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

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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|$$is a norm on \Bbb R^2 for which this is false (consider v_1=(2,0), v_2=(3,3)). 3 Let B_{r,i}(x) denote the open ball of radius r around x according to the ith norm. Suppose the topologies are equivalent. B_{1,1}(0) is open according to ||\cdot||_2, so there is some r>0 s.t. B_{r,2}(0)\subseteq B_{1,1}(0). Therefore, if ||x||_1 =1 then ||x||_2\geq r. So for every x\neq 0, ||\frac{x}{||x||_1}||_2\geq r hence ||x|... 3 You already know that T is linear if T(0) = 0. However, consider the maps$$ 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$$Now substitute equation (1) into equation (2), and solve for C. 2 In the line there is no regular triangle. 2 Here's an elementary way to solve 1. First the idea: The euclidean norms are quite special, since they come from inner products, and isometries in inner product spaces are very well-behaved. Namely, we have the following result: We denote \langle,\rangle inner products and \Vert\cdot\Vert the respective norm. Proposition: If H and K are real vector ... 2 It's not even bounded. Consider the functions (n+1)x^n, n = 1,2,\dots  2 If S=T^*, and f_j\to f weak^* in Y^*, then for any x\in X$$ 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 -...

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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 ...

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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).

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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(...

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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} ...

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$\newcommand\ip[2]{\langle#1,#2\rangle}$ 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 ...

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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 $I_0(x_n) = I(x_n) = x_n \to x \not \in Y$.

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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,...

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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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