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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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