Hot answers tagged

4

Hint: How many elements of $\Bbb R$ are of distance $1$ from $0$?


4

a), Like Omnomnomnom's opinion, the DFT matrix shows a good guidance. Here, DFT matrix $\mathsf{W}$ is defined as below: $$ \mathsf{W}=\cfrac{1}{\sqrt{N}} \begin{bmatrix} \omega_{1,1} & \omega_{2,1} & \omega_{3,1} & \cdots & \omega_{N,1} \\ \omega_{1,2} & \omega_{2,2} & \omega_{3,2} & \cdots & \omega_{N,2} \\ \vdots &...


4

Applying the definition of convolution, where I stressed the fact that the norm is in terms of $x$, and $y$ is a dummy variable \begin{align*}\|f\ast g(x)\|_T &=\|\int_{\mathbb{R}^n}f(y)g(x-y)dy\|_T\\ & \leq\int_{\mathbb{R}^n}\|f(y)g(x-y)\|_Tdy\\ & = \int_{\mathbb{R}^n} |f(y)|\|g(x-y)\|_Tdy\\ & =\int_{\mathbb{R}^n}|f(y)|\|g(x)\|_Tdy\\ \\ &...


3

No. For example $$||(x,y)||=|x|+|x-y|$$is a norm on $\Bbb R^2$ for which this is false (consider $v_1=(2,0)$, $v_2=(3,3)$).


3

Let $B_{r,i}(x)$ denote the open ball of radius $r$ around $x$ according to the $i$th norm. Suppose the topologies are equivalent. $B_{1,1}(0)$ is open according to $||\cdot||_2$, so there is some $r>0$ s.t. $B_{r,2}(0)\subseteq B_{1,1}(0)$. Therefore, if $||x||_1 =1$ then $||x||_2\geq r$. So for every $x\neq 0$, $||\frac{x}{||x||_1}||_2\geq r$ hence $||x|...


3

You already know that $T$ is linear if $T(0) = 0$. However, consider the maps $$ S(x) = T(x) - T(0)\\ R(x) = x + T(0) $$ Then clearly both $R$ and $S$ are isometries with $T = R \circ S$. However, since $S(0) = 0$, $S$ is linear, and by the rank-nullity theorem any injective linear map is surjective. Moreover, the translation $R$ is clearly surjective. ...


3

A more general case is when $B$ is a real normed space. You vcan apply Mazur–Ulam theorem : If ${\displaystyle V}$ and ${\displaystyle W}$ are normed spaces over $\mathbb{R}$ and the mapping ${\displaystyle f\colon V\to W} $ is a surjective isometry, then ${\displaystyle f}$ is affine. So $f$ is affine with $f(0)=0$, so $f$ is linear. So if $B$ ...


3

Hint regarding the usefulness of the completeness of $C[0,1]$: Say $f_n$ is Cauchy in $X$. You can show that in general $||f||_\infty\le ||f||_X$, hence $f_n$ is also Cauchy in $C[0,1]$. So you have $f$ with $||f_n-f||_\infty\to0$. That does not show by itself that $||f_n-f||_X\to0$, but it does give you a handle on things - at least now you have the limit,...


2

Remark that if $\|x\|=1$, $\phi(x)>0$ and $\mid c\mid \leq 1$, $\phi(cx)=c\phi(x)\in (-\phi(x),\phi(x))$ conversely if $d\in (-\phi(x),\phi(x))$, $d=c\phi(x)=\phi(cx)$, $\mid c\mid\leq 1$. We have $\|cx\|\leq 1$ thus $(-\phi(x),\phi(x))\subset \phi(B_E)$. Since $\|\phi\|=1$, for every $0<c<1$, there exists $x_c$ such that $\|x_c\|\leq 1$ and $\...


2

If $(f_n)$ is Cauchy in that norm then you can show that the two sequences $(\int_0^1 f_n)$ and $(\int_0^1 tf(t)\,dt)$ are Cauchy...


2

Consider the sequence $\mathbf x^{(n)}\in\ell^1\subset\ell^2$ defined by $$ x^{(n)}_k=\frac{1}{k}\textrm{ for } k\leq n,\quad x^{(n)}_k=0\textrm{ for } k > n. $$ Then $\mathbf x^{(n)}\overset{d_2}{\to} \mathbf x\in\ell^2\setminus\ell^1$, with $$ x^{(n)}_k=\frac{1}{k}\quad \forall k. $$ So $\ell^1$ is not $d_2$-complete.


2

Normed spaces are sets along with norms. If you want to remove the norm, and just treat it as a space, you're free to do so. We say "normed" if we have a norm in mind. Some spaces are "normable" but we haven't chosen a specific norm. And different norms induce different topologies, in particular different ways of things converging. As you're possibly aware, ...


2

Notice $$ \int_0^1 e^{\tau + t - 3} f(\tau) \, d\tau = e^t \int_0^1 e^{\tau - 3} f(\tau) \, d\tau .$$ So the equation can be rewritten as $$ f(t) = 1 - C e^t ,\tag 1$$ where $$ C = \int_0^1 e^{\tau - 3} f(\tau) \, d\tau .\tag 2$$ Now substitute equation (1) into equation (2), and solve for $C$.


2

In the line there is no regular triangle.


2

Here's an elementary way to solve 1. First the idea: The euclidean norms are quite special, since they come from inner products, and isometries in inner product spaces are very well-behaved. Namely, we have the following result: We denote $\langle,\rangle$ inner products and $\Vert\cdot\Vert$ the respective norm. Proposition: If $H$ and $K$ are real vector ...


2

If $S=T^*$, and $f_j\to f$ weak$^*$ in $Y^*$, then for any $x\in X$ $$ Sf_j(x)=f_j(Tx)\to f(Tx)=T^*f(x)=Sf(x). $$ So $S$ is weak$^*$-continuous. Conversely, assume that $S$ is weak$^*$-continuous. We know that what our $T$ should satisfy if it exists: $Sf(x)=f(Tx)$. So let us use this to define $T$. For any $x\in X$, consider the functional on $Y^*$ ...


2

Assume $a_1, \dots , a_n \ge 0.$ Because the function $x\to x^{r/p}$ is convex on $[0,\infty),$ Jensen implies $$\left (\frac{a_1^p+ \cdots + a_n^p}{n}\right )^{r/p}\le \frac{1}{n}\left(a_1^p)^{r/p}+ \cdots + (a_n^p)^{r/p}\right).$$ The inequality follows from this quite handily.


2

Ideally you need to find points where these suprema are attained. This is possible for $f_2$ as $\exists x \in C[a,b]$ such that $\|x\| = 1$ $$ x(a) = \text{sgn}(\alpha) \text{ and } x(b) = \text{sgn}(\beta) $$ where $\text{sgn}(z) = z/|z|$ (This follows from Urysohn's lemma, if you like, although one can just draw the graph of such a function on $[a,b]$). ...


1

Let $A - B = F$, and suppose that $B$ is diagonalizable with $B = SDS^{-1}$ and $D$ diagonal. Let $\|\cdot\|$ be a vectorial norm such that for every $x\in\mathbb R^n$ and $y$ defined as $y_i = |x_i|\,\,\forall i$, then $\|x\| = \|y\|$. Under these hypotesis, the Bauer-Fike Theorem assure us that for every eigenvalue $\lambda$ of $A$, there exists an ...


1

The standard Holder's inequality (for points in $\mathbb{R}^n$) can be written as $$ \|fg\|_p \leq \|f\|_q \|g\|_r $$ where $p^{-1} = q^{-1} + r^{-1}$; note that this requires $q, r \geq p$. The inequality you wrote above follows by taking $r^{-1} = p^{-1} - q^{-1}$ and $g = \mathbf{1}$. The version above can be proved by taking $\tilde{f} = |f|^p$ ...


1

For $d(x,y)\ne 0$, we have $$\frac{1}{{d\left( {x,y} \right)}} = \frac{{1 + \left\| {x - y} \right\|}} {{\left\| {x - y} \right\|}} = 1 + \frac{1}{{\left\| {x - y} \right\|}} \geqslant 1 + \frac{1}{{\left\| {x - z} \right\| + \left\| {z - y} \right\|}} = \frac{{1 + \left\| {x - z} \right\| + \left\| {z - y} \right\|}} {{\left\| {x - z} \right\| + \left\| {z -...


1

Changing the metric $d$ to $F(d)$ preserves the properties of a metric if: $F(0) = 0$ $F(d) > 0$ when $d > 0$ $F(a+b) \leq F(a) + F(b)$ for all $a,b \geq 0$ This is implied by, and in practice is equivalent to, $F$ being an increasing concave function. It is possible to artificially construct examples of $F$ that are increasing, subadditive and ...


1

Notice that the function $a\mapsto a/(1+a)$ is increasing and use that $d'(x,y)=||x-y||$ is also a metric (satisfies triangle inequality).


1

Write $$\omega_j=e^{\frac{2\pi i\,j}N},\ \ \ j=0,1,\ldots,N-1,$$ the $N^{\rm th}$-roots of unity. For simplicity, I'll renumber the indices of $x$ and $k$ to $0,\ldots,N-1$. So, with $e_0,\ldots,e_{N-1}$ the canonical basis, $$ \mathfrak F(e_j)=\frac1{\sqrt N}\,\sum_{k=0}^{N-1}\,\omega_j^k\,e_k. $$ Then \begin{align} \langle \mathfrak F(e_j),\mathfrak F(...


1

To simplify notation, write $$\omega_j=e^{\frac{2\pi i\,j}N},\ \ \ j=0,1,\ldots,N-1,$$ the $N^{\rm th}$-roots of unity. For simplicity, I'll renumber the indices of $x$ and $k$ to $0,\ldots,N-1$. So, with $e_0,\ldots,e_{N-1}$ the canonical basis, $$ \mathfrak F(x)=\frac1{\sqrt N}\,\sum_{h=0}^{N-1}\sum_{j=0}^{N-1}\,\omega_j^kx_j\,e_k. $$ Then \begin{align} ...


1

$\newcommand\ip[2]{\langle#1,#2\rangle}$ It's false. For example define $T:\Bbb C^2\to\Bbb C^2$ by $$T(x_1,x_2)=(x_2,0);$$then $||T||=1$ while the AM-GM inequality followed by Holder's inequality shows that $$|\ip {Tx}x|\le\frac{||x||^2}{2}.$$ It should probably be noted that it's less trivially false in the complex case than in the real case. As has been ...


1

There's nothing to see, actually. You can find a sequence $(x_n) \subset Y$ such that $x_n \to x \in X \setminus Y$. Then $(I_0(x_n))$ must converge to $I_0(x) \in Y$ (because we want $I_0$ continuous). But $I_0(x_n) = I(x_n) = x_n \to x \not \in Y$.


1

Let $M$ be the space of complex Borel measures on $[0,1].$ By the Riesz Representation theorem (RRT), $C[0,1]^* = M.$ Let $f_n(x) = x^n.$ Suppose $f\in C[0,1]$ and $f_n\to f$ weakly in $C[0,1].$ By RRT, that is the same as saying $\int_{[0,1]}f_n \,d\mu \to \int_{[0,1]}f \,d\mu$ for all $\mu\in M.$ Now for $0\le x \le a< 1,$ $f_n \to 0$ uniformly on $[0,...


1

I suggest the book by Pazy on Semigroups of Linear Operators. It's one of the most elegant and readable books in Functional Analysis I've encountered. On page 14, there is a theorem: Theorem: A Linear operator is dissipative if and only if $$ \|(\lambda I-A)x\| \ge \lambda \|x\|,\;\; x\in\mathcal{D}(A),\; \lambda >0. $$ His setting for ...



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