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0

I'll assume $s \neq 0$, otherwise the problems might be infeasible. For part a), we want to know which point in the halfspace $s^T x \leq r$ is closest to the origin. If $r \geq 0$, then the origin already belongs to this halfspace, so the best choice of $x$ is simply $x = 0$. If $r < 0$, then visually we can see that the shortest way to get from the ...


1

Hint: Use Lagrange multipliers. In both problems, the case $r \ge 0$ is trivially solved by the zero vector. So in what follows, we assume $r < 0$. Also, to ensure feasibility we'll be assuming $s \ne 0$. (a) You can replace $\|x\|$ with $\frac{1}{2}\|x\|^2$ and this won't change the solution (why ?). Now, consider the Lagrangian \begin{eqnarray} L(x, ...


0

This is possible when the Hamel dimensions of $V$ and $W$ are the same infinite cardinal. This is not necessary but only sufficient, since countable dimensional space can be dense in an infinite dimensional space: consider the Hamel span of a countable Hilbert basis. Let us assume this and denote by $B$ and $C$ the bases of $V$ and $W$ so that $C\subset B$. ...


2

More easily you can do this: $a_n(f)=-\frac {b_n(f')}{n}$ and by Riemann-Lebesque lemma we have that $b_n(f')\to 0$ and thus $|b_n(f')|\leq K$ for every $n$ and we have that $|a_n| \leq \frac{K}{n}$.


2

Note that $$\int_{-\pi}^{\pi}f(x) \cos nx dx=\frac{1}{n}f(x)\sin nx\mid_{-\pi}^\pi-1/n\int_{-\pi}^\pi\frac{df}{dx}\sin nx dx=-1/n\int_{-\pi}^\pi\frac{df}{dx}\sin nx dx\\\implies \left|\int_{-\pi}^{\pi}f(x) \cos nx dx\right|=\frac{1}{n}\left|\int_{-\pi}^\pi\frac{df}{dx}\sin nx dx\right|\\\le \frac{1}{n}\max_{x\in ...


0

The left hand side equals to $$\rho^2\|x\|_2^2 + (1 - \rho)^2\|y\|_2^2 + 2\rho(1 - \rho)x \cdot y.$$ While the right hand side equals to \begin{align*} & \rho\|x\|_2^2 + (1 - \rho)\|y\|_2^2 - \rho(1 - \rho)(\|y\|_2^2 + \|x\|_2^2 - 2y\cdot x) \\ = &[\rho - \rho(1 - \rho)]\|x\|_2^2 + [(1 - \rho) - \rho(1 - \rho)]\|y\|_2^2 + 2\rho(1 - \rho)x \cdot y ...


1

If an inverse of any kind exist, $T$ is a bijection. As a consequence of the open mapping theorem, a bijective operator is bounded from below, meaning that there is $c>0$ such that $\|Tx\|\ge c\|x\|$ for all $x$. This and the property $TS=I$ imply that $S$ is bounded.


2

Concerning the usage of "if": In definitions people usually write X is called Y if Z holds, even though they mean "if and only if". Since it is a definition, there is no other object with that name and other probably weaker definitions. Concerning "if", "iff", and $\implies$: "If condition A holds, then statement B is valid" is usually written as $A ...


3

You can approximate the constant function with Gaussians $\phi_{\sigma^2}(x)=\frac{1}{\sqrt{2\pi \sigma^2}} \exp(-x^2/2\sigma^2)$. If I calculated correctly you have $\|\phi_{\sigma^2}\|^2= \frac{1}{2\sqrt{\pi \sigma^2}}$ and since $A\phi_{\sigma^2} = \phi_{\sigma^2+1}$ (the convolution of two Gaussians is Gaussian with the sum of the variances) you get ...


3

Yes, it is true that $\|A\|=1$. The operator $A$ is sometimes called a Fourier multiplier or simply a multiplier because it acts by pointwise multiplication of the Fourier transform. (Alternatively, multipliers are the same as convolution operators). So anyway, we can prove the following: Let $m\in L^\infty(\mathbb{R})$ be a bounded and measurable ...


3

First, notice that $H$ is linear, so you only need to prove linearity at 0. And you have this inequality : $$| H(f) | = |f(1) - f(0) | = \left| \int_0^1 f'(t) dt \right| \leq \int_0^1 | f'(t) | dt \leq \| f\|$$ So clearly, $H(f) \to 0 $ when $f \to 0$ : $H$ is continuous at $0$, and by linearity, everywhere


1

Thr natural thing is to prove the contrapositive. If $T $ is not bounded, there exists a sequence $\{x_n\}_X $ with $\|x_n\|=1$ and $\|Tx_n\|>n^2$. Then $x_n/n $ is a sequence that converges to zero with its image through $T $ unbounded. Conversely, if $x_n\to0$ with $\{Tx_n\} $ unbounded, then $T $ is unbounded.


1

Let $(f_{n_k})$ be a subsequence of $(f_n)$. Pick a subsequence of $(f_{n_k})$, call it $(g_n)$, that converges a.e. to $0$ (this can be done since $(f_n)$ converges to $0$ in $L_1$). Pick $y\in[0,1]$ with $\lim\limits_{n\rightarrow\infty} g_n(y)=0$. Then for any $n$ and any $x\in[0,1]$ $$ |g_n(x)-g_n(y)|\le\biggl|\,\int_y^x g_n'(x)\,\biggr| \le\Vert ...


1

One question per post. For $0\le x\le y\le1$ $$ |f_n(x)-f_n(y)|=\Bigl|\int_x^yf_n'(t)\,dt\Bigr|\le\|f'_n\|_1. $$ Thus $\{f_n\}$ is equicontinuous and has a uniformly convergent subsequence. Since $\{f_n\}$ converges to $0$ in $L^1$, the uniform limit of the aforementioned subsequence must be $0$. This argument can be carried out for any subsequence of ...


0

It's false. The counterexample is simpler than I expected at first; the condition $\sum f_n^2\in L^\infty$ is just too fragile, those convolutions smear the mass of $f_n$ around a bit and it's destroyed. As a general notational convention we're going to say $$\delta=1/\lambda.$$Let $e_\lambda$ be your exponential kernel, so that $$u_n=e_{\lambda_n}*f_n.$$ ...


2

What you're looking for is usually associated with a bilinear form, moreover we say: A scalarproduct on a real vector space $V$ (induced by a bilinear form $B$) is a symmetric, non-degenerated, positive definite bilinear form. A scalarproduct then induces a norm. If we are dealing with a finite dimensional real vector space, we can then also write $$ ...


1

I think the best you can say is that $||\cdot||_A$ is a norm induced by an inner product. (Not all norms are like that.) The matrix $A$ is the Gram matrix of that inner product with respect to the canonical basis of $\mathbb R^n$ (provided $A$ is positive definite).


4

That's only a norm if $A$ is positive definite. If $A$ is positive definite, then I would call what you describe "the $A$ norm".


0

If your question is $$\text{ Is } d(a,b) = \| a - b\|^2 \text{ a norm on } \mathbb R^n?$$ then the answer is no as the triangle inequality can be violated. For example, let $\vec e$ be any unit vector. Then $$d(0\vec e,2\vec e) = 2^2 \color{red}{>} 1^2 + 1^2 = d(0\vec e,1\vec e) + d(1\vec e,2\vec e)$$


1

Define a vector to be the difference: $$ \mathbf{c} = \mathbf{a} - \mathbf{b}.$$ And the take the dot product of the difference, $\mathbf{c} = (c_1, c_2, c_3, \dots, c_n)$, with itself: $$ \| \mathbf{a} - \mathbf{b} \|^{2} = \mathbf{c}\cdot\mathbf{c}, \\ \qquad = c_{1}^{2} + c_{2}^{2} + c_{3}^{2} + \cdots + c_n^{2}. $$


2

The standard (pythagorian) norm for vectors in $\mathbb{R}^n$ is defined as $$||\vec{x}|| \equiv \sqrt{\vec{x}\cdot\vec{x}}$$ where the dot-product of two vectors $\vec{x} = (x_1,x_2,\ldots,x_n)$ and $\vec{y} = (y_1,y_2,\ldots,y_n)$ is defined as $$\vec{x}\cdot\vec{y} \equiv x_1y_1 + x_2y_2 + \ldots + x_n y_n$$ Applying the definitions above with ...


2

as $(\vec{a}-\vec{b}).(\vec{a}-\vec{b})$


-1

For a symmetric matrix you want to maximise $||Ax||$ for $||x||=1$, which is the same as maximising $||Ax||^2 = xA^TAx$ Since we know that if $A$ is symmetric, it is diagnolaisable, so we write $$A = Q^T\Lambda Q$$, with $Q^{-1}=Q^T$then $$xA^TAx= (Qx)^T\Lambda^2(Qx)$$ where $Q$ is orthonormal so $||Qx||=1$ It is straightforward to see what you have ...


3

For any square $A$, $\rho(A)\leq\|A\|_2$ with the equality (not necessarily) if $A$ is normal. Besides the general inequality, $\rho(A)$ and $\|A\|_2$ can be completely unrelated. Consider, e.g., $$ A_\alpha:=\pmatrix{0&\alpha\\0&0} $$ with $\rho(A_\alpha)=0$ but $\|A_\alpha\|_2=|\alpha|$. All the eigenvalues are zero but the 2-norm can be an ...


2

$\def\norm#1{\left\|#1\right\|_1}\def\abs#1{\left|#1\right|}$As you write correctly, we have $$ \norm{Ax} = \norm{\sum_{i=1}^n x^i Ae_i} \le \sum_{i=1}^n \abs{x^i} \norm{Ae_i} $$ Now, note that for every $i$, we have $$ \norm{Ae_i} \le \sup_{1\le j \le n} \norm{Ae_j} $$ Let's call the supremum $S$, then $\norm{Ae_i} \le S$ for all $i$, giving above $$ ...


0

The general way to compute the proximal to function f is to calculate $$ Prox_{f}(x) = \arg\min_y f(y) + \frac{1}{2} ||x-y||^2 $$ The indicator function in convex optimization takes 0 inside and $+\infty$ outside. Thus in your case, the constraint must be satisfied to get the min. The problem is equivalent to $$ Prox_{c}(x) = \arg\min_{y,\ Ay=b} ...


1

For example, consider the norm $$ \|(x,y)\| = x ^2- xy + y^2 $$ We note that $$ \|(2,0)\| > \|(2,1)\| $$ A class of norms that act the way you might expect is the set of "symmetric gauge functions", as referenced here.


1

Take $x = (0, 0), y = (10, 0)$ and $z = (8, 9)$. With $p = 2$ we have a standard Euclidean distance, and $z$ is further away from $x$ than $y$ (their distances are $8 \sqrt{2}$ and $10$, respectively). With $p$ arbitrary large, however, we have the distance between $x$ and $y$ as $10$ while the distance between $z$ and $x$ as $9$.


1

Notice that if $|a_{km}|=max_{i,j}|a_{ij}|$ then $|Ae_m|_{\infty}=|a_{km}|$, where $e_m$ is the column vector with 1 in the $m$ entry and $0$ otherwise. It is clear that $|Ax|_{\infty}\leq |a_{km}|$, if $x\geq 0$ and $|x|_1=1$ . So $max_{|x|_1=1\ ,\ x\geq 0}|Ax|_{\infty}=max_{i,j}|a_{ij}|$


1

Let $\lambda_a \colon b \mapsto a\cdot b$. Then it's easy to see that the operator norm of the matrix is at most $$\max_i \sum_j \lVert \lambda_{a_{ij}}\rVert_{\operatorname{op}}.$$ For a general Banach algebra, it is possible that $\lVert \lambda_a \rVert_{\operatorname{op}} < \lVert a\rVert$ for some $a$. Still, even if we take the operatornorm of the ...


0

I that think the key to answering this question is to recognize that Im(A) is a subspace of ℝ^m. It means that for a given b ∈ ℝ^m, there exists a point of Im(A) closest to b; namely, the orthogonal projection of b onto Im(A). We can write this projection as Ax, for some x ∈ ℝ^n, by the definition of Im(A). Thus, we have the result that for any other point ...


0

Yes, you could solve this using some proximal algorithm such as the Douglas-Rachford method. Let $C = \{ (A,x) \mid D = A + Mx \}$, and let $I_C$ be the indicator function of $C$: \begin{equation} I_C(A,x) = \begin{cases} 0 & \text{if } D = A + Mx, \\ \infty & \text{otherwise.} \end{cases} \end{equation} Your problem can be restated as ...


2

The spectral radius is not a norm; one reason is because there are nonzero matrices all of whose eigenvalues are zero. (Such matrices can't be diagonalizable, but nondiagonalizable matrices exist!) copper.hat gave one example. Your property does hold if you replace $\| \cdot \|$ with the spectral radius. If $A^n \to 0$, then there exists a norm, which can ...


1

If use the norm for matrix as $$\|A\|=\sqrt{\sum\limits_{i,j=1}^{n}|a_{ij}|^2}$$ We can prove $$ \|A\|^n \to 0\implies \|A^n\| \to 0 $$ Converse is not true as a counter example is given by copper.hat. First we prove the following: Lemma: $\hspace{2 mm}\|AB\|\leqslant\|A\|\|B\|$ Prove: \begin{align} ...


1

For $A$ matrix denote by $\rho(A)$ the spectral radius of $A$, $= \max \{ |\lambda_i|\}$, with $\lambda_i$ are the eigenvalues of $A$. We have $A^n \to 0$ if and only $\rho(A)^n \to 0$ if and only if $\rho(A)<1$. For proof one can use the Jordan canonical form. Now if $||\cdot ||$ is any algebra norm on $M_n(\mathbb{R})$ ( for example coming from a ...


14

Let $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ with the $l_1$ norm, then $\|A\| = 1$. Since $A^2 = 0$, we see that $\lim_n \|A^n\| = 0$, but we have $\lim_n \|A\|^n = 1$.


2

Let $X$ be the set of $f:[0,1]\to[0,1]$ such that $f(1)=1$. Then $X$ is complete in the uniform norm, the identity function is in $X$, and it's easy to show that $\Phi$ is a strict contraction on $X$.



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