# Tag Info

0

Since the maps $$y \mapsto x+y$$ and $$y \mapsto Ry$$ are both homeomorphisms, we may assume that $x=0$ and $R=1$. Now check that $$B := \{y\in V : \|x\| \leq 1\}$$ is a closed set [Since the norm is a continuous function from $V$ to $\mathbb{R}$] Now if $y\in B\setminus B(0,1)$, then prove that $y \in Cl(B(0,1))$ [Consider the sequence $y_n := ... -1 Recall that closure of a set is the intersection of all closed sets containing that set or in other words the smallest closed set containing that set. Now what is the smallest closed set containing$B(x,R)$? 0 Yes. I prefer to denote by$A^H$the conjugate transpose of a complex matrix$A$. Note that $$\left< Tv, Tv \right> = (Tv)^{H} Tv = v^{H} T^{H} T v.$$ The matrix$T^H T$is Hermitian and positive-semidefinite. From the equation above, we see that$T$is contraction if and only if$T^H T$has no eigenvalues whose norm is greater than one. Similarly, ... 1 Since$A$is symmetric, there exists an orthonormal basis$v_1, \ldots, v_n$of eigenvectors of$A$. Write$Av_i = \lambda_i v_i$. Since$A$is positive definite,$\lambda_i > 0$. Show that $$\{ x \in \mathbb{R}^n \, | \, ||x||_A = 1 \} = \{ a_1 v_1 + \ldots + a_n v_n \, | \, \lambda_1 a_1^2 + \ldots + \lambda_n a_n^2 = 1 \}$$ which means that$\{ x ...

2

(1) if $f$ is measurable, and $\int_a^b |f(t)|\;dt = 0$, then $f(t) = 0$ for almost all $t$. Proof. Suppose $|\{t : f(t) \ne 0\}| > 0$. (I used $|\cdot|$ for Lebesgue measure.) Then $|\{t : |f(t)|>0\}| > 0.$ But $$\{t : |f(t)|>0\} = \bigcup_{n \in \mathbb N} \left\{t : |f(t)| > \frac{1}{n}\right\}$$ so by countable additivity, there ...

1

Let $a\leq t\leq b$. As $x^2|f(x)|\geq0$, $$0\leq\int_a^t x^2|f(x)|dx\leq\int_a^b x^2|f(x)|dx$$ So, for any $a\leq t\leq b$, $$\int_a^t x^2|f(x)|dx=0$$ Now differentiate it and you get $$t^2|f(t)|=0\quad (a\leq t\leq b)$$ $$|f(t)|=0\quad (a\leq t\leq b, t\ne0)$$ $$\therefore f(t)=0\quad (a\leq t\leq b, t\ne0)$$

1

Hints: For nonexistence of $c$ you could try to construct non-negative continuous functions, which have a small integral, i.e., $\|f\|_1$ is small, but with large $\|f\|_{max}$. The second inequality follows with $c = b - a$.

1

The max norm is not the (absolute value of) the function at one specific point, but at a point that depends on the function itself. In a more general setting where functions are not necessarily continuous, it is called the supremum norm. The integral of $|f|$ cannot be higher than the length of the integration interval multiplied by the maximum of $|f|$ ...

0

$$f(X)=a^T Xb=\mathrm{Tr}(Xba^T)=Tr(ab^TX^T)=Tr(X^Tab^T)=\left<X,ab^T\right>$$ $$\nabla_Xf(X)=ab^T$$

0

For sake of simplicity, let's assume that we're working on the space $L^1([0,1])$, i.e. the space of all (complex) functions defined on $[0,1]$ with bounded $L^1$-norm. We are given that $f_n\rightarrow f$ in $L^1$-norm. Hence for all $\varepsilon>0$, there exists a $N\in \mathbb{N}$ such that for all $n\geq N$, $||f_n-f||<\varepsilon$, or equivalently ...

0

You can use what mathcounterexamples.net already said. However if this is not clear to you you should at lest once write it down in all details Let $a = [a_1,...,a_n]^T$, $b=[b_1,...,b_n]^T,[X]_{i,j} = x_{i,j}$ Then $$\begin{eqnarray*}a^T X b &=& ... 1 f is linear. So its derivative is... 0 What you want to show is false. By the Riemann Lebesgue Lemma,$$ \int_0^{2\pi} \sin(nx) g( x)\,dx \to 0 $$as n\to \infty for every g \in C([0,2\pi]). Thus, if \sin(nx) \to f in L^1, then \int_0^{2\pi} g( x)f(x)\,dx=0 for all g\in C([0,2\pi]), which implies f=0 almost everywhere. But it is not hard to show (you apparently did that already) ... 0 No, the sequence doesn't converge in L^1. You can show that \sin(nx) converge weakly to f=0 on L^1(0,1). If \sin(nx) converge strongly in L^1(0,1), it must converge to f=0 since strong limit is weak limit. Then show \|\sin(nx)\|_{L^1} does not converge to 0. 1 (a) The gradient of a vector valued function is a matrix. For$$f(x_1,x_2,x_3)=\begin{bmatrix}f_1(x_1,x_2,x_3)\\f_2(x_1,x_2,x_3)\end{bmatrix}$$it is$$Df=\begin{bmatrix}\frac{\partial f_1}{\partial x_1}&\frac{\partial f_1}{\partial x_2}&\frac{\partial f_1}{\partial x_3}\\\frac{\partial f_2}{\partial x_1}&\frac{\partial f_2}{\partial ...

0

The function $\phi(x) = \|x\|_2^2= x^T x$ is straightforward to differentiate since $\phi(x+h)-\phi(x) = 2 x^T h + h^T h$, hence $D \phi(x) = 2 x^T$. The function $T(x) = x-x_0$ is affine, hence $DT(x) = I$. The function $s(t) = \sqrt{t}$ has derivative $Ds(t) = {1 \over 2 \sqrt{t}}$. (a) Since $f = s \circ \phi \circ T$, we know $D \log(t) = {1 \over t}$ ...

2

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 $$\frac{d}{dt} [A(t)]^{-1} = - [A(t)]^{-1}\Bigl(\frac{d}{dt} ... 0 Showing a) implies b) is straightforward. Given \epsilon we get a n_0 and then select an n with p(n)>n_0. For these n we have ||x_{p(n)}-x_{p(n+1)}||<\epsilon. For the reverse assume the sequence is not a Cachy sequence and lets construct an increasing sequence p(n) such that ||x_{p(n)}-x_{p(n+1)}||>\epsilon. There must exist an ... 0 Remark that if p is an (strictly) increasing function defined on N, p(n)>n. Suppose that x_n is a Cauchy sequence Let c>0 \exists N such that n,m>N implies that \|x_n-x_m\|<c, in particular \|x_{p(n)}-x_{p(n+1)}\|<c if n>N since p(n),p(n+1)>n>N. done. In the other hands, suppose that for every increasing ... 0 Recall that a matrix A is positive semidefinite if and only if v^\top A v \ge 0 for all vectors v. For (a), see if BAB^\top satisfies this. (Why is v^\top B A B^\top v non-negative?) For (b), consider v^\top A v where v is a standard unit vector. 0 Let$$A = U \Sigma V^*$$be the SVD of A. Then, because  U is orthogonal,$$ ||Ax||= ||\Sigma V^* x|| $$If V^*x = y ,  ||y||=||x||=1 we are left with$$||Ax||^2=||\Sigma y||^2 = \sum _i |\sigma_iy_i|^2 \geq \sum _i |\sigma_qy_i|^2 = \sigma_q^2 $$So, the inequality holds. If x is a right singular vector of A then it is one of the columns of ... 0 Hint: let x \in \mathbb{R}^q and write it as a linear combination of right singular vectors of A. Notice that Ax is a linear combination of the corresponding left singular vectors of A. But these are orthogonal, so you have the Pythagorean theorem, which makes it straightforward to solve the minimization problem. In particular you should find that if ... 1 This problem uses a basic fact about quadratic forms, symmetric matrices and their eigenvalues and eigenvectors, but you don't need to know anything about the latter to solve it. Let v=(\cos\theta, \sin\theta)^T. Then the problem becomes that of finding the maxima and minima of$$(\cos\theta, \sin\theta)\pmatrix{6&-2\\-2&6}\pmatrix{\cos\theta\\ ...

1

In the symmetric case, the answers are given by the unit eigenvectors with the smallest and largest eigenvalue (respectively). This occurs in your case.

0

If $Q$ is an orthogonal Matrix then $Q^T*Q=E$ ($E$ is the unity matrix). This means if $x_i$,$x_j$ are column vectors. Then $x_i^T*x_j^=0$ for $i\neq j$ and $x_i^T*x_j^=1$ for $i=j$. Therefore $Q^T*x =(x_1^T*x_1,x_2^T*x_1,...x_n^T*x_n)^T=(1,0,0,...)^T$

0

Hint: Since $Q$ is an $n\times n$ orthogonal matrix, then $Q^TQ$ is the $n\times n$ identity matrix. Furthermore, if $A$ is any $k\times m$ matrix and $B$ is any $m\times n$ matrix, then letting $b_1,\dots,b_n$ be the columns of $B$ (from left to right), then $Ab_1,\dots,Ab_n$ are the columns of $AB$ (from left to right).

1

It is false. One has $A^{-1}P^{-1}=U^{-1}L^{-1}$; since $P^{-1}$ is a permutation $||A^{-1}P^{-1}||_{\infty}=||A^{-1}||_{\infty}=||U^{-1}L^{-1}||_{\infty}$ and the inequality to show is $||U^{-1}L^{-1}||_{\infty}\geq 1/n\;||U^{-1}||_{\infty}$. A counter-example is: $n=2$, ...

1

(i) (1) follows because the sum of two norms is a norm. (2) is not a norm; $\|f'\|_\infty = 0$ does not imply $f \equiv 0.$ (3) is a norm, even though it the sum of a norm and a non-norm. (4) I do not understand your question "What is $|f(0)|?$" Anyway, this is a norm: Show $|f(0)| +\|f'\|_1 = 0 \implies f' \equiv 0 \implies$ $f$ is constant; the condition ...

0

No, we can't. To construct a simple counterexample, consider $A$ and $B$ scalars and note that a singular value of a scalar is its absolute value. We can say however that $$|\underline{\sigma}(A+B)-\underline{\sigma}(A)|\leq\overline{\sigma}(B),$$ which is a standard perturbation result for singular values. This gives ... 0 Here is a couple of hints. If A is surjective, then the image would contain all continuous functions on [0,1]. In particular, functions that can take arbitrary values at t=0. A is injective if and only if, the only continuous function f such that tf(t)=0 for all t\in[0,1] is f=0 the zero function. 0 Hint: You don't need to find the best possible K; any positive K that works will do. Try using the Triangle Inequality. 0 I was a bit confused by your estimate but I understand it now. |A(x,y)| = |2x + y| \leq |2x| + |y| \leq 2|x| + |y| + |y| \leq 2(|x|+|y|) \leq 2 \|(x,y)\|. So yeah, A is bounded and so A is continuous. 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 ... 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\}. ... 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\|$$You may also want to add that for x=0 the inequality holds because Tx=0. 1 This is not an answer, just an elaboration of a comment. You have (A-zI)^{-1} = (z({A \over z} -I))^{-1} = {1 \over z} ({A \over z} -I)^{-1} = -{1 \over z} (I-{A \over z} )^{-1}. If |z| > \|A\|, we have (A-zI)^{-1} = -{ 1\over z} (\sum_{k=0}^\infty ({A \over z})^k) = -{ 1\over z} (I+{A \over z}\sum_{k=0}^\infty ({A \over z})^k). Note that \|{A ... 1 This is not an answer to your question, but merely an attempt to encourage you to formulate the question in a way that will encourage more attention and answers. You write: By definition of a dual space Lip_0(X)^*, every element is an evaluation function \mu : Lip_0(X) \rightarrow \mathbb{F}. This is not true. The norm-closure in Lip_0(X)^{*} ... 0 We will rescale our interval in order to apply the Remez inequality. We need to find s such that after the map f(x)= (2+s)x-1 the set E will have measure at least 2. We can write 2\le\mu (f(E)) = (2+s)\mu(E), hence we can take s= \frac{2}{\mu(E)}-2. Rescale p by its supremum on E:$$q(x)=\frac{p(x)}{\sup_{E}|p(x)|}.$$With this value of ... 0 You probably mean the inner product \langle \boldsymbol{x}, \boldsymbol x \rangle. If you have two vectors \boldsymbol a = (a_1,a_2,\dots ,a_n) and \boldsymbol{b} = (b_1,b_2,\dots,b_n), then the inner product \langle \boldsymbol{a},\boldsymbol{b}\rangle is defined as$$\langle \boldsymbol{a},\boldsymbol{b}\rangle = \sum^n_{i=1}a_ib_i = a_1b_1 + ...

0

You can see a vector in $\mathbb{R}^n$ as an oriented segment in $n$-dimensional space that start from the origin and goes to the point $X$ of coordinates $(x_1,x_2,\cdots,x_n)$. So the inner product $\langle x,x\rangle$ is : $$\langle x,x\rangle= x_1^2+x_2^2+\cdots+x_n^2$$ and $\sqrt{\langle x,x\rangle}$ is the usual Pythagorean lenght of the segment ...

0

Say $A=\{(x_1, x_2) \in \mathbb R^2: x_1 + x_2 > r\}$, then $A$ is the whole area above the line $l$ defined by $y=x+r$ (the identity line moved vertically $r$ units.) Take a point $x=(x_1,x_2)\in A$, then $(x_1,x_2)$ must be above line $l$. Hence $\epsilon = d(x, l) > 0$ (the minimum distance from the point to the line) Thus, the ball ...

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, ...

0

In your case $\langle x,x\rangle$ is the scalar product of the vector $x$ with itself.

0

In general inner product space it is not true that there exists such a $x_0$ such that $Tx_0=y$ and $||x_0||\le||x||$ for all $x$ that satisfies $Tx=y$. Consider the inner product space $V=l^2$ where a linear operator $T$ is defined by $$Te_k=k\cdot e_1$$ where $e_k=(\delta_{nk})_{n=1}^{\infty}$. It is not hard to see that $x=\frac1k e_k$ is a solution to ...

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 ... 0 The Moore-Penrose generalised inverse of T applied to y gives the least norm vector x that minimises ||Tx-y||. Note that every linear map has such an inverse, even if the dimensions of the domain and codomain are different. The usual way to compute such and inverse is via the SVD. 0 For a solution to exist we must require that y is in the range of T. Next we note that adding an element of the null space \mathcal{N} of T to x does not change things. Thus the allowable x are \begin{equation*} x=u+v,\;u\perp \mathcal{N},\;v\in \mathcal{N} \end{equation*} Note that u is unique since if u_{1} and u_{2} are both \perp ... 0 \begin{align} \|f+g\|^2 &= \int |f+g|^2 \\ &\le \int |f||f+g| + \int |g||f+g| \\ &\le \|f\|\|f+g\| + \|g\| \|f+g\| & \text{Cauchy-Schwarz} \\ &= (\|f\|+\|g\|)\|f+g\|. \end{align} 0 WLOG, x_i\geq 0 for every i. From \sum_{i>1}x_i\leq \sqrt{(n-1)\sum_{i>1}x_i^2}=\sqrt{(n-1)\left(1-x_1^2\right)}, we have x_1\sum_{i=1}^nx_i\leq x_1^2+x_1\sqrt{(n-1)\left(1-x_1^2\right)}. By AM-GM, \sqrt{ax_1^2\cdot b\left(1-x_1^2\right)}\leq \frac{b}{2}-x_1^2, where 0\leq a<b satisfy ab=n-1 and b-a=2. Note that a=\sqrt{n}-1 and ... 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 $$The integral term is only needed to make this a norm rather than a seminorm. The fact that these are mutually nonequivalent follows by considering f_{a,n}(x)=\max(0,1-n|x-a|) which satisfies$$ ...

Top 50 recent answers are included