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16

Let's call $f$ an $n$-SOP if we can write $$f(x,y) = \sum_{k = 1}^n g_k(x)\cdot h_k(y)$$ with continuous functions $g_k, h_k \colon [0,1] \to \mathbb{R}$. If $f$ is an $n$-SOP, for every family $x_1 < x_2 < \dotsc < x_r$ of $r > n$ points in $[0,1]$, the set $$\left\{ \begin{pmatrix} f(x_1,y) \\ f(x_2,y) \\ \vdots \\ f(x_r,y)\end{pmatrix} : y ...


12

Take your open cover of a ball of radius 5. Shrink it by a factor of 5. It's an open cover of the ball of radius 1. Therefore it has a finite subcover. Now reflate the subcover by a factor of 5. Et Voila! It covers the ball of radius 5.


12

Note that $C_c \subset C_0$, but $C_c \neq C_0$. For example, $f(x) = \dfrac{1}{x^2+1}$ belongs to $C_0$ but not $C_c$. What you seem to be assuming is that $\lim_{|x|\to\infty}f(x) = 0$ implies that there is some $N > 0$ with $f(x) = 0$ for all $|x| > N$. This is not true, as the above example demonstrates. That is, a function can limit to zero at ...


10

Generally, without completeness, you can't deduce that a weak$^\ast$ convergent sequence is bounded. Let $X = c_{00}$ be the space of sequences with only finitely many nonzero terms, endowed with the $\lVert\,\cdot\,\rVert_\infty$-norm. Its completion is $c_0$, the space of sequences converging to $0$, and its dual therefore isometrically isomorphic to ...


9

In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the "discrete metric" $d(x,y)=0$ if $x=y$ and $1$ otherwise, all balls $B_r(x)$ for $r \leq 1$ are the same (they are just $\{ x \}$) while all balls $B_r(x)$ for $r>1$ are also the same ...


9

Pick an enumeration of the rationals $r_1,r_2,r_3,...$. Let $d(f,g) = 0$ if $f = g$. Else let $r_n$ be the first rational in the enumeration for which $f(r_n) \neq g(r_n)$. Define $d(f,g) = f(r_n) - g(r_n)$. This is well defined since if $f \neq g$, there must be some rational number for which $f(r) \neq g(r)$, since $f$ and $g$ are continuous, and the ...


9

It is certainly not dense. The linear functional $$ T(x_n) = \sum_{n=0}^{\infty} \frac{x_n}{n} $$ Is continuous on $l^2$. Thus, the set $T^{-1}(0)$ must be a closed subspace of $l^2$. If it were dense, we'd have $l^2 = T^{-1}(0)$. But the sequence ${\frac{1}{n}}$ obviously isn't in this space.


8

When $d=1$ it is of course possible to embed large chunks of $S^d$ isometrically in ${\mathbb R}^n$. When $d\geq2$ it is not possible to embed even tiny pieces of $S^d$ isometrically in ${\mathbb R}^n$. Proof. Take any three points $x_1$, $x_2$, $x_3\in S^d$ forming a small equilateral triangle in the metric of $S^d$. This then is an "ordinary" spherical ...


8

Here's what I think the claim actually said (or at least, was meant to say): For $M \subset \Bbb X$ where $\Bbb X$ is a normed space, the following are equivalent: $M$ is closed For all sequences $(u_n)\subset M$, $u_n \to u$ as $n \to \infty$ implies that $u \in M$. This is quite different from what you've stated.


7

If something is non-invertible, there's two (non-disjoint) possibilities: it fails to be injective, or it fails to be surjective. In finite dimension, these are the same, but in infinite-dimensional spaces, weird things can happen. If it fails to be injective, there's $x \ne y$ such that $(T - \lambda I)(x) = (T - \lambda I)(y)$. So $(T - \lambda I)(x - y) ...


6

For finite-dimensional vector spaces, injectivity and surjectivity are equivalent. That's not the case for an arbitrary Hilbert space. The classic examples are the left- and right-shift operators $L, R:\ell^2 \to \ell^2$, given by \begin{align*} L(x_1, x_2, \dots) &= (x_2, \dots) \\ R(x_1, x_2, \dots) &= (0, x_1, x_2, \dots). \end{align*} The map $L$ ...


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

The inclusion $\mathscr{C}^1[a,b] \hookrightarrow \mathscr{C}^0[a,b]$ is continuous, so $F$ is also closed in $\mathscr{C}^1[a,b]$. Thus $F$ is a Banach space with respect to the $\mathscr{C}^0$-norm and with respect to the $\mathscr{C}^1$-norm. Hence these two norms are equivalent on $F$ (since they are comparable). Say we have $$\lVert ...


6

If $A$ is hermitean and its eigenvalues are $\{\lambda_1,\dots,\lambda_n\}$, then $\mathcal A$ is also hermitean and its eigenvalues are $\{-\lambda_1,+\lambda_1,\dots,-\lambda_n,+\lambda_n\}$. (This is true also allowing repeated eigenvalues; the multiplicities behave as you would guess.) If $v\in\mathbb C^n$ is an eigenvector of $A$ with eigenvalue ...


6

First note that the sequence is bounded: $$ |\phi_n(f)|=n^{-1}\,\left|\sum_1^nf(j)\right|\leq\max\{|f(j)|:\ j=1,\ldots,n\}\leq\|f\|_\infty. $$ This shows that $\|\phi_n\|\leq1$ for all $n$, so the sequence $\{\phi_n\}$ lies in the unit ball of $(\ell_\infty)^*$. In the weak$^*$-topology, the unit ball of the dual is compact, and so every sequence within it ...


6

In fact, continuity of only one of the two functions is sufficient. Suppose $\varphi$ is continuous. Pick $x_0\in\Bbb R$ and $\epsilon>0$. Let $S$ and $T$ be the periods of $\varphi$ and $\psi$ respectively. Take $\delta>0$ such that if $|x-x_0|<\delta$ then $|\varphi(x)-\varphi(x_0)|<\epsilon/2$ and let $N\in\Bbb N$ be such that if $x>N$, ...


5

No it is not possible, but your question is very thoughtful. Here's a fun counterexample: suppose two continuous functions $f$ and $g$ satisfy $f(1-x)=g(x)$ for all $x$, then $\int_0^1 f(x)dx=\int_0^1 g(x)dx$ (can you show this?) To understand intuitively why it is not possible to solve for $f$ given $a,b,c$, interpret the integral as the area under the ...


5

I'll assume real scalars for convenience. If $y = \int_D f\; dP \notin C$, then by the Separation Theorem there is $r \in \mathbb R$ and a continuous linear functional $\phi$ such that $\phi(y) > r$ while $\phi \le r$ on $C$. But $\phi(y) = \int_D \phi(f)\; dP \le \int_D r \; dP = r$, contradiction.


5

Let $$ f=g=h=\frac1{t^{1/3}}\,1_{(0,1]}^\vphantom{2}(t). $$ Then $$ \|f\|_2=\|g\|_2=\|h\|_2=\left(\int_0^1\frac1{t^{2/3}}\,dt\right)^{1/2}=\sqrt3, $$ and $$ \int_{\mathbb R}fgh=\int_0^1\frac1{t}\,dt=+\infty. $$


5

EDIT: The following works on any non-reflexive infinite-dimensional Banach space. Suppose $f$ is a continuous linear functional on $X$ such that $\|f\| = 1$ but $f(x) < 1$ for all $x \in B$, i.e. $f$ does not assume its supremum on $B$. Take $F(x) = x/(1-f(x))$ for $x \in B$, and $F(x) = F(x/\|x\|)$ on its complement. For example, on the space $c_0$ of ...


5

Strictly positive scaling is a homeomorphism on closed balls. So either none is compact or all.


5

We often find metrics on function spaces from norms, i.e., $d(f,g):=\|f-g\|$. As norms you can take for example $\|f\|_\infty :=\sup\{\,f(x)\mid x\in I\,\}$ if $I$ is a compact interval and we consider the space of continuous functions on $I$. Or $\|f\|_2:=\int |f(x)|^2\,\mathrm dx$ for square-integrabvle functions. And many more.


5

I believe that A. Ya. Helemskii, Lectures and Exercises on Functional Analysis, Translations of Mathematical Monographs , vol. 233 (2006). in conjunction with some standard book such as G. K. Pedersen, Analysis Now, Springer-Verlag, (1989). should suit you well. Note that Pedersen's book itself contains lots of basic information about Banach and ...


4

Let $x \in \mathcal{H}$ with $\lVert x\rVert = 1$. Show that $x$ cannot be in the orthogonal complement of $\operatorname{span} \{ f_n : n \in \mathbb{N}\}$. Since $\{ e_n : n \in \mathbb{N}\}$ is a Hilbert basis of $\mathcal{H}$, we can write $$x = \sum_{n = 0}^\infty c_n\cdot e_n.$$ Consider now $$y = \sum_{n = 0}^\infty c_n \cdot f_n.$$ Then we have ...


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 $\alpha =1$ ...


4

I don't think this is the place to ask that kind of question. But since you asked, if you're having difficulty with this book and you're having trouble finding motivation to read it, why don't you just leave it for a bit, and come back to it after you're done with your other books? Or perhaps it is a too complex book for your level? I don't really know...


4

You can justify the interchange of orders of integration by letting $$ G(x,y) = e^{-(x-y)^{2}/2} $$ and noticing that $G(x,y)f(x)\overline{g(y)}$ is jointly measurable in $x,y$, and is bounded by \begin{align} |G(x,y)f(x)\overline{g(y)}| & = |G(x,y)^{1/2}f(x)||G(x,y)^{1/2}\overline{g(y)}| \\ & \le ...


4

If $\delta_y$ denotes the Dirac delta "function" then no, those don't give a basis for $L^2$; they're not even elements of $L^2$. In any case, any two bases have the same cardinality. Except that there are different flavors of "basis" here. "Basis" could mean basis in the sense of straight linear algebra; any element is a unique (finite) linear combination ...


4

You can approximate the constant function with Gaussians $\phi_{\sigma^2}(x)=\frac{1}{\sqrt{2\pi \sigma^2}} \exp(-x^2/2\sigma^2)$. If I calculated correctly you have $\|\phi_{\sigma^2}\|^2= \dfrac{1}{2\sqrt{\pi \sigma^2}}$ and since $A\phi_{\sigma^2} = \phi_{\sigma^2+1}$ (the convolution of two Gaussians is Gaussian with the sum of the variances) you get ...


4

Yes, it is true that $\|A\|=1$. The operator $A$ is sometimes called a Fourier multiplier or simply a multiplier because it acts by pointwise multiplication of the Fourier transform. (Alternatively, multipliers are the same as convolution operators). So anyway, we can prove the following: Let $m\in L^\infty(\mathbb{R})$ be a bounded and measurable ...



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