# Tag Info

3

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

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 ...

0

Using the dual basis, the matrix representation of $T^+$ is given by the transpose, $T^t$.

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.

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 $\tilde{f} = |f|^p$ ...

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

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 -... 0 Besides other great answers and references, I tried to follow your thoughts to complete the proof. \begin{aligned} d(x,z)+d(z,y) &= \frac{||x-z ||}{1+|x-z||}+\frac{||z-y ||}{1+||z-y||} \\ & = \frac{||x-z ||}{1+|x-z||} * \frac{1+||z-y||}{1+||z-y||} +\frac{||z-y ||}{1+||z-y||} * \frac{1+|x-z||}{1+|x-z||} \\&=\frac{||x-... 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). 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\\ \\ &... 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 &...

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} ...

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 $Y^*$ ...

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 $B$ ...

2

It's not even bounded. Consider the functions $(n+1)x^n, n = 1,2,\dots$

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 $||x|... 0 Your analysis is correct. Also your$\tilde f$is the only such extension. By the Riesz representation theorem, every bounded linear functional g on$\mathbb{R}^3$(with Euclidean norm) is of the form$g(x) = \left<x, y \right>$for some$y \in \mathbb{R}^3$. If$g$is defined by$y = (\alpha_1,\alpha_2,\alpha_3)$for some$\alpha_3 \neq 0$, then$...

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

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 $[0,... 1 There's nothing to see, actually. You can find a sequence$(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$. 0 Let$f\equiv0$. Then$f_n-f\xrightarrow{\lVert\cdot\rVert_\infty}0$on$[0,1)$. For for any$T\in\mathrm{dual}(\mathscr C[0,1],\lVert\cdot\rVert_\infty)$, we have that$T(f_n-f)\to0$on$[0,1)$, since$T$is continuous. Therefore since$\{f_n\}$is bounded, and since$T(f_n)\to T(f)$on$[0,1)$which is dense in$[0,1]$, we have that$f_n$is weakly ... 0 Any bounded linear functional defined on a subspace of a Hilbert space admits a unique norm preserving extension. The proof is given here. More on Banach spaces with unique extension property for functionals you can find in this discussion 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 ... 4 Hint: How many elements of$\Bbb R$are of distance$1$from$0$? 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$\...

0

This is a fairly standard result, here is a proof that uses the Borel Cantelli lemma. A slightly more general result is true, and the more general result hints at a direction of proof. It is not difficult to show that if $f_n\to f$ (in $L^1$) then $f_n$ converges to $f$ in probability, that is for all $\epsilon>0$, $\lim_n \lambda \{ x | |f_n(x)-f(x)| \... 0 Inner products In$\mathbb C:\;\,\qquad \langle x, y\rangle=\bar yx$In$\mathbb C^2:\qquad\langle x, y\rangle=\bar y^\top x$Norms In$\mathbb C:\;\,\qquad ||x||_1^2=x\bar x$In$\mathbb C^2:\qquad||x||_2^2=||x_1||_1^2+||x_2||_1^2Calculation $$||e^{\theta}x-e^{i\phi}y||_2^2=(e^{i\theta}x_1-e^{i\phi}y_1)\overline{(e^{i\theta}x_1-e^{i\phi}... 2 If (f_n) is Cauchy in that norm then you can show that the two sequences (\int_0^1 f_n) and (\int_0^1 tf(t)\,dt) are Cauchy... 2 Consider the sequence \mathbf x^{(n)}\in\ell^1\subset\ell^2 defined by$$ 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, ... 1 Yes. Your argument is valid. In a nutshell, your argument is this: Take any f \in X^*. Since X is finite dimensional, f is bounded. So, f \in X'. So, X^* \subseteq X', as desired. When X is infinite dimensional, we can always construct an unbounded linear function f on X. Finding an explicit construction of such an X is tricky. ... 0 Here's the issue with your boundedness proof. Your argument boils down to the following calculation: \begin{align*} \|Bf\|_{L^2}&=(\int_0^\infty\left|\frac{1}{x}\int_0^xf(t)dt\right|^2dx)^{1/2}\\ &\leq(\int_0^\infty\left(\frac{1}{x}\int_0^\infty|f(t)|\mathbf1_{[0,x]}dt\right)^2dx)^{1/2}\\ &\leq(\int_0^\infty\frac{1}{x^2}\|f\|_{L^2}^2xdx)^{1/2}\\ &... 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,... 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. $$His setting for ... 0 Hints:$$(x_n\to\xi\implies A(x_n)\to A(\xi))\implies(x_n\to x \implies A(x_n) = A(x_n-x+\xi+x-\xi) = A(x_n-x+\xi) + A(x-\xi)\to\cdots)A$bounded in a neighbourhood of zero implies$A\$ continuous in zero.

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