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6

Let $\{a_n\}$ be a sequence of positive real numbers that increases to $1$, with the property that the sequence of products $$ a_1,\ a_1a_2,\ a_1a_2a_3,\ a_1a_2a_3a_4,\ \ldots$$ converges to a positive value. It's not hard to write down a specific example. Let $\cal H = \ell^2(\mathbf R)$ and define $T : \cal H \to \cal H$ by $$T(x_1,x_2,x_3,\ldots) = (0, ...


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

Note, that $\|\cdot\|$ is somehow the composition of the $2$-norm and the $3$-norm on $\mathbb R^2$. We have $$\|x\| = \|(\|(x_1,x_2)\|_2,x_3)\|_3$$ and with this at hand the proof is easy (but a bit ugly). We have by the triangle inequality for the $2$-norm $$a:= \|(x_1+y_1,x_2+y_2)\|_2 \leq \|(x_1,x_2)\|_2 + \|(y_1,y_2)\|_2 =:b$$ and since for all ...


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 $$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$


3

When you say $K$ is continuous I assume it is with respect the norm on $C([0,1])$, that is, $|f\|_\infty$. The two norms are not comparable. Let $K$ be the identity. Then $$\sup_{\|f\|_\infty\leq 1} \|Kf\|_\infty=1$$ but $$\sup_{\|f\|_2\leq 1} \|Kf\|_\infty=\infty.$$ To see it consider a sequence $\{f_n\}$ of functions such that $f_n$ is supported on ...


3

$y=(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...) \notin c_{00}$, $y_1=(1,0,0,...)$, $y_2=(1,\frac{1}{2},0,0,...)$ and so on. $y_n \in c_{00}$ and $\|y_n -y \|_{\infty}=\frac{1}{n+1}$ which tends to $0$ when $n$ tends to $\infty$.


3

Hint: For the upper bound, observe that $$ \left(\sum_{i=1}^n |v_i|^p\right)^{1/p}\leq\left(\sum_{i=1}^n \max|v_i|^p\right)^{1/p}=n^{1/p}\max|v_i|. $$ For the lower bound, observe that $$ \left(\sum_{i=1}^n |v_i|^p\right)^{1/p}\geq\left( \max|v_i|^p\right)^{1/p}=\max|v_i|. $$ Now, take limits.


3

The $\ell^p$ spaces are a special case of the $L^p$ spaces obtained by using the counting measure on the set of natural numbers. If you squint closely at the integral it looks like a sum or indeed as Forever Mozart points out: summation is just integration with the trivial measure on $\mathbb{N}$.


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

Yes, the inequalities hold. Let $B$ be the closed unit ball for the norm $\|\cdot \|$. The assumptions imply that $\pm e_j$, the standard basis vectors, are in $B$. Hence, their convex hull is contained in $B$. This convex hull is the unit ball for the $\ell^1$-norm, which implies $\|\cdot \|\le \|\cdot \|_1$. Similarly, we need to prove that $B$ is ...


2

Plugging in $e_j$ (the $j^{\text{th}}$ standard basis vector) which has norm 1, we get $$\| A \|_1 = \max_{\|x \|_1 =1} \|Ax\|_1 \ge \| A e_j \|_1 = \sum^n_{i=1} \lvert A_{ij} \rvert.$$ Since this holds for all $j$, we have $$\|A\|_1 \ge \max_{1\le j \le n} \sum^n_{i=1} \lvert A_{ij} \rvert.$$ Conversely, for any $x \in \mathbb R^n$ with $\|x\|_1 = 1$, we ...


2

The homogenous solutions constitute a subspace. Choose $x_p$ to be in its orthogonal complement (this is always possible). Then the cross terms on the right side of your equation vanish.


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

No. You have to look at a set of arbritray matrices in a neighbourhood of a given $P$ and show that any matrix in that set is orthogonal. You won't succeed with this, though, since that set is actually not open but compact. (It's bounded and closed as the counterimage of a point under a continuous map) (Edit: Actually it is a closed smooth submanifold of ...


2

You might note that $ac-bd$ and $ad+bc$ are the real and imaginary parts of $(a+ib)(c+id)$, so the formula says $|zw|^2 = |z|^2 |w|^2$ where $z= a+ib$ and $w=c+id$. For the second part, it might help that $25988 = 2^2 \times 73 \times 89$. Can you write $2$, $73$, $89$ each as the sum of two squares?


2

The prime factorization of $25988$ is $2^2 \cdot 73 \cdot 89$. Observe that $73 = 3^2 + 8^2$ and $89 = 5^2 + 8^2$. Using the norm formula, we have \begin{equation*} \begin{aligned} 25988 &= 2^2(3^2 + 8^2)(5^2 + 8^2) \\ &= 2^2((3 \cdot 5 - 8 \cdot 8)^2 + (3 \cdot 8 + 8 \cdot 5)^2) \\ &= 2^2(49^2 + 64^2) \\ &= 98^2 + 128^2, \end{aligned} ...


2

1) If $|f| \ge |g|$ then $$(|f| + |g|)^p \le (2|f|)^p = 2^p |f|^p \le 2^p (|f|^p + |g|^p).$$ You can derive the same inequality if $|g| \le |f|$. 2) Once the inequality in 1) is established you have $$\int |f+g|^p \le 2^p \int |f|^p + 2^p \int |g|^p < \infty.$$ Thus by definition $f+g \in L^p$. 3) Since $p' = \frac{p}{p-1}$ it follows that $$ ...


2

You correctly observed that $\|x\|_2\le \sqrt{\|x\|_1\|x\|_\infty}$ is a special (easy) case of Hölder's inequality, which essentially amounts to $|x_i|^2\le \|x\|_\infty |x_i|$. To prove the second part, first use the arithmetic-geometric means inequality. After that, use Hölder's inequality again: $$\|x\|_1\le \sqrt{n}\|x\|_2$$ and observe that ...


2

I think I found the solution. Continuing with my reasoning, applying the Cauchy formula, we have: $$1 = a_n = {1 \over 2\pi i}\oint_{|z| = 1} {P(t) \over t^{n+1}} dt$$ Therefore, we have: $$1 = |a_n| = |{1 \over 2\pi i}\oint_{|z| = 1} {P(t) \over t^{n+1}} dt| \le {1 \over 2\pi}\max_{|z| = 1}|P(z)| l(|z| = 1) = \max_{|z| = 1}|P(z)|$$ Here $l(\gamma)$ is the ...


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

Let's define for $a\in A$ the left multiplication operator and the right multiplication operator $$ L_a : A \rightarrow A, L_a(x)=ax \quad \text{and} \quad R_a: A \rightarrow A, R_a(x)=xa.$$ As multiplication is assumed to be continuous we get that $L_a$ and $R_a$ are both continuous for all $a\in A$. Let $B=B_1(0,A)$ denote the unit ball in $A$. Then ...


2

Consider $\left< , \right> :C^1 [a,b]\times C^1 [a,b] \to \mathbb{R} $ defined by $$\left< f, g\right> =\int_{[a,b]} f(u) g(u) du +\int_{[a,b]} f'(u) g'(u) du $$


1

Note that $$ \left(\frac{x}{e^x}\right)'=\frac{e^x(1-x)}{e^{2x}} $$ vanishes at $x=1$ (which is a maximum of the function). Hence, $$ p(f)=\frac1e\sup_{\|v\|=1}f(v)=\frac1e\|f\|^*. $$


1

The differential is given by, $$Df(x_1,x_2) = \left( \frac{x_1}{\|x\|}, \frac{x_2}{\|x\|} \right)$$ which has defined for $x \in \mathbb{R}^2 \setminus \{0\}$


1

Note that $\langle Tw, Tw\rangle=\langle w, w\rangle$. Taking $w=u-v$ gives $\langle T(u-v),T(u-v)\rangle=\langle(u-v),(u-v)\rangle\implies \left| u-v\right|^2=\left|Tu-Tv\right|^2$


1

Diagonalization will help here. (When in doubt and working with normal matrices, try utilizing diagonalization!) Write $T = UDU^*$, then $T^* = UD^* U^*$, giving that $T^*T^2 = UD^*D^2U^*$. However $T^3 = UD^3 U^*$. Since unitary conjugation does not change the operator norm, this boils down to considering $D^*D^2$ and $D^3$. Here $Dg(x) = f(x)g(x)$ for ...


1

Since $f$ is continuous (at $0$), there is a neighbourhood $U$ of $0$ such that $f(U)\subset(-1,1)$. Choose $\delta>0$ such that $\{x\in X|\|x\|\leq\delta\}\subseteq U$. Then, if $x\in X$ is such that $\|x\|\leq \delta$, we have $x\in U$, and hence, $|f(x)|\leq 1$. Since $\|\frac{\delta x}{\|x\|}\|=\delta$, it follows that for all $x\in X$ we have $$ ...


1

Normally we talk about normal matrices in a space with inner product, and we consider the norm induced by the inner product (unless explicitly said that another norm should be considered). So I will assume we are supposed to use the norm induced by the inner product. Your proof for the case where $A$ is normal is correct. The result is not valid is $A$ is ...


1

You can check this directly: $$(ac-bd)²+(ad+bc)²=(ac)²+(bd)²-2abcd+(ad)²+(bc)²+2abcd$$ $$=a²(c²+d²)+b²(c²+d²)=(a²+b²)(c²+d²).$$ $25988=2²\times 73\times 89=2²(8²+3²)(8²+5²)=2²[(8²-15)²+(40+24)²]=98²+128²$



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