# Tag Info

0

First of all, I think you mean $H_0^1$ is the closure of $C^1$, not $H^1$. Secondly, your PDE will solved by using second existence theorem or directly by Lax-Milgram theory by using Bilinear operator. For more information, I suggest you to read our bible Evans book, chapter 6. Look for First existence theorem and second existence theorem

2

Let $\left\{ \mathbf{x}^{n}\right\}$ be a Cauchy sequence in $\mathbb{R}^{k}$. Then for all $\epsilon>0$, there exists $N$ s.t. $$\left\Vert \mathbf{x}^{n}-\mathbf{x}^{m}\right\Vert _{\infty}<\epsilon$$ for all $n,m>N$. This implies that $$\left|x_{i}^{n}-x_{i}^{m}\right|<\epsilon$$ for each $i$. So each $\left\{ x_{i}^{n}\right\}$ is a ...

1

One of the proofs is the following that uses quadratic function which I like: $0\leq \displaystyle \sum_{j=1}^n(a_{1j}-\lambda x_j)^2 = \displaystyle \sum_{j=1}^n a_{1j}^2 - 2\lambda\displaystyle \sum_{j=1}^n a_{1j}x_j + \lambda^2\displaystyle \sum_{j=1}^n x_j^2 = f(\lambda), \forall \lambda \in \mathbb{R} \Rightarrow \triangle' \leq 0 \Rightarrow ... 0 You can't say anything... Take$a_k = (s,0,\ldots,0)$with$s\in \{10^{-132},1,10^{345}\}$for every$k$. Then you will face all three situations ($\leq,\geq$and$=$). 0 For any matrix$A$,$\|A\|_F^2=\sum_{i=1}^r \sigma_i^2$, where$\sigma_i$is the$i$th singular value, and$r$is the rank of$A$. For a rank-$1$matrix$A=ba^T$, the singular value decomposition becomes trivial: $$A = \left( \frac{b}{\| b\|_2}\right)\left( \|b\|_2\|a^T\|_2\right)\left( \frac{a^T}{\| a^T\|_2}\right)$$ So clearly$\| A\|_F^2 = \left( ...

1

No, it cannot be true. Take $A=B$ but $a\ne b$. Then $$\|aA-bB\|_1 = |a-b| \|A\|_1 \not\le \max(a,b) \|A-B\|_1 = 0.$$

1

Obviously, the norm is at least what you anticipate by setting $f = e_n$ where $n$ almost realizes the supremum of $|\lambda_n - \lambda|^{-1}$. Conversely, write $f$ as $f = \sum_n c_n e_n$. Then $$\|(K - \lambda I)^{-1} f\|^2 = \sum_n |c_n|^2 |\lambda_n - \lambda|^{-2} \leq (\sup_n |\lambda_n - \lambda|^{-2}) (\sum_n |c_n|^2).$$ Taking the square root ...

1

Let $$E_n(f)=a,\qquad E_n(g)=b$$ There exist two polynomials $p_1$ and $p_2$ such that $$\|f-p_1\|<a+\epsilon,\qquad \|g-p_2\|<b+\epsilon$$ so $$E_n(f+g)\le\|f+g-p_1-p_2\|\le a+b+2\epsilon$$ for all epsilon, so $$E_n(f+g)\le a+b=E_n(f)+E_n(g)$$

0

Hint: if $p$ is a good approximation to $f$ and $q$ is a good approximation to $g$, try approximating $f+g$ by $p + q$.

3

We have $$(x,y)\in B(0,1)\iff N(x,y)=\sup_{t\in\mathbb{R}}\frac{|x+ty|}{t^2+t+1}\le1\iff |x+ty|\le t^2+t+1, \forall t\in\Bbb R\\\iff -t^2-t-1\le x+ty\le t^2+t+1,\forall t$$ the second inequality gives $$t^2+(1-y)t+1-x\ge0,\forall t\iff \Delta_2=(1-y)^2+4x-4\le0$$ and the first inequality give $$t^2+(1+y)t+x+1\ge0,\forall t\iff \Delta_1=(1+y)^2-4x-4\le0$$ ...

1

The specific problem you have appears to be trivial for state-of-the-art MIQP solvers (i.e., branch and bound/cut). It is solved to 5-6 digits of accuracy in 0s. There might be smarter ways to solve the problem, but I reckon not much smarter, as it is a combinatorial problem by nature. Heuristics might work very well though, in case you don't care about ...

1

Ah, I didn't realize that $X$ was required to be positive as well. Are you sure a clean solution exists? Since $BX$ is monotonic in all entries of $X$, dynamic programming can solve the problem in time pseudopolynomial in the magnitudes of the entries of $A$; this approach will only be practical for small $A$ however. Branch and bound codes like MIQPBB ...

0

If $A$ and $B$ are $\color{blue}{\rm HPD}$, then $$\begin{split} \color{red}{\lambda_{\max}(AB)} &= \lambda_{\max}(B^{1/2}AB^{1/2}) \\&= \max_{x\neq 0}\frac{x^*B^{1/2}AB^{1/2}x}{x^*x} \\&= \max_{x\neq 0}\frac{x^*Ax}{x^*B^{-1}x} \\&= \max_{x\neq 0}\left(\frac{x^*Ax}{x^*x}\right)\left(\frac{x^*x}{x^*B^{-1}x}\right) \\&\color{red}{\leq} ... 0 Let A=B=\operatorname{diag}(-2,1), then AB = \operatorname{diag}(4,1), but \lambda_\max (AB) = 4, \lambda_\max (A) = \lambda_\max (B)= 1. If the matrices are positive semi-definite, then \|A\|=\lambda_\max(A) (since A is unitarily diagonalisable) and the spectral norm is submultiplicatve, hence the result holds. 3 Submultiplicative norms for non-square matrices obviously do not make sense, but with square matrices, I don't think what you originally wrote was wrong. More specifically, we have the following Proposition. Suppose A,B\in M_n(\mathbb C). If \rho(AB)=\|AB\|_\square\le1 for some submultiplicative norm \|\cdot\|_\square, then there exists some ... 1 @ ziutek , a matrix norm ||.|| is defined on a space M_{n,m}. If ||.|| is sub-multiplicative, then necessarily m=n and A cannot be rectangular. Moreover there is a big mistake in your line 5. Indeed if \rho(BA)\leq 1, then, we have only the following: for every \epsilon>0, there is an induced norm N() s.t. N(BA)\leq \rho(BA)+\epsilon. ... 0 Another approach that extends to more general settings is to use the connection between the norm and the inner product,$$||x||^2 = (x,x).We have the finite difference, \begin{align} ||x+sh||^2 - ||x||^2 &= (x+sh,x+sh) - (x,x) \\ &= (x,x) + 2s(x,h) + s^2(h,h) - (x,x) \\ &= 2s(x,h) + s^2(h,h). \end{align} The gradient acting in the direction ... 2 Usually the triple bar notation is used for the subordinate norm for the linear transformation i.e. if f: (E,||.||_E)\to (F,||.||_F) is a linear transformation then we definite the subordinate norm of f by|||f|||=\sup_{x\in E\setminus\{0_E\}}\frac{||f(x)||_F}{||x||_E}$$1 Given the dual quaternion defined by z = (a+a_0\epsilon)+(b+b_0\epsilon)i+(c+c_0\epsilon)j+(d+d_0\epsilon)k = (a+bi+cj+dk)+(a_0+b_0i+c_0j+d_0k)\epsilon, for a,a_0,b,b_0,c,c_0,d,d_0 \in \Bbb R, i,j,k defined by i^2=j^2=k^2=ijk=-1, and \epsilon defined by \epsilon \ne 0 and \epsilon^2 = 0, the conjugate \overline{z} is defined by : ... 1 I think it is. Let 1<p<\infty be given and notice that the mapping T from W^{1,p}(\Omega)\to L^p(\Omega, R^{N+1}) via$$ T[u]\to (u,\nabla u) $$is isomorphic and closed. Together with the fact that L^p(\Omega, R^{M}) is uniformly convex for any M\geq 1, here we are interested in the case M=N+1, hence we know that W^{1,p} is uniformly ... 0 I found this post when I had the same question. It seems to me that the answer is no. Let's look at the \ell_2-norm: \|\boldsymbol{x}\|_2 = \sqrt{x_1^2 + \cdots + x_n^2} and recall the definition of a non-decreasing function: A function, f(x), is non-decreasing on an interval I if f(b) \geq f(a) for all b > a where a,b \in I. Without ... 4 In general, any symmetric positive definite W induces the norm \|x\|_W=\sqrt{x^TWx}=\|W^{1/2}x\|_2 on \mathbb{R}^n. The matrix norm induced by \|\cdot\|_W can be related to the spectral norm as$$ \|A\|_W =\max_{x\neq 0}\frac{\|Ax\|_W}{\|x\|_W} =\max_{x\neq 0}\frac{\|W^{1/2}Ax\|_2}{\|W^{1/2}x\|_2} =\max_{y\neq ...

2

Assuming $v\neq 0$, note that $$2\|v\|\frac{d}{dt} \|v\| = \frac{d}{dt} \|v\|^2=\frac{d}{dt} v\cdot v=2v\cdot \dot\alpha.$$ It follows from the Cauchy-Schwarz inequality that $$\frac{d}{dt}\|v\|\leq \|\dot \alpha\|.$$ If you do not assume $v\neq 0$, it may be that $\| v\|$ is not classically differentiable. Assuming that $v$ is once continuously ...

0

I think that the inequality is false. For $\Omega=D_1(0) \subset \mathbb{R}^2$ and $v(x)=|x|^n$, then $\nabla v(x)= n|x|^{n-2}x$, so the left term of the inequality gives $\sqrt{2 n^2 \pi}$, but the right one gives $C \sqrt{n \pi}$. I hope not to be wrong with integrals!

2

There are no elements in $\Bbb F_p((t))$ with infinitely many negative powers of $t$; all of them are of the first form you described. (Indeed there can be no multiplicative structure on the set of infinite sums of the second form you described, since multiplication of them is ill-defined in general.)

2

You should use $\max x_i$ or $\max_i x_i$. It is not a norm, so you shouldn't use $\|x\|_\square$ or anything like that. Maybe you could define a notation like $(x)_{max}$, but I don't think there already is one.

2

$\max_{1\leq i \leq N} x_i$ is the standard notation, and note that the symbol $||\cdot||:\mathbb{R}^n \rightarrow \mathbb{R}$ usually denotes a norm, which max is not.

0

You are correct.When f=$0$, its norm is also 0.But not converse,failing the first property of norm.

-1

Case 1: Assume without loss of generality, that $x_1$ is your maximum and assume that $|x_1| > |x_i|\ \forall i=2,..,n$. Then we get $$\left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p} = \sqrt[p]{|x_1|^p\cdot(1+ \frac{|x_2|^p}{|x_1|^p}+...+\frac{|x_n|^p}{|x_1|^p}) } \to \sqrt[p]{|x_1|^p} =||x||_∞ \ \mathrm{for }\ p\to\infty ,$$ because $$\frac{|x_i|^p}{|x_1|^p} ... 2 The constraint \|\mathbf x\|_2^2 \leq \alpha \|\mathbf y\|_2^2 is inactive when \alpha is large enough so that the solution to the unconstrained problem is the solution to the constrained problem. Then, the unconstrained problem does not depend on \alpha. 0 (Note that the norm is not relevant here, this is really a question about continuous non-negative integrable functions.) It is not true. Let t(x) = \max(0,1-|x|), and note that t is continuous and \int t = 1. Then let f(x) = \sum_{n \in \mathbb{Z}} t(n^2(x-n)). It is easy to verify that f is continuous and \int f = \sum_{n \in \mathbb{Z}} {1 ... 1 There is a little caveat you should consider. Except for that, what you say is true, independently of the norm you put on F. More generally, every integrable function g from the real line to itself (notice this: by considering the norm of F you're just considering a function from R to R) has most of its mass in a compact. In fact, you can prove ... 1 Not really. For any two matrices A and B and any \epsilon>0, put T=\frac\epsilon{2\|A-B\|}I. Then \|TA-TB\|=\frac\epsilon2<\epsilon. So you would be able to do this for any two matrices. If you require this to be true for any \epsilon>0, then this would imply TA=TB which of course implies A=B. EDIT: The best bound we can do is ... 1 it is not true, say on the real line take x_0=3, r=2, x=1.2, then \dfrac r 2 x-x_0= -1.8\not\in B(x_0,r). Hmm, edit, now the assumption says ||x||=1. On the real line take x_0=2.5, r=2, x=1, then \dfrac r 2 x-x_0= -1.5\not\in B(x_0,r). 1 1) As Zouba said A\mapsto\text{Tr}(A) is positive, so it achieves its norm at the unit. Thus \|\text{Tr}\|=n. 2) Assume \|A\|_p\leq1. Then in particular s_j(A)\leq1 for all singular values of A.$$ |\text{Tr}(A)|\leq\text{Tr}(|A|)=\sum_{j=1}^ns_j(A)\leq\left(\sum_{j=1}^ns_j(A)^p\right)^{1/p}\,\left(\sum_{j=1}^ns_j(A)^q\right)^{1/q}\leq n^{1/q}. ...

1

We can show a slightly different result, which gives a slightly stronger version of the statement in question when $t\geq 0$ with $\|\cdot\|=\|\cdot\|_{\infty}$. Let $$\|A\|_{\mathrm{max}}:=\max\limits_{i,j}|a_{ij}|$$ be the $\max$-norm of $A$ and $\|A-B\|_{\max}\leq t$. It is tempting to write $\|T(A-B)\|_{\max}\leq \|T\|_{\max}\|A-B\|_{\max}$, which is ...

1

1) The map is positive, hence $\|tr \|=tr(I)=n$.

5

Not necessarily. For example, we can take $$A = \pmatrix{1&0\\0&1}, \quad B = \pmatrix{1&1\\0&1}$$ the only eigenvalue of $A-B$ is $0$, so $A \geq B$. However, $\|A\|_2 = 1 < \|B\|_2$. Regarding your update: when $A,B$ are symmetric, this amounts to showing that if $A,B,$ and $A-B$ are positive semidefinite, then $\|A\|_2 \geq ... 1 There is a value of$\vec{x}$for which$||\vec{y}+A\vec{x}||$is minimized. Think of$\vec{y}$as a vector in 3-space,$A\vec{x}$is all the points in a plane through the origin, and$y+Ax$is the set of points in a plane through the point$y$. There is one point in this plane closest to the origin. Call it$y+Ax_0$. As soon as$\alpha$is large enough ... 1 Thanks for asking this question. I have problems understanding the same thing. Particularly as follows. Although the nuclear norm is a relaxation for the rank, I don't understand how the nuclear norm can work as surrogate for minimising rank unless it is an upper bound for the rank. As was pointed out in the original post, the nuclear norm does not ... 1 Hint: Existence : Without loss of generality we can suppose x=0 (we could simply translate by x the set A). Uniqueness : Suppose there were$a_0,b_0∈A$such that ||$a_0$∥=∥$b_0$∥=d and using the parallelogram law. 5 For each$p\ge 1$, we have$\|z\|_p=1$so$|x|^p+|y|^p=1$. Geometrically, this is what happens as$p$varies: (In the limit as$p\to\infty$, the edges sharpen to get a square.) 3 You can split your problem into $$\min \|x\| \quad {s.t.} A_1 x = b_1, \ l_1\le x\le u_1,$$ and $$\min \|y\| \quad {s.t.} A_2 y = b_2, \ l_2\le y\le u_2.$$ There is no coupling between both optimization variables. 0 You can use $$I+BA^T=I+AA^T+(B-A)A^T.$$ Hence $$\|(I+BA^T)(I+AA^T)^{-1}\|_2\leq 1+\|B-A\|_2\|A^T(I+AA^T)^{-1}\|_2.$$ By using the SVD of$A$, one can [from the fact that$x\mapsto x/(1+x^2)$is for$x\geq 0$bounded by$1/2$] get $$\|A^T(I+AA^T)^{-1}\|_2\leq \frac{1}{2},$$ so $$\|(I+BA^T)(I+AA^T)^{-1}\|_2\leq 1+\frac{1}{2}\|B-A\|_2.$$ You can ... 0 Use the chain rule and the assumptions on$F$to estimate the Sobolev norm of$\nu \circ F$, with respect to the Sobolev norm of$\nu$and some constants depending on$F$. 3 Let$p(x)=(x-1)(x-2)\cdots(x-n-1)$. Then the polynomial$p$has degree$n$and$\|p\|=0$while$p\neq 0$so the function in 1 is not a norm. Let$p=c$,$c$is a non-zero constant. Degree of$p$is$0$and$p\neq 0$but$\|p\|=\sup_{x\in [0,1]}|p'(x)|=0$so 4 is not a norm. EDIT: Thanks for warnings. Degree of$p$is$n+1$so it is not in$P_n(\mathbb R)$. ... 0 You are on the right track. For the induced norm$\|A\|_\infty=\text{max}\frac{\|Ax\|_\infty}{\|x\|_\infty}$we can also write the induced norm$\|A\|_\infty=\text{max}_{x \neq 0} (\|Ax\|_\infty)$where$\|x\|_\infty = 1$. We know that the definition of the infinity norm on matrices is choosing the largest element. So follow these steps where ... 1 Let$x^{(M)}\in S_F$be a sequence where$x_n^{(N)} = 1/n^2$if$n\leq N$and zero elsewhere. Note that$\sum_n 1/n^2 < \infty$. If$M>N$, then$||x^{(M)}-x^{(N)}|| = \sum_{k=N+1}^M 1/k^2\to 0$as$N,M\to \infty$, so the sequence$(x^{(N)})_{N\in \mathbb{N}}$is cauchy. Now we need to show that no point in$S_F$can be the limit. Suppose$y\in S_F$... 0 You need to find a Cauchy sequence that is not convergent. In your case you could take an infinite sequence$\mathbf a$such that$\|\mathbf a\|_1<\infty$, and approximate it by elements in$S_F\$.

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