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6

Suppose $A$ and $B$ are positive operators on a finite-dimensional inner product space $V$ and $\|A + B\| = \|A\| + \|B\|$. Because $A$ and $B$ are positive, $A+B$ is also a positive operator. Thus there exists $x \in V$ such that$$\|x\| = 1 \quad \text{and} \quad\|A+B\| = \langle (A+B)x, x \rangle. $$ Now \begin{align*} \|A+B\| &= \langle (A+B)x, x ...


5

It fails for the sequence $1,0,0,\dots $


4

Notice that for each vector $x$, one has $$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$ Thus $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$


4

You're correct. But note that in the BV definition, there is no benefit in taking $\|x\|<1$, so the definition may as well have stipulated that $\|x\|=1$.


4

The answer is yes. Let $$J = \left( {\begin{array}{*{20}c} \hfill {J_1} & \hfill {} & \hfill {} & \hfill {} \\ \hfill {} & \hfill {J_2} & \hfill {} & \hfill {} \\ \hfill {} & \hfill {} & \hfill \ddots & \hfill {} \\ \hfill {} & \hfill {} & \hfill {} & \hfill {J_n} \\ \end{array}} \right)$$ be ...


4

Use paralleogram law for $\frac{x}{2}$ and $\frac{y}{2}$ to obtain $||\frac{x+y}{2}||^2 + ||\frac{x-y}{2}||^2 = \frac{1}{2}||x||^2 + \frac{1}{2}||y||^2$ and so you get $1$ in the right hand side. Since the LHS is a sum of two non-negative terms, you get the desired inequality since $x\neq y$ .


3

The trick is to first observe that $$ ||x||_{\infty}^p=\sup_n|x(n)|^p\leq \sum_{n}|x(n)|^p=||x||_p^p $$ hence $||x||_{\infty}\leq ||x||_p$. If $p<q<\infty$, then $$||x||_q^q=\sum_n|x(n)|^q\leq ||x||_{\infty}^{q-p}\sum_n|x(n)|^p=||x||_{\infty}^{q-p}||x||_p^p\leq ||x||_p^q$$ with the last inequality using the $\infty$ case. Taking $q$th roots shows that ...


3

Note that if $\|x\| < 1 $ and $z^T x \ge 0$, then $z^T {x \over \|x\|} \ge z^T x$. Hence $\sup \{z^T x : \|x \| \le 1 \} = \sup \{z^T x : \|x \|= 1 \}$.


3

In general: continuity can be defined only for functions between topological spaces. It seems that you refer to a definition of continuity in terms of limits, and this is usually done in a metric space (that is also a topological space) (see here) . A vector space of matrices as, for example, $M(n, \mathbb{R})$ (that is a vector space) becomes a metric ...


3

Well you need a notion of norm in order to speak of convergence,i.e. $$A_jB_j\longrightarrow AB\mathrm{\ \ iff\ \ }\forall\epsilon>0,\exists N>0\mathrm{\ s.t.\ }j>N\Rightarrow\Vert AB-A_jB_j\Vert<\epsilon$$ Anyway here you don't need to go back to the definition to prove that the product is continuous. You basically need properties which are ...


3

So this is my final answer for any interested party answer and thank you @joy $\|\frac{x+y}{2}\|^2 \leq \|\frac{x+y}{2}\|^2 + \|\frac{x-y}{2}\|^2 = \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2$ and since $x \neq y$: $\|\frac{x+y}{2}\|^2 < \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2=\frac{1}{2}r^2+\frac{1}{2}r^2=r^2$ Hence: $\|\frac{x+y}{2}\| < r$


2

In the paragraph above Lemma 2.6, the authors explicitly state "[we write] $\mathcal L^\infty$ for the space of measurable bounded functions"


2

I assume you want this $\forall K>0$, not all $t$. Since $f(t)^Tf(t)\geq0$, you can take $\beta=(\int_0^{\infty}f(t)^Tf(t)\,dt)^{1/2}.$


2

Suppose $x=0$. Then, it is easy to show that $X=\{0\}$. Therefore $\mathrm{dim}X=0$. Suppose $x\neq 0$. Assume WLG that $\|x\|=1$. Then, there is at least one vector $u\in X$, $u\neq x$. Take $u^\prime=u-\langle u,x\rangle x$. It is easy to show that $0=\langle u^\prime,x\rangle$. Thus, $u^\prime=0$. Therefore $u-\langle u,x\rangle x=0$. That is ...


2

Sure, for example, take $ A $ to be a rotation matrix, in other words any matrix of the form $$ A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} $$ We can check directly that the norm is conserved. Let $ x = (a, b)^T $, then we have $$ Ax = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & ...


2

The first matrix norm is called the Frobenius norm, it's natural as in that it's the default Euclidean norm if the matrix were interpreted as a vector in $\mathbb{R}^{m \times n}$. The second matrix norm for $A \in \mathbb{R}^{m\times n}$ is the operator norm given by the linear operator $x \mapsto Ax$, it is naturally induced by the norms you choose for ...


2

Hint: Show that if $E$ is a vector space finite dimensional then all norms in $E$ are equivalents, i.e., if $\|\cdot\|_1$ and $\|\cdot\|_2$ are norms in $E$ then there are $c_1,c_2 >0$ constants such that $$c_1 \|x\|_1\le\|x\|_2\le c_2\|x\|_1$$ for every $x\in E.$


2

This is not possible For simplicity, just take $r=2$ and the nonzero singular values of $C$ are $1,1$. So that the F norm of C is $\sqrt2$. Now you may put the nonzero singular values of $A$ as $1/N, N$ and those of $B$ as $N, 1/N$ aligned accordingly Then $A,B$ have the same F-norm $\sqrt{N^2+1/N^2}$, but $N$ can be as big as you like


2

You are right. And it has to be that way, because the norm that the Hermitian product induces on $\mathbb C^n$ is the same as the Euclidean norm on $\mathbb R^{2n}$. Therefore, for a fixed $x\in\mathbb C^n$, the set $$ \{ y\in\mathbb C^n \mid \|x+y\|^2 = \|x\|^2+\|y\|^2 \} $$ has dimension $2n-1$ over $\mathbb R$ -- and therefore it cannot be a linear ...


1

In fact, we can show more - that every matrix $A$ which preserves the lengths of vectors in $\mathbb{R}^2$ is a rotation or a reflection.


1

Functional analysis has a lot to do with spaces of functions, as its name suggests (more so historically, but it still does). Using single bars for the norm of a function is ambiguous because $|f|$ also means the function $|f|(t) = |f(t)|$. Double bars $\|f\|$ eliminate the ambiguity. By the way, in some function spaces it is important to observe that ...


1

In a finite dimensional real (or complex) vector space all norms are equivalent. That means (following your notation) there are $c,d\in \mathbb R$ such that $$c||x||\leq ||x||_2\leq d||x||$$ Now the properties you want to show follows from this inequality.


1

First, notice that if we have two vectors $$ v = \begin{pmatrix} v_1 \\ v_2 \\v_3 \end{pmatrix} \ \ w = \begin{pmatrix} w_1 \\ w_2 \\w_3 \end{pmatrix} $$ we can rewrite the dot product $$\langle v, w\rangle = v^H \times w = \begin {pmatrix} v_1^* & v_2^* & v_3^*\end{pmatrix} \times \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix}$$ Now, we wish ...


1

The statement is true. More generally, for every $\alpha\in (0,1)$ the metric space $(\mathbb{R}^n, \|x-y\|^\alpha)$ can be isometrically embedded into a Hilbert space (you only need $\alpha=1/2$ here). This is Corollary 4.8 in the book Embeddings and extensions in analysis by Wells and Williams, who attribute this result to John von Neumann. The ...


1

First at all $c$ is not equal to $d$. If $x_n\to x$ under $||\cdot||_1$ then for all $\epsilon>0$ there is a $N$ such that for all $n>N$ we have $||x_n-x||<\epsilon$. Now from $$c\|x_n-x\|_2 \leq \|x_n-x ||_1 \leq d\|x_n-x\|_2$$ it is clear that $$\|x_n-x\|_2 \le \frac{\epsilon}{c}$$ One can deduce that $x_n\to x$ under $\|\cdot\|_2$ and ...


1

The first inequality is clear by the definition of the matrix norm. The second is clear by the definition and $\frac{|ABx|}{|x|}=\frac{|ABx|}{|Bx|}\frac{|Bx|}{|x|}$.


1

The first inequality is immediate, since $$ \|A\| = \sup_{x \in \mathbb{R}^n \setminus \{0\}} \frac{|Ax|}{|x|}\ge \frac{|Ax|}{|x|}. $$ For the second note that for $x\ne0$ with $Bx\ne0$, we have $$ \|AB\| = \sup_{x\ne0} \frac{|ABx|}{|x|}=\sup_{Bx\ne0}\left( ...


1

Pick $a_k = \frac{1}{2^k}$. Then $$ b_n = \frac{1}{n}\sum_{k=1}^n a_k = \frac{1}{n}\sum_{k=1}^n \frac{1}{2^k} = \frac{1}{n}\bigg(1 - \frac{1}{2^{n+1}}\bigg) > \frac{1}{2n}, $$ so $\sum b_n$ diverges


1

Just note that $\|f\|_K^2=\int_0^K f(t)^Tf(t) \, dt\leq \int_0^\infty f(t)^Tf(t) \, dt$ for all $K$, since $f(t)^Tf(t)$ is always nonnegative.



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