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0

If one of the members of the product was NOT invertible,it would map a sequence of vectors, that doesn't converge to zero, to a sequence converging to zero.But then the product would also do that, making the product non-invertible.


1

No. It is not true. Let $A$ be the unit ball centered in the origin, let $D$ be the points with rational coordinates. Let $U$ be the complement of $A$.


0

It is only true that $U\cap D$ and $A\cap D$ are non-empty. $A\cap U$ can happen to be the empty set. Why $U\cap D$ is nonempty: Because $U$ is open, then for each $u\in U$ there is a ball $B(u,r)$ with center $u$ and some positive radius $r>0$, such that $B(u,r)\subset U$. Now, because $D$ is dense in $\mathbb R^n$ it follows that for each ...


0

Take $V=C[0,1]$ with the uniform norm, $A$ is the subspace of all polynomials (convex), $D=C[0,1]\setminus A$ (dense). But $A\cap D=\emptyset$.


0

First, $U\cap D\cap V=U\cap D$ because $U\subset V$. Second, any non-empty open subset of $V$ intersects $D$, precisely because $D$ is dense in $V$ (this is a definition of dense subset). So yes, $U\cap V\cap D$ is non-empty. Note that we have only used the topology of $V$. There's no need of linearity, convexity, norm, etc.


0

Yes. It suffices to consider the case $n=2$ and here to show that there exists a path from $(0,0)$ to $(1,0)$ (assuming wlog. that these points are in the set). The obvious attempt is the curve $\gamma\colon [0,1]\to\mathbb R^2$, $t\mapsto (t,0)$, but for each disk it intersects we need to take a detour along the boundary of the disk. So try this $t\mapsto ...


4

Let $Q=S-T$. By assumption, $\langle Qv,v \rangle =0$ for all $v=\alpha x + \beta y$. Now, $$ 0=\langle Q(\alpha x+y),\alpha x+y \rangle = |\alpha|^2 \langle Qx,x \rangle + \langle Qy,y \rangle + \alpha \langle Qx,y \rangle + \bar{\alpha} \langle Qy,x \rangle \\ = \alpha \langle Qx,y \rangle + \bar{\alpha} \langle Qy,x \rangle. $$ Choosing first $\alpha =1$ ...


1

By considering $S-T$, it suffices to show that if $(Sz,z)=0$ for all $z$, then $S=0$. To show that $S=0$, it suffices to show that $(Sx,y)=0$ for all $x,y\in X$. Now think about what $(Sz,z)=0$ tells you when $z=ax+by$ is a linear combination of $x$ and $y$. By varying $a$ and $b$, can you show that $(Sx,y)$ must vanish?


0

The vector space of all polynomials of degree at most $n$ is isomorphic with $\mathbb{R^{n+1}}$ and is complete for every natural number.


1

As pointed out in a comment, the question doesn't quite make sense because $f\in BMO$ is only defined almost everywhere. But it seems to me that in fact there does exist a continuous $f\in BMO(\mathbb R)$ with small norm mapping $\mathbb R$ onto $S$. I'm going to write $||f||_*$ for the BMO seminorm. First a reduction to something a little simpler: Lemma: ...


0

The first axiom for a norm space is N1: $\left\Vert \mathbf{x}\right\Vert =0\implies \mathbf{x}=\mathbf{0}$. Now, if we take this away and have the other two viz. homogenity and the triangle inequality, then we're left with what is called a seminorm. Any vector space over $\mathbb{R}$ or $\mathbb{C}$ can be converted into a seminorm space as follows: take ...


0

We have by definition $$ \mathbf{x} = \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots\\ x_n \end{array} \right] $$ and $$ \mathbf{x}^T = \big[ x_1, x_2, \cdots, x_n \big] $$ so we get $$ \mathbf{x}^T \mathbf{x} = \big[ x_1, x_2, \cdots, x_n \big] \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots\\ x_n \end{array} \right] = x_1^2 + x_2^2 + \cdots + ...


0

The question does not make completely sense as it is now, but I think this should help. \begin{align} \sum\limits_{i=1}^n (x_i-c_i)^2 &= \sum\limits_{i=1}^n (x_i^2 + c_i^2 -2x_i c_i)\\ &=\sum\limits_{i=1}^n x_i^2 +\sum\limits_{i=1}^n c_i^2 - 2\sum\limits_{i=1}^n x_i c_i \\ &=\mathbf{x}^T \mathbf{x} + \mathbf{c}^T\mathbf{c} - ...


0

The left hand side equals to $$\rho^2\|x\|_2^2 + (1 - \rho)^2\|y\|_2^2 + 2\rho(1 - \rho)x \cdot y.$$ While the right hand side equals to \begin{align*} & \rho\|x\|_2^2 + (1 - \rho)\|y\|_2^2 - \rho(1 - \rho)(\|y\|_2^2 + \|x\|_2^2 - 2y\cdot x) \\ = &[\rho - \rho(1 - \rho)]\|x\|_2^2 + [(1 - \rho) - \rho(1 - \rho)]\|y\|_2^2 + 2\rho(1 - \rho)x \cdot y ...


1

If an inverse of any kind exist, $T$ is a bijection. As a consequence of the open mapping theorem, a bijective operator is bounded from below, meaning that there is $c>0$ such that $\|Tx\|\ge c\|x\|$ for all $x$. This and the property $TS=I$ imply that $S$ is bounded.


0

All norms on a finite dimensional vector space are equivalent so the norm on $(\Bbb{R}^n)^{\ast}$ is BOUNDED BY (thanks Ian) a constant times any other norm on $(\Bbb{R}^n)^{\ast}$. Practically there's little need to know exactly what the constant is, but the dual norm of any $l^p$ norm is the $q$ norm where $\frac1p+\frac1q=1$. Practically though, it's ...


1

I think the Theorem works better if we write $L=\{Tx:\ \|x\|<1\}$. Note that $B_x(\delta)=x+B_0(\delta)=x+\delta\,B_0(1)$. So $$ TB_x(\delta)=Tx+\delta\,TB_0(1)= y+\delta L. $$ By the Theorem, there exists $r>0$ with $B_0(r)\subset L$. Then $$B_y(\delta r)=y+\delta B_0(r)\subset y+\delta L=TB_x(\delta).$$ So $V$ is open.


2

Actually, for any locally compact Hausdorff $X$, and Radon measure $\mu$ on $X$, the set of all continuous compactly supported function $C_c(X)$ is dense in $L^p(X)$ for all $1\leq p<\infty$ (theorem 3.14 of "Real and complex analysis" by Walter Rudin). Hence $C_c(X)\subset L^1(X) \cap L^p(X)$, and so $L^1(X) \cap L^p(X)$ is dense in $L^p(X)$, for all ...


2

It's not. Consider $f_{n}(x)=\frac{1}{x^{1+1/n}}\in L^1([1,\infty))\cap L^2([1,\infty))\subset L^2([1,\infty))$. Then: $$\|f_n(x)-\frac1x\|_2=\int_1^{\infty}\left(\frac{1}{x^{1+1/n}}-\frac{1}{x}\right)^2dx=\frac{1}{2/n+1}-\frac{2}{1/n+1}+1\to 0$$ as $n\to\infty$. However $\frac1x\not\in L^1([1,\infty))$.


3

First, notice that $H$ is linear, so you only need to prove linearity at 0. And you have this inequality : $$| H(f) | = |f(1) - f(0) | = \left| \int_0^1 f'(t) dt \right| \leq \int_0^1 | f'(t) | dt \leq \| f\|$$ So clearly, $H(f) \to 0 $ when $f \to 0$ : $H$ is continuous at $0$, and by linearity, everywhere


1

Thr natural thing is to prove the contrapositive. If $T $ is not bounded, there exists a sequence $\{x_n\}_X $ with $\|x_n\|=1$ and $\|Tx_n\|>n^2$. Then $x_n/n $ is a sequence that converges to zero with its image through $T $ unbounded. Conversely, if $x_n\to0$ with $\{Tx_n\} $ unbounded, then $T $ is unbounded.


2

The separable case still requires some form of the Axiom of Choice, although it's less clear why. Say $(x_n)$ is a dense sequence of elements of $X$. Say $Z_n$ is the span of $Z$ and $x_1,\dots,x_n$. Now you simply extend your functional to $Z_1,$ then to $Z_2$, etc. You find you've extended it to the union of the $Z_n$, which is dense in $X$, and now ...


1

We have $=$ if and only if $x$ and each row of $A$ are linearly dependent. This follows directly from Cauchy Schwarz inequality. Proof: Let $a_i^T$ denote the $i$-th row of $A$. Then, we have $$ \| Ax \|^2 = \sum_{i=1}^m (a_i^T x)^2 \le \sum_{i=1}^m \|a_i\|^2 \|x\|^2 = \|A\|_{\mathrm F}^2 \| x \|^2, $$ where the inequality follows from Cauchy Schwarz. Now, ...


1

No it doesn't hold in $L^1$. Take $f(x)=g(x)=\frac{1}{\sqrt{x}}$ for $x \in (0,1)$ and $f(x)=g(x)=0$ elsewhere. $\Vert f \Vert_1=\Vert g \Vert_1=2$ but $\int fg =+\infty$.


2

Let me show you another way of proving that $(\ell^{\infty}, d_{\infty})$ is not separable. The proof is specific to $\ell^{\infty}$ but it is a cute trick, nonetheless. Assume, to the contrary, that there is a countable dense set $S$ in $\ell^{\infty}$. Enumerate $S$ into a list, i.e. $S=\{x_1, x_2, x_3, …\}$. Since $x_i\in\ell^{\infty}$, we can write ...


2

If $X$ is separable, then all uncountable subsets of $X$ contain a point of accumulation. As a result, $A$ contains a point of accumulation, say $x_0$. Consider $B_d[x_0, \epsilon]$. Then, $\left( B_d[x_0, \epsilon] \setminus \{x_0\}\right) \cap A \neq \varnothing$ (since $x_0$ is a point of accumulation). Hence, there's $y_0 \in A$, $y_0 \neq x_0$, such ...


1

Hint: if $D$ is dense in $X$, given $x \in A$, we must have $D \cap B(x, \epsilon/2) \neq \varnothing$. Take a point there and call it $f(x)$. We have a map $f: A \to D$, then. Prove that $f$ is injective, and conclude that $D$ is not countable. Since $D$ was an arbitrary dense set, $X$ is not separable.


0

If $T=0$, then the adjoint $T^\times$ is also equal to zero and the inequality is trivial. If $T \neq 0$, there exists $x_0$ with $Tx_0 \neq 0$. Theorem 4.3.3 is then applied to $Tx_0$.


1

For the first part, see this question: Assume $T$ is compact operator and $S(I- T) = I $.Is this true that $(I- T)S =I$? For the second, let $A = I - (I-T)^{-1}$, then $A(I-T) = -T$ and so $A = -(I-T)^{-1}T$. Since $T$ is compact, so is this product.


0

Note that: $$I-(I-T)^{-1} = I-(I-T+T)(I-T)^{-1} = I-I-T(I-T)^{-1} = -T (I-T)^{-1}$$ A compact operator composed (on the right or on the left) with a bounded operator is still compact. But $-T$ is compact, and $(I-T)^{-1}$ is bounded...


1

Here are a few things that fail: Distributive law: In general, $(\alpha + \beta)A \ne \alpha A + \beta A$. Examples: $A=\mathbb N$, $\alpha=1$, $\beta=1$: $(\alpha + \beta)A = 2\mathbb N = \{0,2,4,6,8,\ldots\}$ $\alpha A + \beta A = \mathbb N + \mathbb N = \mathbb N$ $A=\mathbb N$, $\alpha=1$, $\beta=-1$ $(\alpha + \beta)A = 0\mathbb N = \{0\}$ ...


2

First, you can forget $S,T,U$, and $V$. Say $A$ is that $2\times 2$ matrix with operator entries. Suppose you could prove $$||\alpha F||\le||\alpha||\,||F||\quad(i)$$for all $F\in L^p\oplus L^p$. Then it would follow that $$||(\alpha A)F||=||\alpha(AF)||\le ||\alpha||\,||AF||\le||\alpha||\,||A||\,||F||,$$which is exactly what you want. So you only need to ...


0

If you use the operatornorm with respect to the Euclidean norm on $\mathbb{C}^m$, you will end up with isoemtrically isomorphic spaces: Both $M_{kn}(\mathbb{C})$ and $M_n(M_k(\mathbb{C}))$ are obviously complete in this norm and the norms satisfy the $C^\ast$-property: $\|x^\ast x\|=\|x\|^2$ in both cases. But such a norm is uniquely determined.


0

Let $f(x)=y$ where $y(t) = x\left( \dfrac{t-a}{b-a} \right)\cdot\dfrac 1 {\sqrt{b-a}}$ for $a\le t\le b$. Then $$ \|y\|_2^2=\int_a^b (y(t))^2\,dt=\int_a^b \left(x\left( \frac{t-a}{b-a} \right)\right)^2 \frac{dt}{b-a} = \int_0^1 (x(u))^2\,du = \| x \|_2^2 $$ Let $g(x)=z$ where $z(t)= x\left( \dfrac{t-a}{b-a} \right)$ Then $$ \|z\|_\infty = \sup_{a\le t\le ...


3

Consider a function $g\in C[0,1]$. Let $f[g]\in C[a,b]$ be defined by $f[g](x)=\frac{1}{\sqrt{b-a}}g(\frac{x-a}{b-a})$. You should be able to prove that's an isometry just by u-substitution! The basic idea is, you translate/stretch the function so that it covers the new interval, then renormalize it to give it the same norm as before. You can apply the same ...


1

You have that $$d(x,z)^2 = \langle x-z, x-z \rangle = \langle x-y+y-z, x-y+y-z \rangle$$ $$= \langle x-y, x-y \rangle + 2 \langle x-y, y-z \rangle + \langle y-z, y-z \rangle$$ $$ = d(x,y)^2 + 2 \langle x-y, y-z \rangle + d(y,z)^2$$ Then by Cauchy-Schwarz, $$ \leq d(x,y)^2 + 2 d(x,y)d(y,z) + d(y,z)^2$$ $$ \leq \left( d(x,y) + d(y,z) \right)^2$$


3

Hint. For a) Develop $P(x_1,\dots, x_n)$ for the first variable knowing that $$x_1=\sum_{i_1=1}^{n_1} x_1^{i_1} e_{i_1}$$ and using linearity of $P$ for the first variable. You get $$\vert P(x_1,\dots, x_n) \vert= \left\vert P(\sum_{i_1=1}^{n_1} x_1^{i_1} e_{i_1},x_2, \dots,x_n) \right\vert =\left\vert \sum_{i_1=1}^{n_1} x_1^{i_1} P(e_{i_1}, x_2, \dots, ...


2

$\def\norm#1{\left\|#1\right\|_1}\def\abs#1{\left|#1\right|}$As you write correctly, we have $$ \norm{Ax} = \norm{\sum_{i=1}^n x^i Ae_i} \le \sum_{i=1}^n \abs{x^i} \norm{Ae_i} $$ Now, note that for every $i$, we have $$ \norm{Ae_i} \le \sup_{1\le j \le n} \norm{Ae_j} $$ Let's call the supremum $S$, then $\norm{Ae_i} \le S$ for all $i$, giving above $$ ...


1

For example, consider the norm $$ \|(x,y)\| = x ^2- xy + y^2 $$ We note that $$ \|(2,0)\| > \|(2,1)\| $$ A class of norms that act the way you might expect is the set of "symmetric gauge functions", as referenced here.


0

Let $\phi:V\to\mathbb{K}$ (where $\mathbb{K}$ is the base field) be a discontinuous linear functional$^{(1)}$. Let $x\in W\setminus 0$. Define $T:V\to W$ by $Tv=\phi(v)x$. Let's show that $T$ is discontinuous: Since $\phi$ is discontinuous, it is unbounded, so there exists a bounded sequence of vectors $\{v_n\}$ in $V$ such that $\{\phi(v_n)\}$ is ...


0

The function $\phi .f(x)$ is a continuous function, as a product of continuous functions. Since it is defined on a compact set, it takes on a maximum $M$, then $T_{\phi} \leq \int_0^1 Mdx =M(1-0)=M$ Then $T_{\phi}$ is a bounded linear operator, and so it is continuous. EDIT: As pointed out in the comments, to make this more rigorous ( I took shortcuts ) we ...


0

Hard to beat the simple solution of @Keith: We'll give a proof for $X$ infinite dimensional Banach space (extra condition). First, show that there exists a sequence $x_n$ in $X$ such that $||x_n|| =1$ and $d(x_n, \langle x_1, \ldots x_{n-1}\rangle ) \ge \frac{1}{2}$. One constructs the sequence inductively. Once $x_1$, $\ldots x_{n-1}$ are obtained, take ...


1

Another way, first we show linearity: $K(f+g)=\int_0^1 k(x,y)f(x,y)dy+ \int_0^1 k(x,y)g(x,y)dy=\int_0^1 k(x,y)(f(x,y)+g(x,y))dy $ , by linearity of the integral. Then we use 1st countability of $C[0,1]$ (since it is a metric space), and show sequential continuity, which is equivalent to continuity: Assume $f_n \rightarrow f$ in $C[0,1]$ , so that $Sup ...


3

let $f\in C[0,1]$ and $M=\max_{[0,1]\times [0,1]} k$. Then $$\left|\int_0^1 k(x,y) f(y) dy\right| \le M \int_0^1 |f|\le M |f|_{C[0,1]}$$ Now take $\sup$ over $x\in [0,1]$. (Note that the map in question is obviously linear)


0

First, since $V$ is finite-dimensional, the image of all transformations in $L(V, W)$ have finite rank, and are therefore bounded, so we do have the operator norm here. Otherwise I'd insist that we use $B(V, W)$ instead. Additionally, we may assume without loss of generality here that $V$ is $\mathbb{R}^n$ with the $1$-norm, for some $n$, since every two ...


6

Every infinite-dimensional normed space has a non-closed subspace. Let $X$ be an infinite-dimensional normed space, let $a$ be a nonzero vector. Assume by induction that we have found vectors $x_1, x_2, \dots, x_{n-1}$ for which $|x_i - a| < 1/i$ and $a \not\in V_{n-1} = \Sigma_{i=1}^{n-1} \mathbf{R}x_i$. We will extend this sequence by finding an $n$th ...


0

Classically, the approach is to show that $T(B_V)$, where $B_V$ is the unit ball of $V$, is bounded in $W$. That is, $T(B_V) \subseteq MB_W$ for some $M$. Then, if we have any bounded set $A$, that is $A \subseteq NB_V$, then, $$T(A) \subseteq T(NB_V) = NT(B_V) \subseteq NM B_W,$$ i.e. $T(A)$ is bounded, which is what we want to prove. So, by continuity at ...


1

Let $r>$ be such that: $ A \subset B_r(0) $. With $B_r(0)$ the ball of radius $r$ centered in $0$. Suppose $T(A)$ is not bounded. Then for each $n \in N$ there exists a $x_n \in A$ such that: $||Tx_n|| > n+1 $. The sequence $\frac{x_n}{n+1}$ converges to $0$ because $A$ is bounded. But clearly $|| T \left( \frac{x_n}{n+1} \right)|| >1$. This ...



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