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1

Here are some references that might help: Wikipedia page: https://en.wikipedia.org/wiki/Nash%E2%80%93Moser_theorem "An Inverse Function Theorem in Frechet Spaces" by Ivar Ekeland: https://www.ceremade.dauphine.fr/~ekeland/Articles/InverseFunctionTheorem.pdf "On the Nash-Moser Implicit Function Theorem" by Lars Hormander: ...


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Regard the transform: $$s_0(\varphi,\psi):=s(\overline{\Lambda_{-s}(N)}^0\varphi,\overline{\Lambda_{-s'}(N)}^0\psi)$$ By unitarity one gets: $$|s_0(\varphi,\psi)|\leq\|s\|\cdot\|\overline{\Lambda_{-s}(N)}^0\varphi\|_s\cdot\|\overline{\Lambda_{-s'}(N)}^0\psi\|_{s'}=\|s\|\cdot\|\varphi\|_0\cdot\|\psi\|_0$$ By Lax-Milgram one has: ...


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For any $z\in\mathbb C^n$ and $h\in H$ we have $$ \langle Bz, h \rangle = \sum_{j=1}^n z_j \langle h_j, h \rangle = z^H (\langle h_j, h \rangle )_{1\le j\le n}. $$ Thus, it follows $$ B^*(h) = \begin{bmatrix} \langle h_1, h \rangle \\ \vdots \\ \langle h_n, h \rangle \end{bmatrix}, $$ which is linear and continuous in $h$, as the inner product is bilinear ...


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By the previous thread: $$\mathcal{D}_0:=\bigcup_{R>0}\mathcal{R}E(B_R):\quad\mathcal{H}^{s+1}=\overline{\mathcal{D}_0}^{s+1}\subseteq\overline{\mathcal{D}\Lambda_a(E)\cap\mathcal{D}N}^{s+1}\subseteq\mathcal{H}^{s+1}$$ Calculations give:* ...


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Your approach sounds like it should work. This should get you started: using the standard basis for $\mathbb{C}^n$ and linearity of the inner product on $H$, we note that $$ \langle Bz,h\rangle = \langle z,B^*h\rangle = \sum_{j=1}^n z_j\langle e_j,B^*h\rangle. $$ Use your expression for $Bz$ to determine the $\langle e_j,B^*h\rangle$. (Alter things as you ...


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Restrictions Denote for shorthand: $$\mathcal{D}_0:=\bigcup_{R>0}\mathcal{R}E(B_R)\subseteq\mathcal{D}\Lambda_{a}(N)$$ Regard restrictions: $$\Lambda_{\Delta s}^0(N):\mathcal{D}_0\to\mathcal{D}_0:\varphi\mapsto\Lambda_{\Delta s}(N)\varphi$$ By measurable calculus: $$E(B_{R_\varphi})\Lambda_{\Delta s}(N)\varphi=\Lambda_{\Delta ...


0

They are norms since: $$\Lambda_s(\lambda)\neq0\quad(\lambda\in\mathbb{C})\implies\mathcal{N}\Lambda_s(N)=(0)$$ Parallelogram identity: $$\nu_{\varphi+\psi}(A)+\nu_{\varphi-\psi}(A)=\|E(A)(\varphi+\psi)\|^2+\|E(A)(\varphi-\psi)\|^2\\ =2\|E(A)\varphi\|^2+2\|E(A)\psi\|^2=\nu_\varphi(A)+\nu_\psi(A)$$ For negative ones: ...


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Since $Ax$ and $y$ are both in $Y$, the first norm should indeed be the norm in $Y$, not in $X$. Compact operator is (often, at least) implicitly assumed to be linear and the notation $Ax$ instead of $A(x)$ also suggests linearity. Let us calculate the derivative of $F$ at $x\in X$. I will assume that your Hilbert spaces are over the reals; the complex ...


4

Because $A$ is normal, then $\mathcal{N}(A)=\mathcal{N}(A^{\star})$ is invariant under $A$ and $A^{\star}$, as is its orthogonal complement $$ \overline{\mathcal{R}(A)}=\mathcal{N}(A)^{\perp}=\mathcal{N}(A^{\star})^{\perp}=\overline{\mathcal{R}(A^{\star})}. $$ That allows you to reduce to the case where $\mathcal{N}(A)=\{0\}$, and where the ranges ...


0

This is only possible if you know that $(u_n')$ is uniformly bounded in $L^2(0,T;V^*)$. Then write $$ \left|\int_0^T \langle u_n'-u',w\rangle \right| \le \left|\int_0^T \langle u_n'-u',w-w_j\rangle \right|+\left|\int_0^T \langle u_n'-u',w_j\rangle \right|\\ \le \|u_n'-u'\|_{L^2(0,T;V^*)}\|w-w_j\|_{L^2(0,T;V)}+\left|\int_0^T \langle u_n'-u',w_j\rangle ...


1

Note: I really like the other answer. This is just a side note to add a little intuition as to why convolution has a useful function in Mathematics. To see why it gets so much attention, consider the product of two power series \begin{align} \sum_{n=0}^{\infty}f(n)z^{n}\sum_{n=0}^{\infty}g(n)z^{n} & = ...


2

Citing from jeff560.tripod.com: CONVOLUTION. Expressions that would now be described as “convolutions” appear in Laplace’s earliest work on sums of independent random variables, “Mémoire sur l’inclinaison moyenne des orbites des comètes, sur la figure de la terre, et sur les functions,” Mém. Acad. R. Sci. Paris (Savants Étrangers), 7, (1773), ...


2

Nothing would go wrong per se, but in an incomplete normed space, there are simply much fewer weakly compact sets. Let $X$ be a normed space, and $\tilde{X}$ its completion. Then we have a canonical isometric isomorphism $\rho \colon \tilde{X}' \to X'$, namely the restriction of a functional to $X$. Thus we may identify the two spaces (that is habitually ...


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Approximate Unit Regard ball cone: $$\mathcal{B}_+:=\{A\in\mathcal{A}:\|A\|<1:A\geq0\}$$ Order elements: $$E,E'\in\mathcal{I}\cap\mathcal{B}_+:\quad E\leq E'$$ Then one has: $$I\in\mathcal{I}:\quad\|I-IE\|,\|I-EI\|\stackrel{E\to1}{\longrightarrow}0$$ (That is the hard part!) Quotient Norm Note that it holds: $$1-\sigma(E)\geq0\implies\|1-E\|\leq1$$ ...


0

Smoothness certainly does not imply boundedness. The function $f(x) = e^x$ is smooth (analytic, even) but is unbounded along with all of its derivatives. In the link you provided the assumption is that the functions are smooth and compactly supported. The key is the last assumption. We know from real analysis that any continuous function with compact ...


1

You consider $f(x)=F(g(x),h(x))$ where $F$ may be one of several "standard operations" and wonder if it is possible to have $f,g$ differentiable at $a$ and $h$ not. For $F(u,v)=u+v$ your conjecture is ture because we obtain that $h(x)=f(x)-g(x)$ is the difference of differentiable functions. Similarly for $F(u,v)=u-v$. For $F(u,v)=uv$ we have the trivial ...


1

What you can say is that $f$ may not be differentiable. For example, take $g(x)=0$, and $h(x)=|x|$. Then clearly $f(x)=g(x)h(x)=0$ is differentiable. What you can say, though, is the following: If $g$ is differentiable in $x$ and $g'(x)\ne 0$, then $g(x)h(x)$ is differentiable if and only if $h(x)$ is. If you want something more general than the ...


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For any Banach space of dimension $>2$, $X \backslash \{0\}$ is simply connected.


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Since $\nu$ is a $\sigma$- finite measure, you can find a sequence of disjoint sets $E_n$ such that $\Bbb{R} = \cup_n E_n$ and $\nu(E_n) < \infty.$ Define now $$\nu_n(C)= \nu(C \cap E_n).$$ Note that $\nu(C) = \sum_n \nu_n(C)$ Let $f_n = \frac{d \nu_n}{d\mu} $ and consider the $\sigma$ algebra $\mathcal{B}_k$ generated by the dyadic decomposition of ...


0

It seems the following. For each point $x=(x_n)\in\ell^\infty$ put $x_*=\underline{\lim}_{n\to\infty} x_n$ and $x^*=\overline{\lim}_{n\to\infty} x_n$. Then both values of $x_*$ and $x^*$ are finite, $x_*\le x^*$, and $x_*= x^*$ iff $x\in c$. The set $c$ is closed in $\ell^\infty$. Indeed, let $x=(x_n)\in\ell^\infty\setminus c$. Then $\varepsilon= ...


1

Nearly every Physical system has a linear regime where if you superimpose two small causes, the resulting effect is the superposition of the two separate causes, and if you double a small cause, then the effect is doubled. The state of the system is characterized by a collection of numbers (a vector) and the principle of superposition is stated in terms of ...


2

From Hölder: $$ I=\int_{a}^{b}(x-x_1)^2\ldots(x-x_n)^2 dx \leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^4 \ldots (x-x_n)^4dx \right)^{1/2} \leq \\ \leq \left(\int_{a}^{b}(x-x_1)^4 dx \right)^{1/2}\left(\int_{a}^{b}(x-x_2)^8 dx \right)^{1/4}\left(\int_{a}^{b}(x-x_3)^{8}\ldots(x-x_n)^8 dx\right)^{1/4} \leq \ldots \\ \leq ...


1

If $\varphi$ is a bounded linear functional on $L^1([0,1]),$ then there exists $g\in L^\infty([0,1])$ such that $$\varphi (f) = \int_0^1 fg$$ for all $f\in L^1.$ Thus $$\varphi (\chi_{[0,x]}) = \int_0^xg.$$ Since $g$ is bounded, that integral is Lipschitz, hence is absolutely continuous with room to spare.


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Each $f_n(s)= \int_0^\infty x^{s/2-1}e^{-\pi n^2x}\,dx$ is defined on $(0,\infty).$ Each $f_n \to \infty$ at $0$ and $\infty.$ You make the change of variables $x=y/(\pi n^2)$ and you get the identity $$\int _0^{\infty }x^{s/2-1}e^{-\pi n^2x}dx=\frac{1}{n^s}\pi^{-s/2}\Gamma(s/2)$$ -as observed by Olivier Oloa. It follows that we need $s>1$ for even ...


2

You may observe that, using a standard integral representation of the gamma function, we have $$ \int _0^{\infty }x^{s/2-1}e^{-\pi n^2x}dx=\frac{1}{n^s}\pi^{-s/2}\Gamma(s/2) $$ then your series rewrites $$ \sum_{n=1}^{\infty} \int_{0}^\infty x^{\frac{s}{2}-1}e^{-\pi n^{2}x}dx=\pi^{-s/2}\Gamma(s/2)\sum_{n=1}^{\infty} ...


0

The triangle inequality is clear from $\lVert 0\rVert = 0$ if $x = 0$ or $y = 0$, so we need only consider the case $x\neq 0 \neq y$. Let's denote $B := \{ z \in N : \lVert z\rVert \leqslant 1\}$. By the homogeneity and non-negativity of $\lVert\,\cdot\,\rVert$, we have $$\biggl\lVert \frac{1}{\lVert x\rVert}\cdot x\biggr\rVert = \biggl\lvert ...


2

Yes there is! It is easy to construct absurd examples, but I guess you are looking for more interesting examples, such an example is provided by Zonal spherical functions where the translation operator is induced from a group and the measure is induced from a Haar measure. Also, if I remember correctly M.A. Naimark (see his book Normed Rings) ...


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This is theorem 5.4 page 115 from Theory of the integral, S. Saks. It is proved via Vitali's covering theorem.


0

It suffices to prove that $C_0(X,\mathbb{R})$ separates points and vanishes nowhere. Let $\bar X$ be the one-point compactification of $X$ so that $\bar X$ is compact Hausdorff, hence $T_4$, in which Urysohnn's lemma is applicable. Now, define $\mathscr{C}\triangleq \{f\in C(\bar X,\mathbb{R}): f(\infty)=0\}$. Note that $\mathscr{C}$ is an ...


1

Every normed linear space is Hausdorff, hence singletons are closed sets in a normed linear space. $T:V\to W$ is bounded iff it is continuous, Now $\{0\}$ is closed in $W$ and $T$ is continuous. So, by continuity of $T$, $$ T^{-1}(\{0\})$$ is closed. Hence $$\ker(T) = \{x\in X: Tx =0\}$$ is closed.


1

Note that your statement can be written as $A$ surjective $\Leftarrow$ $R(A)_\perp =\{ w\in W: \ \langle Av,w\rangle = 0 \ \forall v\in V\}= \{0\}$ The desired implication fails if $A$ has dense range but is not surjective (provided $V,W$ are normed spaces, $A$ continuous, $W^*$ continuous dual). If the range of $A$ is dense then $$ \langle ...


2

It should be closed. Let $T:V\to W$ be a bounded linear operator between normed linear spaces. We have that $$ \ker(T) = \{x\in X: Tx =0\} = T^{-1}(\{0\}). $$ Since bounded operators are continuous and $\{0\}$ is a closed set in the norm topology, we see that the kernel is closed. The kernel is not open unless $T$ is the zero operator. Assume $T\neq 0$. We ...


1

Note that all of the work is in Exercise 8, which says that if $0\leq x\leq y$ in a C*-algebra and $x$ is invertible, then $y^{-1}\leq x^{-1}$. You will apply this with $x=1+\alpha a$ and $y=1+\alpha b$, and the rest follows easily. For a solution to Exercise 8, see Inversion in a unital C* algebra or positive invertible operators or Inverse of ...


1

I think both $H$ and $H^0$ are the same set, the difference in notation only emphasizes the structure under consideration. Namely, $H$ is an Adobe space, while $H^0$ is a linear space, that is an affine space with a chosen zero element. The automorphism groups are different because a diffeomorphism of a linear space is required to fix the origin. But since ...


0

As Ian suggested, the statement follows by applying the Rellich-Kondrachov theorem to the gradients. Generally, any time you apply the Sobolev embedding on a bounded domain and give up a bit in the exponent of the target space, the embedding is compact. An intuitive reason is that when exponents are related in this way, small scale features affect the ...


0

The main thing: $y = y_1 + \ldots + y_m$ and the $y_i$'s are paiwise orthogonal ( an orthogonal decomposition) then: $$||y||^2 = ||y_1||^2 + \cdots + ||y_m||^2 $$ It's easy to check that if $e_1$, $\ldots$, $e_n$ is an $\it{orthonormal}$ family then $x - \sum_{i=1}^n \langle x, e_i \rangle e_i$ is perpendicular to all the $e_i$'s and so we have the ...


1

$Q1.$ According to my school's EE graduate course catalog, there are a couple of classes covering Hilbert, Banach and L^p spaces; and linear functionals. Is it possible to cover these topics without a formal mathematical background? $A1.$ It is the responsibility of the school to provide a curriculum for the students that is both meaningful an doable. ...


0

Special thanks to Simon B. By uniform extension: $$\varphi\in U'\implies\overline{\varphi}\in\overline{U}'$$ This happens uniquely: $$\overline{U}=E\implies\ker\pi=(0)$$ Especially one has then: $$\|\pi[\overline{\varphi}]\|=\|\overline{\varphi}\circ\iota\|=\|\varphi\|=\|\overline{\varphi}\|$$ Conversely assume: $$U=\overline{U}\neq E$$ By Riesz lemma: ...


1

Yes, this property is called the Baire property, and every complete metric space is a Baire space https://en.wikipedia.org/wiki/Baire_category_theorem


2

Your reasoning is correct, for a continuous (bounded) everywhere defined operator $A$ on a Banach (in particular on a Hilbert) space, the denseness of the range of $\lambda\mathbb{I} - A$ together with the boundedness of the inverse already implies the surjectivity of $\lambda\mathbb{I} - A$, and an equivalent definition of the resolvent set in this setting ...


4

Since $X_n\xrightarrow{p} X$ there exists a subsequence $X_{n_k}\xrightarrow{a.s.}X$. However, the event $\{X\le x\}=$$\{\lim X_{n_k}\le x\}$ is a tail event so that $X$ must be constant by Kolomogorov's $0-1$ law.


0

Nate has already answered this perfectly well, but perhaps it is worth noting the question becomes a little more difficult if we require $X^+$ to be large enough to generate $X$, i.e. $X=X^+-X^+$. For an example in this case, let $$X^+=\{(x_n)\in c_0:\forall n(0\leq nx_{2n}\leq x_{2n+1})\},$$ which is complete w.r.t. the supremum norm. Then $X=X^+-X^+$ ...


1

For $1\le j\le n$, $$ \begin{align} \left\langle x-\sum_{k=1}^n\langle x,e_k\rangle e_k,x\right\rangle &=\langle x,x\rangle-\sum_{k=1}^n\langle x,e_k\rangle\langle x,e_k\rangle\\ &=\|x\|^2-\sum_{k=1}^n\langle x,e_k\rangle^2\tag{1} \end{align} $$ and $$ \begin{align} \left\langle x-\sum_{k=1}^n\langle x,e_k\rangle e_k,e_j\right\rangle &=\langle ...


1

\begin{align} \left\|x-\sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i\right\|^2&=\langle x-\sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i\,,x-\sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i\ \rangle \\ &=\langle x,x\rangle+\underbrace{\langle \sum \limits_{i=1}^{n}\langle x,e_i\rangle e_i,\sum \limits_{i=1}^{n}\langle x,e_i\rangle ...


0

There must be a typo; you mean $\eta_h(x)=\frac1h\eta(\frac xh)$, right? Assuming so: I have no idea why you're interested in that particular function, nor what its integral is. Say $$\alpha=\int\eta.$$ Young's inequality says in part that $||f*g||_p\le||f||_p||g||_1$; since $\eta\ge0$ this shows that $$||f*\eta_h||_p\le\alpha||f||_p.$$ And standard ...


2

The spectrum of the $C^*$-algebra $C_b(\mathbb R)$ of bounded continuous functions on $\mathbb R$ corresponds to the Stone-Čech compactification $\beta \mathbb R$ of $\mathbb R$. Any finite positive Borel measure on $\beta \mathbb R$ gives you a positive linear functional on $C_b(\mathbb R)$. However, the points of $\beta \mathbb R \backslash \mathbb R$ ...


2

In the complex case, you have $(Su,v)=\overline {(Sv,u)} $ for all $u,v$. Then $$ (Su,v)=\overline {(Sv,u)}=(u,Sv)=(S^*u,v). $$ So $(\, (S-S^*)u,v)=0$ for all $v $, which implies $(S-S^*)u=0$. As this occurs for all $u $, $S-S^*=0$. For the real case, just remove the bars.


3

We have $$f(X)\subseteq\{0\}\cup f(\mathrm{supp}(f)),$$ which is compact in $\mathbb{R}$ (since $\mathrm{supp}(f)$ is compact and $f$ is continuous), hence bounded.


1

If you do know the Weierstraß theorem, then you can prove it like that: Let $f \in \mathcal{K}(X)$ and denote by $K$ the support of $f$. Then $f|_{K^c}=0$ by the very definition of the support. Moreover, by the Weierstraß theorem, $f|_K$ is bounded. Combining both facts, proves that $f$ is bounded. If you do not know the Weierstraß theorem, then have e.g. ...


1

A closed densely-defined linear operator on a Hilbert space can have empty spectrum. For example, let $H=L^{2}[0,1]$ and let $A=\frac{d}{dx}$ on the domain consisting of absolutely continuous $f \in L^{2}$ for which $f(0)=0$ and $f' \in L^{2}$. To show that $A$ has no spectrum, it is enough to prove that the resolvent $R(\lambda)=(A-\lambda I)^{-1}$ exists ...



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