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6

This is a really interesting question, and here is a partial answer. The $1$ and $\infty$ norms do not come from inner products. For a norm to have an associated inner product actually gives you a lot of structure. For example (if the scalars are real for convenience), $$\left\| x - y \right\|^2 = \langle x - y, x -y \rangle = \langle x, x \rangle - 2 ... 4 Every inner product on \mathbb{R}^n can be written as \langle x, y \rangle = x^t A y, where A is a (symmetric) positive definite matrix. These matrices can be orthogonally diagonalized, i.e. there is an orthogonal matrix M so that A = M^t \operatorname{diag}(\lambda_1, \ldots, \lambda_n) M. This means in particular \langle x, y \rangle = (Mx)^t ... 4 Hint. A_k u is by definition the map x \mapsto u(x + 1/k) (where u \colon \mathbf R \to \mathbf R is a given measurable, p-integrable map). You only have a sequence of maps A_k \colon L^p(\def\R{\mathbf R}\R) \to L^p(\R) here, neither the domain nor its subsets consist of sequences, L^p(\R) is the space of p-integrable maps \R \to \R. Do not ... 3 Note that by definition, A is nilpotent means that there exists a finite k such that A^k = 0. The property that you're referring to, it seems, is that A should satisfy \lim_{k \to \infty} \|A^k\| = 0. Indeed, any matrix A such that \|A\|<1 will satisfy \lim_{k \to \infty}\|A^k\| = 0, since \|A^k\| \leq \|A\|^k. However, a matrix A ... 3 Let's look at general p<q. Since the inequality$$ \lVert x \rVert_p \geqslant \lVert x \rVert_q $$is homogeneous (i.e. it is unchanged by x \mapsto \lambda x for scalar lambda), we can assume \lVert x \rVert_p = 1 (the inequality's trivial for x=0, of course). Therefore every component, x_i, of x, satisfies \lvert x_i \rvert<1, so ... 3 Note that \|x\|_1 = |x_1| + |x_2|. So if |x_1| + |x_2| = 1, it's described by the four equations:$$ \begin{cases} x_1 + x_2 = 1 & \text{if } x_1 \ge 0, x_2\ge 0,\\ x_1 - x_2 = 1 & \text{if } x_1 \ge 0, x_2<0,\\ -x_1 + x_2 = 1 & \text{if } x_1<0, x_2 \ge 0,\\ -x_1 - x_2 = 1 & \text{if } x_1< 0, x_2<0, \end{cases}$$which ... 3 Equivalence of norms means more than just that they induce the same topology (although, for topological vector spaces, the equality of the induced topology implies that "more"). It means that the identity \operatorname{id} \colon (X,\lVert\,\cdot\,\rVert_1) \to (X,\lVert\,\cdot\,\rVert_2) is a bi-Lipschitz map. Thus equivalent norms induce the same ... 3 Just calculate$$ \rho^2=\rho\rho^*=\left(\frac{1}{N}\text{id}+\sum_{k=1}^{N^2-1}\nu_kF_k\right) \left(\frac{1}{N}\text{id}+\sum_{m=1}^{N^2-1}\nu_mF_m\right)^*=\\ =\frac{1}{N^2}\text{id}+\sum_{k,m=1}^{N^2-1}\nu_k\bar\nu_mF_kF_m^*+\text{terms with only $F_k$ or $F_m^*$} $$and use the fact that \{\text{id},F_k\} is an ON basis (w.r.t. the Frobenius norm)$$ ...

3

No. It's not an induced norm. Let $\|\cdot \|$ be your norm. If it were is an induced norm then you'd have $\|A ^2 \| \leq \|A\| ^ 2$ for every matrix $A$. But consider $A = \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right)$ to get $2 = \|A ^2 \| \nleq \|A\| ^ 2 = 1^2 = 1$. Thus, your norm can't be an induced one.

3

Suppose to the contrary that there exists an $x\neq 0$ such that $(I-A)x = 0$. Assume without loss of generality that $\|x\| = 1$. Then $\|Ax\| = \|x\| = 1$. Hence, $\|A\| = \max_{\|x\|=1} \|Ax\| \geq 1$, which is a contradiction.

3

The Ball of unity that is induced by the "norm" is not convex. The points $(1, 0)$ and $(0, 1)$ are in the Ball, but their midpoint isn't. This can be visualized in this plot.

3

Counterexample to the first question: A nilpotent matrix. The second part is true, however: Since $A$ is bounded linear there is a uniform bound $c$ on $||A^1||,\dots, ||A^n||$, and then we have $|A^{kn+i}v|\le c||A^n||^k|v|\to 0$.

3

Hint: $$\| x\|_1 = |x_1| + \cdots + |x_n| = (1, 1, \cdots, 1)\cdot (|x_1|, |x_2|, \cdots, |x_n|).$$

2

I give to you the case $r=1$ and $p=2$, and I leave to you to think of the general case. To show that there does not exist a constant $c$ such that $$\|f\|_2\leq c\|f\|_1,\quad\forall f\in C[0,1],$$ it suffices to show that given $A>0$ we can find a positive continuous function $f$ such that $$\frac{\int_0^1 (f(x))^2\,dx}{\Bigl(\int_0^1 ... 2 Mathematicians might phrase this question by asking how to show that all matrix norms, defined by \| A \| = \max ( \|Ax\|/\|x\|), are actually norms, in the sense usually stated in the definition of a norm of a vector space. Except that this addresses only one aspect of that. For norms of vectors, you know that for every scalar \alpha and every vector ... 2 The comment by Geoff Robinson was really an answer: You have enough knowledge from class to know that I-A^{N} is invertible, and$$I - A^{N}= (I-A) ( I + A + A^{2} + \ldots + A^{N-2} + A^{N-1})$$2 The definition of the norm \|x\| is (x,x), where the latter is the standard inner product. The inner product (x,y) is given by \sum_i x_iy_i, i.e., the sum of the products of the components. If you treat x and y as n\times 1 matrices, then this expression is simply x^Ty. From all of this you get \|x\|=x^Tx. (By the way, the reason for ... 2 The multiplication didn't turn to a transpose - you are still multiplying x^T and x. The value is (x^T)\cdot x. If you had assumed that x was a row vector, then you'd have \|x\|^2 = xx^T = x\cdot (x^T). So you are treating that internal T as a binary operation of some sort, when it is not - it is a unary operation. If z is a complex number, ... 2 For n\ge2 let$$ f_n(x)=\begin{cases}n\,x & \text{if }0\le x\le1/n, \\ 2-n\,x & \text{if }1/2<x<2/n, \\0 & \text{if }2/n\le x\le1.\end{cases} $$Then f_n\to0 in L^1, pointwise, but not uniformly. 2 Hint: notice that by the FTC we have$$|f(x)| \le |f(y)| + \int_y^x|f'(t)|\,dt \le |f(y)| + \int_0^1|f'(t)|\,dt.$$2 It's enough to show that E|X|^p is continuous in p. Why not use dominated convergence: p\mapsto |X|^p is continuous, and if p\in[1,p_0] then |X|^p\le 1+|X|^{p_0}. (Think of p_0>1 as large but fixed.) In this way you show that p\mapsto E|X|^p is continuous on [1,p_0] for each p_0>1. 2 Hint: The claim is trivial when v=v^\prime. Otherwise,$$\frac{\left|Tv-Tv^{\prime}\right|}{\left|v-v^{\prime}\right|}=\frac{\left|T\left(v-v^{\prime}\right)\right|}{\left|v-v^{\prime}\right|}\leq\sup_{x\neq0}\frac{\left|Tx\right|}{\left|x\right|}=\ldots$$2 You actually only need to use the property that \|\lambda (Ax)\| = |\lambda|\|Ax\| Then, take any x\neq 0, and you know that x = \|x\|\cdot \frac{x}{\|x\|}, where the norm of \frac x{\|x\|} is 1. Using these two facts, you can prove that the sets$$S_1=\{\|Ax\| | \|x\|= 1\}\\S_2 = \{\frac{\|Ax\|}{\|x\|}| x\neq 0\}$$are the same set. 2 Using the so-called 1-trick (a special application of the Cauchy-Schwarz inequality):$$||x||_1^2 = \left( \sum_{i=1}^n x_i \right)^2 = \left( \sum_{i=1}^n 1 \times x_i \right)^2 \le \left( \sum_{i=1}^n 1^2 \right) \left( \sum_{i=1}^n x_i ^2 \right) = n \times ||x||_2 ^2$$Where x=(x_1,...,x_n). 2 some hints: First step: prove the "\geq" part of the equality \|u\| = \sup \{ | \varphi(u) |: \varphi \in X', \|\varphi\| \leq 1 \}. For this you only need to show that \|u\| \geq \varphi(u) if \varphi has the conditions mentioned. Second step: show that there is a \varphi \in X' with  \varphi(u) = \| u \| using Hahn-Banach (start by defining ... 2 No, simply consider two vectors x \ne y that have the same norm. However, the inequality | \| x\| - \|y\|| \le \|x - y\| holds [it is often called reverse triangle inequality]. By the triangle inequality, the following inequalities hold:$$ \| x \| = \|x - y + y\| \le \| x - y \| + \|y\| \| y \| = \|y - x + x\| \le \| y - x \| + \|x\|$$The ... 2 Your use of Cauchy-Schwarz to get the first inequality is fine, it shows that |x^Ty| \leq 1 if ||x||_2 \leq 1 and ||y||_2 = 1. For the other direction, go with what a comment suggested and use y = x / ||x||_2. 2 It depends on the meaning of "matrix norm". In linear algebra literature, a "matrix norm" usually refers to a submultiplicative norm, but in other areas of mathematics, submultiplicativity is not an implicit assumption. Edit. Your definition of f_X(M) can be extended to cover M_n(\mathbb R) if we rewrite the old definition as ... 1 The major implication is that you know the space of continuous linear functionals on the space (for 1<p<+\infty), and you can, therefore, apply all the properties of \ell_{q}. Among other things this allows you to introduce weak topology, which, in turn, sometimes is the natural topology for the space of solutions of an equation that you study. ... 1 There is a metric space structure on a normed linear space (X, \|\cdot \|): define the distance between x, y\in X by$$d(x, y) = \|x-y\|.$$On the other hand, whenever you have a metric on X, you can use that to define a topology on X: A set U \subset X is called open if for all x\in U, there is \epsilon >0 so that$$ d(x, y) < ...

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