# Tag Info

6

Yes, $$C([0,1]) = \bigcup_{n = 1}^\infty \underbrace{\{ f \in C([0,1]) : \lVert f\rVert_\infty \leqslant n\}}_{A_n},$$ and $A_n$ is closed for each $n$ - if $\lVert g\rVert_\infty > n$, then there is a $\delta > 0$ and a non-degenerate interval $[a,b] \subset [0,1]$ such that $\lvert g(x)\rvert \geqslant n+\delta$ for all $x\in [a,b]$, and hence ...

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 ...

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 $\alpha =1$ ...

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

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)

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 $$\vert P(x_1,\dots, x_n) \vert= \left\vert P(\sum_{i_1=1}^{n_1} x_1^{i_1} e_{i_1},x_2, \dots,x_n) \right\vert =\left\vert \sum_{i_1=1}^{n_1} x_1^{i_1} P(e_{i_1}, x_2, \dots, ... 3 Consider a function g\in C[0,1]. Let f[g]\in C[a,b] be defined by f[g](x)=\frac{1}{\sqrt{b-a}}g(\frac{x-a}{b-a}). You should be able to prove that's an isometry just by u-substitution! The basic idea is, you translate/stretch the function so that it covers the new interval, then renormalize it to give it the same norm as before. You can apply the same ... 2 Actually, for any locally compact Hausdorff X, and Radon measure \mu on X, the set of all continuous compactly supported function C_c(X) is dense in L^p(X) for all 1\leq p<\infty (theorem 3.14 of "Real and complex analysis" by Walter Rudin). Hence C_c(X)\subset L^1(X) \cap L^p(X), and so L^1(X) \cap L^p(X) is dense in L^p(X), for all ... 2 It's not. Consider f_{n}(x)=\frac{1}{x^{1+1/n}}\in L^1([1,\infty))\cap L^2([1,\infty))\subset L^2([1,\infty)). Then:$$\|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$$as n\to\infty. However \frac1x\not\in L^1([1,\infty)). 2 \def\norm#1{\left\|#1\right\|_1}\def\abs#1{\left|#1\right|}As you write correctly, we have$$ \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} $$Let's call the supremum S, then \norm{Ae_i} \le S for all i, giving above$$ ...

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 ...

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 ...

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 ...

2

a topological vector space that is not a metric space: take $V=C(\Bbb{R})$ where the topology is given by convergence on compact sets. A basis for this topology is given by sets of the form $$U_{K,f,\varepsilon} = \{ g : \sup_K |g-f| < \varepsilon \}$$ where $f \in V$ is continuous, $\varepsilon >0$ is a positive real number, $K \subset \Bbb{R}$ is a ...

1

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

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.

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 $\int fg =+\infty$.

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?

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

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.

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.

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 $A$.

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.

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: ...

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 $Sup ... 1 Take$V = \mathbb R^2$,$A$a line segment,$D = V \backslash A$. 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. 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. 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 ...

Only top voted, non community-wiki answers of a minimum length are eligible