R}:$$\exists c \in [a,b]: f'(c)=\frac{f(b)-f(a)}{b-a}$$Now: ||f'||_{L^{\infty}} is the infinity norm, which is the supremum value of |f'| on some interval. So, your formula is saying that the absolute value of the ... 2 The homogenous solutions constitute a subspace. Choose x_p to be in its orthogonal complement (this is always possible). Then the cross terms on the right side of your equation vanish. 4 Note, that \|\cdot\| is somehow the composition of the 2-norm and the 3-norm on \mathbb R^2. We have$$\|x\| = \|(\|(x_1,x_2)\|_2,x_3)\|_3$$and with this at hand the proof is easy (but a bit ugly). We have by the triangle inequality for the 2-norm$$a:= \|(x_1+y_1,x_2+y_2)\|_2 \leq \|(x_1,x_2)\|_2 + \|(y_1,y_2)\|_2 =:b$$and since for all ... 2 Yes, the inequalities hold. Let B be the closed unit ball for the norm \|\cdot \|. The assumptions imply that \pm e_j, the standard basis vectors, are in B. Hence, their convex hull is contained in B. This convex hull is the unit ball for the \ell^1-norm, which implies \|\cdot \|\le \|\cdot \|_1. Similarly, we need to prove that B is ... 2 No. You have to look at a set of arbritray matrices in a neighbourhood of a given P and show that any matrix in that set is orthogonal. You won't succeed with this, though, since that set is actually not open but compact. (It's bounded and closed as the counterimage of a point under a continuous map) (Edit: Actually it is a closed smooth submanifold of ... 2 You correctly observed that \|x\|_2\le \sqrt{\|x\|_1\|x\|_\infty} is a special (easy) case of Hölder's inequality, which essentially amounts to |x_i|^2\le \|x\|_\infty |x_i|. To prove the second part, first use the arithmetic-geometric means inequality. After that, use Hölder's inequality again:$$\|x\|_1\le \sqrt{n}\|x\|_2$$and observe that ... 3 When you say K is continuous I assume it is with respect the norm on C([0,1]), that is, |f\|_\infty. The two norms are not comparable. Let K be the identity. Then$$\sup_{\|f\|_\infty\leq 1} \|Kf\|_\infty=1$$but$$\sup_{\|f\|_2\leq 1} \|Kf\|_\infty=\infty.$$To see it consider a sequence \{f_n\} of functions such that f_n is supported on ... 0 This is the Holder inequality with p=1 and q=\infty. In addition, we can prove this inequality using the following way:$$\sum_{i=1}^n \vert x_i\vert^2=\sum_{i=1}^n \vert x_i\vert \vert x_i\vert\leq \max_{i}\vert x_i\vert\sum_{i=1}^n \vert x_i\vert=\Arrowvert x \Arrowvert_1 \Arrowvert x\Arrowvert_\infty.$$0 Yes, you are right, the statement is not true. You can sketch the simple following example on \mathbb{R}^2. Let x_0=(1,1)^T and$$A= \begin{pmatrix} 1 & 0 \\ \frac{2}{5}&\frac{1}{5} \end{pmatrix}. Then it is easy to see for p=1, the minimum is attained at (2,1)^T, however for p=2, the minimum is attained at ... 0 \begin{aligned} \parallel u \parallel_{*} & = \enspace \max\limits_{\parallel x \parallel \leq 1} <u, x>\\ \leq & \enspace \max\limits_{\parallel x \parallel \leq 1} \frac{<u, x>}{\parallel x \parallel}\\ \leq & \enspace \max\limits_{\parallel x \parallel \neq 0} \frac{<u, x>}{\parallel x \parallel}\\ = & ... 2 I think I found the solution. Continuing with my reasoning, applying the Cauchy formula, we have:1 = a_n = {1 \over 2\pi i}\oint_{|z| = 1} {P(t) \over t^{n+1}} dt$$Therefore, we have:$$1 = |a_n| = |{1 \over 2\pi i}\oint_{|z| = 1} {P(t) \over t^{n+1}} dt| \le {1 \over 2\pi}\max_{|z| = 1}|P(z)| l(|z| = 1) = \max_{|z| = 1}|P(z)|$$Here l(\gamma) is the ... 0 Your method is probably having difficulty since a function like z^n is in fact bounded by 1 on the interior of the disk. Thus, choosing any "special" values on the interior - or any set of special values - cannot suffice as proof. It's perhaps possible that this exercise could be completed purely algebraically, as you are trying to do, but I wouldn't say ... 0 Couple of Tips: ||Ax||^2=x^T(A^TA)x. Try to see yourself how this comes out. Now see the wiki for the derivative you are looking for. If you are not particular on the lagrangian route for this problem, be informed that$$\min_{||x||_2=1}x^TBx\,=\,\lambda_{min}(B)$$for all symmetric matrices B and \lambda_{min}(.) is the least eigenvalue. This ... 1 Normally we talk about normal matrices in a space with inner product, and we consider the norm induced by the inner product (unless explicitly said that another norm should be considered). So I will assume we are supposed to use the norm induced by the inner product. Your proof for the case where A is normal is correct. The result is not valid is A is ... 1 You know that a norm is completely described by its unit ball. Seeing the way unit balls of ||.||_p converge, help to have a precise understanding of the why and how of the limit. See (classical) pictures below of the unit balls of ||.||_1, ||.||_2, ||.||_3 and ||.||_9 in \mathbb{R}^2. It is clear that these balls are getting more and more square ... 3 Hint: For the upper bound, observe that$$ \left(\sum_{i=1}^n |v_i|^p\right)^{1/p}\leq\left(\sum_{i=1}^n \max|v_i|^p\right)^{1/p}=n^{1/p}\max|v_i|. $$For the lower bound, observe that$$ \left(\sum_{i=1}^n |v_i|^p\right)^{1/p}\geq\left( \max|v_i|^p\right)^{1/p}=\max|v_i|. $$Now, take limits. 2 Let's define for a\in A the left multiplication operator and the right multiplication operator$$ L_a : A \rightarrow A, L_a(x)=ax \quad \text{and} \quad R_a: A \rightarrow A, R_a(x)=xa.$$As multiplication is assumed to be continuous we get that L_a and R_a are both continuous for all a\in A. Let B=B_1(0,A) denote the unit ball in A. Then ... 3 y=(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...) \notin c_{00}, y_1=(1,0,0,...), y_2=(1,\frac{1}{2},0,0,...) and so on. y_n \in c_{00} and \|y_n -y \|_{\infty}=\frac{1}{n+1} which tends to 0 when n tends to \infty. 0 Your estimate holds only in a neighborhood of x. Hence, you need can not control whether you may choose such a C independent of x. You need to rely on the linearity of your operator in the following way: By continuity of f in 0, there exists \delta>0 such that for all x\in X with \Vert x \Vert_X <\delta holds$$\vert f(x) - f(0) \vert ...

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Since $f$ is continuous (at $0$), there is a neighbourhood $U$ of $0$ such that $f(U)\subset(-1,1)$. Choose $\delta>0$ such that $\{x\in X|\|x\|\leq\delta\}\subseteq U$. Then, if $x\in X$ is such that $\|x\|\leq \delta$, we have $x\in U$, and hence, $|f(x)|\leq 1$. Since $\|\frac{\delta x}{\|x\|}\|=\delta$, it follows that for all $x\in X$ we have $$... 3 The \ell^p spaces are a special case of the L^p spaces obtained by using the counting measure on the set of natural numbers. If you squint closely at the integral it looks like a sum or indeed as Forever Mozart points out: summation is just integration with the trivial measure on \mathbb{N}. 0 A fancy approach to Lagrange's identity: differentiation.$$\frac{\partial}{\partial a}(a^2+b^2)(c^2+d^2) = 2a(c^2+d^2),\frac{\partial}{\partial a}\left((ac-bd)^2+(ad+bc)^2\right)=2c(ac-bd)+2d(ad+bc) $$they match, and the same happens for \frac{\partial}{\partial b},\frac{\partial}{\partial c},\frac{\partial}{\partial d}, so (a^2+b^2)(c^2+d^2) and ... 1 You can check this directly:$$(ac-bd)²+(ad+bc)²=(ac)²+(bd)²-2abcd+(ad)²+(bc)²+2abcd=a²(c²+d²)+b²(c²+d²)=(a²+b²)(c²+d²).25988=2²\times 73\times 89=2²(8²+3²)(8²+5²)=2²[(8²-15)²+(40+24)²]=98²+128² 0 The first part of the problem reduces the complexity of the second, by writing AB as a sum of two squares if A and B are, but does not lead to a mechanical solution. To start the process one needs to find factors of 25988 that can be written as sums of two squares. There is a theorem of Fermat that only numbers with an even power of every prime of ... 2 The prime factorization of 25988 is 2^2 \cdot 73 \cdot 89. Observe that 73 = 3^2 + 8^2 and 89 = 5^2 + 8^2. Using the norm formula, we have \begin{equation*} \begin{aligned} 25988 &= 2^2(3^2 + 8^2)(5^2 + 8^2) \\ &= 2^2((3 \cdot 5 - 8 \cdot 8)^2 + (3 \cdot 8 + 8 \cdot 5)^2) \\ &= 2^2(49^2 + 64^2) \\ &= 98^2 + 128^2, \end{aligned} ... 2 You might note that ac-bd and ad+bc are the real and imaginary parts of (a+ib)(c+id), so the formula says |zw|^2 = |z|^2 |w|^2 where z= a+ib and w=c+id. For the second part, it might help that 25988 = 2^2 \times 73 \times 89. Can you write 2, 73, 89 each as the sum of two squares? 0 Hint valid for any inner product space: the norm is a composition of functions:x\longmapsto(x,x)\longmapsto\langle x,x\rangle\longmapsto\sqrt{\langle x,x\rangle} = \|x\|$$and \langle\cdot,\cdot\rangle is bilinear. 1 The differential is given by,$$Df(x_1,x_2) = \left( \frac{x_1}{\|x\|}, \frac{x_2}{\|x\|} \right)$$which has defined for x \in \mathbb{R}^2 \setminus \{0\} 6 Let \{a_n\} be a sequence of positive real numbers that increases to 1, with the property that the sequence of products$$ a_1,\ a_1a_2,\ a_1a_2a_3,\ a_1a_2a_3a_4,\ \ldots$$converges to a positive value. It's not hard to write down a specific example. Let \cal H = \ell^2(\mathbf R) and define T : \cal H \to \cal H by$$T(x_1,x_2,x_3,\ldots) = (0, ...

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Explicit continuous functions that work on the unit interval (i.e. demonstrate the failure of the parallelogram law holding) are $f(x) = 1-x$ and $g(x) = x$. $2||f||^2_\infty = 2$ and $2||g||^2_\infty = 2$, while $||f+g||^2_\infty = 1$ and $||f-g||^2_\infty = 1$.

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$\kappa_2(A) = |\sigma_{\max}|/|\sigma_{\min}|$, your formula only holds if $A$ is normal. In this case we have that $\sigma_{\max} = 1.9896$ and $\sigma_{\min} = 0.4974$ and $$\dfrac{1.9896}{0.4974} \approx 4$$

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There is a simple reason why you can't. Take two opposite vectors with arbitrary little norm. The difference between the two can be as small as you wish but the difference between their normalized counterparts will always be of norm 2. For instance in $\mathbb R^2$ : $x1=(\frac 1 n,0),x_2=(-\frac 1 n,0),n\in \mathbb N$, the difference has norm $\vert\vert ... 1 Note that$\langle Tw, Tw\rangle=\langle w, w\rangle$. Taking$w=u-v$gives$\langle T(u-v),T(u-v)\rangle=\langle(u-v),(u-v)\rangle\implies \left| u-v\right|^2=\left|Tu-Tv\right|^2\$

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