# Tag Info

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".

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

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 ...

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

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 $$... 2 The standard (pythagorian) norm for vectors in \mathbb{R}^n is defined as$$||\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$$Applying the definitions above with ... 2 as (\vec{a}-\vec{b}).(\vec{a}-\vec{b}) 2 What you're looking for is usually associated with a bilinear form, moreover we say: A scalarproduct on a real vector space V (induced by a bilinear form B) is a symmetric, non-degenerated, positive definite bilinear form. A scalarproduct then induces a norm. If we are dealing with a finite dimensional real vector space, we can then also write$$ ...

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 $A ... 2 More easily you can do this:$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 $$\int_{-\pi}^{\pi}f(x) \cos nx dx=\frac{1}{n}f(x)\sin nx\mid_{-\pi}^\pi-1/n\int_{-\pi}^\pi\frac{df}{dx}\sin nx dx=-1/n\int_{-\pi}^\pi\frac{df}{dx}\sin nx dx\\\implies \left|\int_{-\pi}^{\pi}f(x) \cos nx dx\right|=\frac{1}{n}\left|\int_{-\pi}^\pi\frac{df}{dx}\sin nx dx\right|\\\le \frac{1}{n}\max_{x\in ... 1 Thr natural thing is to prove the contrapositive. If T is not bounded, there exists a sequence \{x_n\}_X with \|x_n\|=1 and \|Tx_n\|>n^2. Then x_n/n is a sequence that converges to zero with its image through T unbounded. Conversely, if x_n\to0 with \{Tx_n\} unbounded, then T is unbounded. 1 Hint: Use Lagrange multipliers. In both problems, the case r \ge 0 is trivially solved by the zero vector. So in what follows, we assume r < 0. Also, to ensure feasibility we'll be assuming s \ne 0. (a) You can replace \|x\| with \frac{1}{2}\|x\|^2 and this won't change the solution (why ?). Now, consider the Lagrangian \begin{eqnarray} L(x, ... 1 I think the best you can say is that ||\cdot||_A is a norm induced by an inner product. (Not all norms are like that.) The matrix A is the Gram matrix of that inner product with respect to the canonical basis of \mathbb R^n (provided A is positive definite). 1 Let (f_{n_k}) be a subsequence of (f_n). Pick a subsequence of (f_{n_k}), call it (g_n), that converges a.e. to 0 (this can be done since (f_n) converges to 0 in L_1). Pick y\in[0,1] with \lim\limits_{n\rightarrow\infty} g_n(y)=0. Then for any n and any x\in[0,1]$$ |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 ... 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. 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: $$\| \mathbf{a} - \mathbf{b} \|^{2} = \mathbf{c}\cdot\mathbf{c}, \\ \qquad = c_{1}^{2} + c_{2}^{2} + c_{3}^{2} + \cdots + c_n^{2}.$$ 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 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 ... 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}|$1 Basically, because$f$is linear. Let's set $$A = \sup\{|f(x)| : x \in X, \|x\| \le 1\}$$ and $$B = \sup\left\{\frac{|f(x)|}{\|x\|}, x \in X, x \ne 0\right\}.$$ Suppose$x \in X$with$\|x\| \le 1$. If$x=0$then$f(x) = 0$, so$|f(x)| \le B$. If$x \ne 0$then$|f(x)|\le \frac{|f(x)|}{\|x\|}$since the denominator is at most 1. So$|f(x)| \le B$in ... 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\$.

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