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9

I'm not sure if this is 100% standard, but I've always understood this notation to mean: $f$ is a function from $\Omega$ to $\mathbb{R}$ which has compact support (the subscript $0$), and is smooth/infinitely differentiable (the superscript $\infty$). I just checked, for example, that Stein and Shakarchi use this notation in their Functional Analysis. I ...


8

Let's take $L^p$ space for example: how can it be in both real and functional analysis? If one uses "$L^p$" as notation for the set of functions that are Lebesgue integrable to the $p$th power, and proceeds to study the properties of individual functions in this set, this is Real Analysis. The mode of thinking here is not much different from studying ...


7

Let's go ahead and prove Eric Wofsey's claim (see his comment under the question), that if $H$ is an infinite-dimensional Hilbert space with an orthonormal basis $U$ of cardinality $\kappa$, then any Hamel basis $B$ of $H$ has cardinality $|B| = \kappa^{\aleph_0}$. First remark: $|B| = |H|$. Since $H$ is infinite-dimensional, we have $|B| \geq 2^{\aleph_0} ...


6

This is not true. Simply take $f = \chi_{[-1,1]}$ (the indicator function of the interval $[-1,1]$). Then $f \in L^p$ for all $p \in (0,\infty]$, but if $\widehat{f} \in L^1$ was true, then Fourier inversion would imply that $$ f = \mathcal{F}^{-1} \widehat{f} \in C_0 $$ would be (almost everywhere equal to) a continuous function. This is clearly not the ...


6

Make a change of variables $u = e^x$ $$\int_0^1 f(x) e^{nx}dx = \int_{1}^e g(u) u^{n}du = 0$$ where $g(u) = \frac{f(\log(u))}{u}$ is a continuous function. You can now apply Weierstrass approximation theorem. Since the above holds for all $n$ we have that $$\int_{1}^e g(u) P(u)du = 0$$ for any polynomial $P(u)$. Now pick the polynomial to approximate ...


5

Consider a convergent sequence $a_n\in X+Y$. Each $a_n=x_n+y_n$ for some $x_n\in X$ and $y_n\in Y$. Since $Y$ is compact there is a convergent subsequence $y_{n_k}\to y\in Y$. Therefore $x_{n_k}=a_{n_k}-y_{n_k}$ is convergent. Since $X$ is closed $x_{n_k}\to x$ for some $x\in X$. Therefore $a_{n_k}\to x+y\in X+Y$. Since $a_n$ is convergent, $a_n\to x+y\in ...


5

I will use your post to add the explanations inline. [I] Theorem: The space $\ell^p$ is complete; here $p$ is fixed and $1 \leq p < \infty$. Proof: Let $(x_n)$ be any Cauchy sequence in $\ell^p$, where $x_m = ( \xi^{(m)}_1, \xi^{(m)}_2, \dots )$. Then, for every $\varepsilon > 0 $ there is an $N$ such that for all $m, n >N$, ...


5

It seems that Nigel Kalton gave an answer to your question. He proved in [1], theorem 3.5, that for every compact metric space $K$, $C(K)$ is a $2$-absolute Lipschitz retract. [1]: Kalton, N. J. Extending Lipschitz maps into C(K)-spaces. Israel J. Math. 162 (2007), 275–315. Link to article Added by OP: For completeness, I'll outline the steps of ...


5

Response to questions 1 and 2: Take $A=B = L^2(\mathbb{Z})$ and $C=0$. Let $S((a_i)) = (a_i - a_{i-1})$ and $T$ be the zero map. Since $\sum_i (a_i- a_{i-1})^2 \leq \sum_i (2 a_i^2 + 2 a_{i-1}^2) = 4 \sum_i a_i^2$, the operator $S$ is bounded and hence closed. Note that any sequence $(b_i)$ in the image of $S$ obeys $\sum_{i=-\infty}^{\infty} b_i =0$, so ...


5

Here are some basic facts about strongly continuous semigroups that will be useful: Let $A$ be the infinitesimal generator of $S(\cdot)$. Then $D(A)$ (the domain of $A$) is dense and $A$ is closed. For any $x\in D(A)$ we have $S(t)Ax=\frac{d}{dt}S(t)x$. If two strongly continuous semigroups have the same infinitesimal generator, then in fact they are the ...


5

No. Let $T:\ell^2\to \ell^2$ be defined by $T(x_1,x_2,x_3,\ldots)=\left(x_1,\frac12x_2,\frac13x_3,\ldots\right)$. Then $T$ is injective and bounded, but its range is not closed. Let $X=\ell^2$ and $Y=T(\ell^2)$ to get a counterexample for your question with $T$ being a bounded bijection from a Banach space to a normed space. For your more general ...


5

Hint: If an operator can be approximated by finite rank operators, then it is compact. Try to show that $\|T-T_k\| \to 0$ for a suitable chosen sequence $\{ T_k\}$ of finite rank operators. Try with $$ T_k: (x_1, x_2, \ldots)\mapsto (x_1, \frac{x_2}{2}, \ldots, \frac{x_k}{k}, 0, 0, \ldots ). $$ It is obvious that $rk(T_k)=k$, i.e., it is finite rank ...


5

By using the Cauchy-Schwarz inequality, $(u\cdot v)^2 \leq ||u||^2 ||v||^2$, on the vectors $u=(1,1,\ldots,1)$ and $v=(x_1,x_2,\ldots,x_n)$ we get $$\left(\sum_{k=1}^n x_k\right)^2 \leq \sum_{k=1}^n x_k^2\sum_{k=1}^n 1 = n \sum_{k=1}^n x_k^2$$ Now applying the same inequality, but on the vectors $u=(1,1,\ldots,1)$ and $v=(x_1^2,x_2^2,\ldots,x_n^2)$ we get ...


4

This is true for all powers, even or odd. Proposition 1.11e on page 234 says that $\|a\|=r(a)$ for any Hermitian $a$, where $r(a) $ is the spectral radius of $a$, that is $r(a) = \lim_{k\to\infty} \|a^k\|^{1/k}$. Since $a^{n}$ is also Hermitian, we have $$ \|a^{n}\| =r(a^n) = \lim_{k\to\infty} \|a^{kn}\|^{1/k} = \left(\lim_{k\to\infty} ...


4

Well, I think that by "closure" your text-book means that addition and scalar multiplication are well-defined. Indeed, call $V$ your candidate vector space of all the sequences with finite support (i.e., with a finite number of non-zero terms), then you can think to $V$ as a subset of the cartesian product $\mathbb K^{\mathbb N}$ (the space of all sequences ...


4

It is useful to recall the following lemma: If $f$ is a twice differentiable function over $I=[a,b]$ and $$M_0=\sup_{x\in I}|f(x)|,\quad M_1=\sup_{x\in I}|f'(x)|,\quad M_2=\sup_{x\in I} |f''(x)|,$$ then $M_1^2\leq 4M_0 M_2$. It is a well-known exercise from baby Rudin's: you can find a proof of it here. By the Cauchy-Schwarz inequality, it gives: ...


4

For suitable $f$, we have $$ \mathcal{F}(f')(\xi)=2\pi i\xi\mathcal{F}(f)(\xi) $$ Therefore, if one exists, it would be $$ \mathcal{F}(f'/f)(\xi)=2\pi i\xi\mathcal{F}(\log(f))(\xi) $$ which would lead to $$ \mathcal{F}(\log(f))(\xi)=\frac1{2\pi i\xi}\mathcal{F}(f'/f)(\xi) $$ if $f'/f$ has a Fourier Transform. To show a bit more care, we can make the ...


4

For $K \subset \mathbb{R}^n$ compact, consider the space $$\mathcal{D}_K := \{ \varphi \in \mathcal{D} : \operatorname{supp} \varphi \subset K\}$$ endowed with the seminorms $$\lVert\varphi\rVert_k = \sup \{ \lvert D^k\varphi(x)\rvert : x \in \mathbb{R}^n\}$$ for $k \in \mathbb{N}^n$. It is straightforward to show that $\mathcal{D}_K$ is then a Fréchet ...


3

Note that $g = \max(f_1,f_2,\ldots,f_n)$ is convex. Let its minimum on $[0,1]$ be at $x=p$. There are three cases to consider: $p=0$, $p=1$ and $0 < p < 1$. I'll do the third case. Note that $g$ has left and right one-sided derivatives $D_-g(p) \le 0$ and $D_+g(p) \ge 0$ at $p$. There exist $J$ such that $g(p) = f_{J}(p)$ and $D_-g(p) = D_- ...


3

Any metric compatible with the product topology on $X\times Y$ will work. Three of the simpler are $$\begin{align*} &d(\langle x_0,y_0\rangle,\langle x_1,y_1\rangle)=\max\{d_X(x_0,x_1),d_Y(y_0,y_1)\;,\\ &d(\langle x_0,y_0\rangle,\langle x_1,y_1\rangle)=d_X(x_0,x_1)+d_Y(y_0,y_1)\;,\text{ and}\\ &d(\langle x_0,y_0\rangle,\langle ...


3

The first condition you will want to impose is that if $N$ is a null-set, then so is $g^{-1}(N)$. To see this, note that in general $\chi_{M} \circ g = \chi_{g^{-1}(M)}$, where $\chi_{M}$ is the characteristic function/indicator function of the set $M$. Hence, if there was a null-set $N \subset [a,b]$ such that $g^{-1}(N)$ is not a null-set, then you would ...


3

When $\|u\|<1$, then $\mathrm{id}+u$ is invertible with inverse $$\sum_{n=0}^{+\infty}(-1)^nu^n\,.$$ This relation holds in any unital Banach algebra, in particular the unital Banach algebra of endomorphisms of a Banach space. Thus, $S=T+(S-T)=T(\mathrm{id}+T^{-1}(S-T))$ is invertible as soon as $\|T^{-1}(S-T)\|<1$. But ...


3

The zero subscript usually means vanishing at infinity, in a very general context. Vanishing at infinity means that for every $\varepsilon$, there is a compact set $K$ such that the function is smaller than $\varepsilon$ outside $K$. In other words, $C_0(X)$ is the closure of $C_c(X)$ (compactly supported continuous functions) under uniform convergence ...


3

Try something like $f_n(x) = ne^{-nx}$. Then $$\|f_n\|_1 = \int_0^1 ne^{-nx} \, dx = 1 - e^{-n}$$ for all $n$ but $\|f_n\|_\infty = n$.


3

Alternatively, let $A$ be the $C^*$-algebra generated by $a$. Since $a$ is hermitian (or more generally normal), $A$ is commutative and isometrically isomorphic under the Gelfand transform to $(C(X),\|\cdot\|_\infty)$, where $X=\sigma(a)$ is the spectrum of $a$, and $a$ corresponds to the identity function of $X$; thus ...


3

It is due to Arzela Ascoli (in a sense). We assume that the claim is false. Then there is a compact subset $K$ and (check this!) Some $\epsilon>0$ and some sub sequence $(f_{n_k})_k$ so that for each $k$, there is some $x_k \in K$ with $$|f_{n_k}(x_k)-f(x_k)| >\epsilon \qquad (\dagger).$$ But Arzela Ascoli yields a further sub sequence ...


3

If a set does not meet some ball, then it cannot be dense in a line through the origin passing through the center of this ball.


3

If you're asking about cardinality, it turns out they both have the same cardinality as $\mathbb{R}$. For $\mathbb{R}^2$ this has been proven many ways. An overkill approach is to use space-filling curves. For $C^k(\mathbb{R},\mathbb{R})$, here's a sketch of a proof based on the Cantor-Schroder-Bernstein theorem. Verifying one half of the hypotheses of the ...


3

I would say "Yes" to all your questions. The so-called weak* topology is an example of a weak topology. Weak convergence of measures can be considered a weak* topology. To clarify a new example of convergence of this kind, you should master that $\sigma (\mathcal{F}, X)$ notation. (I think it goes back to Mackey, and was popularized by Bourbaki.)



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