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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 ...


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 ...


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

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 ...


4

Let $J $ be an index for the cardinality of an orthonormal basis of $H $. Then $H $ is isometrically isomorphic to $\ell^2 (J) $, so it is enough to discuss the problem on this latter space. Define the product $fg $ pointwise, i.e. $fg (j):=f (j)g (j) $. The question is whether this product stays in $\ell^2$, and whether the norm is submultiplicative. We ...


3

Another Banach-algebra structure on the Hilbert space $\mathsf{hs}(H)$ of Hilbert-Schmidt operators on a Hilbert space $H$ is just operator multiplication (composition). There is a natural involution on this algebra but it does not make it a C*-algebra.


3

Let $Af = \int_{0}^{x}f(t)dt$ in $L^{2}[0,1]$. Then $A : L^{2}\rightarrow L^{2}$ is bounded. Let $W$ consist of all continuously differentiable $g \in L^{2}[0,1]$ for which $g(0)=g(1)=0$. $W$ is dense in $L^{2}[0,1]$ because $\{ \sin(n\pi x) \}_{n=1}^{\infty}\subset W$ is an orthogonal basis of $L^{2}[0,1]$. However, $A^{-1}W$ is not dense because $f \in ...


3

You need to bound the quantity $$\int_{-\infty}^{+\infty} \left| \int_{-\infty}^{+\infty} K(x,y) f(y) \,dy \right|^2\,dx.$$ First note that it is certainly bounded by $$\int_{-\infty}^{+\infty} \left| \int_{-\infty}^{+\infty} |K(x,y) f(y)| \,dy \right|^2\,dx.$$ Rewrite as $$\int_{-\infty}^{+\infty} \left| \int_{-\infty}^{+\infty} \left( \frac{|f(y)|}{w(y)} ...


3

Edit Here is a proof that works if $\mu$ is semifinite. Suppose $\phi \notin L_\infty(X, \mu).$ Then given any $R>0$ there is a set of positive measure $E \subset X$ such that $\vert \phi \vert > R$ on $E.$ Let $f \in L_1(X,\mu)$ such that $f$ vanishes outside of $E$. Taking absolute value, we can assume $f$ is real and non-negative. Then $$\Vert Mf ...


3

This is false. Consider $$\mu = \sum_{n=-\infty}^\infty \infty \delta_n + \lambda$$ where $\infty \delta_n$ is the measure that has infinite point mass at $n$ and $\lambda$ is the Lebesgue measure on $\mathbb{R}$. Let $\phi(n) = n,$ for $n\in \mathbb{Z}$ and $\phi(x)=1$ for $x$ not an integer. Then $f \in L^1(\mu)$ implies $f(n) = 0$ for all $n\in ...


3

Thank you for the comments. I was able to use them to provide an answer to When does $\sum_{i=1}^{\infty} X_i$ exist for random sequences $\{X_i\}_{i=1}^{\infty}$?. I'll restate the result from there as it provides an example: Consider the measure space $L^2([0,1])$ with uniform lebesgue measure. Define the functions (Haar functions) $f_{2^i + k}$ by ...


3

Uniform closure, of course not. Closure in the topology of pointwise convergence yes, as has been pointed out, but this is trivial (immediate from the fact that a polynomial can take any values you want on any finite set) and I don't see how it's good for anything. There's a closure for which the answer is yes, and which also actually allows one to prove ...


3

I think the authors are somewhat inconsistent in what they mean by the domain of an operator. Let's ignore the boundary conditions for now and focus on the order of smoothness. The domain of $(-\Delta)^s$ as an operator into $L^2$ is a Sobolev space of order $2s$. Indeed, $s$ fraction of the Laplacian is like $2s$ derivatives. The domain of ...


3

You're thinking too Euclidean! It is entirely possible in an arbitrary normed linear space to have points other than antipodal points that satisfy this property. As an example, consider the points $(1, 0)$ and $(0, 1)$ in the unit sphere under the $1$-norm on $\mathbb{R}^2$. Good question though!


3

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 ...


3

Because of the subspaces being selfadjoint, $B(H)V_1\subset V_2$ implies that $V_1B(H)\subset V_2$. If $V_1\ne0$, then $B(H)V_1B(H)\subset V_3$ contains all finite-rank operators, and thus $V_3$, being closed, contains the compact operators. If $V_3$ contains a non-compact operator, then the ideas in this answer show that $I\in V_7$ (I didn't count ...


3

The weak topology has the property that the dual is the same as the dual with respect to the original topology, if $(E,\tau)$ is a topological vector space, and $E^\ast$ its (topological) dual, then $$(E,\sigma(E,E^\ast))^\ast = E^\ast.$$ For the topology of pointwise convergence on $C(S)$, the dual is easily described: if $\lambda \colon C(S) \to ...


2

It may be the case that these open balls are actually equal to each-other despite having different radii. For instance, if we equip $\mathbb{Z}$ with the usual metric, then $B_{1/2}(0) = B_{1/3}(0)$, for instance.


2

$\int_0^1\vert f_n(x)-1\vert dx=\int_0^{1/n}\vert f_n(x)-1\vert dx+\int_{1/n}^1\vert f_n(x)-1\vert dx$ Since $f_n(x)=1$ for $x\geq\frac{1}{n}$, the second term is equal to $0$. You can also remove the absolute value in the first term since for $0\leq x\leq\frac{1}{n}$, $f_n(x)=nx\leq 1$.


2

Let $\{ x_n \}_n $ be a sequence, bounded say by $M>0$. If $T^*T$ is compact, there exist a subsequence $\{ T^*T(x_{n_k})\}_k$ that converges and hence is Cauchy. Let $k<k' \in \mathbb{N}$, then \begin{align*} \|T(x_{n_k}) - T(x_{n_{k'}}) \|^2 & =\| T(x_{n_k}-x_{n_{k'}}) \|^2 \\ & = \langle T(x_{n_k}-x_{n_{k'}}), T(x_{n_k}-x_{n_{k'}}) \rangle ...


2

It may be of some value to write an explicit answer (which is pretty similar to the comment by user251257, only containing more details). Let $e:U\to\mathbb{R}$ be differentiable, and let $e':U\to(\mathbb{R}^n)^*$ denote the differential of $e$. Let $u\in U$, and let $h\in\mathbb{R}^n$. By differentiability, we ...


2

Let's prove it. Suppose $k$ is in the kernel of $A$ and $x_0$ is a specific solution, so that $Ax_0=b$. We have $$A(x_0+k)=Ax_0+Ak=Ax_0=b.$$ Conversely, suppose that $x_0$ and $y_0$ are two solutions. Then $$A(x_0-y_0)=0$$ implying that $x_0-y_0=:k$ is in the kernel. Therefore any two solutions differ by an element of the kernel.


2

There is nothing wrong about your argumentation. Here is a "real" counterexample: Set $$f(x) := 1_{[0,1/2]}(x) \qquad \text{and} \qquad g(x) := 1_{[1/2,1]}(x).$$ Then $$\|f\|_1 = \|g\|_1 = \frac{1}{2}$$ and therefore $$2 (\|f\|_1^2+ \|g\|_1^2) = 1.$$ On the other hand, it is not difficult to see that $$\|f+g\|_1 = \|f-g\|_1 = 1.$$ Hence, ...


2

First for the eigenvalues: By the spectral theorem (and since $A$ is a positive (hence self-adjoint) operator), there is a measure space $(X, M, \mu)$ and some (measurable) function $g : X \to [0,\infty)$, such that $A$ is unitarily equivalent to the multiplication operator $$ M_g : L^2(\mu) \to L^2 (\mu), h \mapsto g\cdot h. $$ Thus, it suffices to prove ...


2

For convenience, scale $\phi$ so that $\|M_{\phi}\|_{\mathcal{L}(X)}=1$. For $\delta >0$, the $\chi_{\delta}$ be the characteristic function of the set where $|\phi| \ge 1+\delta$. Then, for any $f \in L^{1}$, $$ (1+\delta)\int |f\chi_{\delta}|d\mu \le \int |f\chi_{\delta}| |\phi|d\mu \le \int |f\chi_{\delta}|d\mu. $$ Thus $\|f\chi_{\delta}\|=0$ ...


2

I know a very special case, and I guess it is extensible. Suppose that $H$ is Hilbert space such that $H\cong (H_1\hat{\otimes}(H_2)^*)^*$, where $\hat{\otimes}$ is projective tensor product, and $H_1,H_2$ is some Hilbert spaces. For any Hilbert space $\mathcal{H}$, always $\mathcal{H}^{**}=\mathcal{H}$. Also for two Banach space $E,F$ always ...


2

As noted in my comment, $E$ has to be a Hilbert space. To see this, fix a linear functional $\varphi \in E'$ with $\Vert \varphi \Vert = 1$. For $z \in E$, define $$ A_z : E \to E, x\mapsto \varphi(x) \cdot z. $$ It is not hard to see that $E \to B(E), z \mapsto A_z$ is linear and isometric. Thus, if $B(E) = H$ is a Hilbert space, we see that $E$ is ...


2

What you need to show is that $\phi $ is one-to-one, and that it is a $*$-homomorphism. That lets you define the norm on $M_n (A) $ and the two properties you want are inherited automatically from the codomain. Then you need to show that $M_n (A) $ is closed. This is done by showing that norm convergence is equivalent to entrywise norm convergence, and as ...


2

Consider in $\Bbb{R}^2$,$B(v_1)=3v_1$ and $B(v_2)= 12 v_2$. Let $A(v_1) = v_2$ and $A(v_2)= 4v_1$ where $v_1 = (1,0)$ and $v_2=(0,1)$ Note that $$|A(v)|=\frac{1}{3}|B(v)|$$ So we conclude that $$a_1\|Bu\| \leq \|Au\| \leq a_2\|Bu\| $$ with $a_1 = \frac{1}{3}$ and $a_2 = \frac{1}{3}$ Now note that $A*(v_1)= 4v_2$ $$|A^*(v_1)|\nleq \frac{1}{3} B(v_1)$$ ...


1

$$\sum_{j=1}^d |n_j|^{2k} \geq \max_{j=1,\ldots,d} |n_j|^{2k}$$ implies, using the mentioned inequality, $$(2\pi)^{2k} \sum_{j=1}^d n_j^{2k} \geq (2\pi)^{2k} \max_{j=1,\ldots,d} |n_j|^{2k} \geq \left( \frac{2\pi}{\sqrt{d}} \right)^{2k} \|n\|_2^{2k}$$



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