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$K$ is defined by the relation $$\int_\mathbb{R} f(x)\mathbf{1}_{[n,n+1)}(x)\,dx = 0$$ for all $n \in \mathbb{Z}.$ That looks a lot like we're defining basic elements of $K^\perp.$ In particular, this seems to define $$K^\perp = \{g \in L^2(\mathbb{R}) : \forall n \in \mathbb{Z}, g \text{ is constant on } [n,n+1)\}$$ which can be naturally identified with $... 0 Let$C$be the weak-* closure of the convex hull of$\pm S$. Then$C$is a closed convex subset of$B_{X^*}$. Let$f \in B_{X^*} \backslash C$. By Hahn-Banach in the locally convex space$X^*$with its weak-* topology (such that only evaluations at a point of$X$are linear forms), there is some$x \in X$and some$a > 0$such that, for all$g \in S$,$|...

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Yes, there are. The first systematic study of examples are the "batman" wavelets of Belogay/Wang (1999): "Construction of compactly supported symmetric scaling functions". One can construct them for any number of bands, making the scaling functions smooth is an optimization task. For instance, with a 3-channel decomposition with one scaling and two wavelet ...

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As others have said, the question doesn't make sense as stated, but what we can prove is that if $f: (0,1) \rightarrow \mathbb{R}$ is continuous and $|f^{-1}(x)|\leq x^2$, then $f$ is differentiable a.e.. For that it will suffice to show that $f$ is differentiable a.e. on an interval of the form $[1/n, 1-1/n]$. Note that $|f|$ is bounded on such an ...

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As far as I understand for $L^{\infty}$ solutions, the initial data is taken in $L^1_{loc}$ sense. Please correct me if I am wrong.

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Choose any smooth $f$ on $\mathbb R$ with both $\int_{\mathbb R}f^2, \int_{\mathbb R}(f')^2$ finite and nonzero. Set $f_n(x)=f(nx).$ Then $$\int_{\mathbb R}f_n^{\,2} = \frac{1}{n}\int_{\mathbb R}f^2 \to 0.$$ But $$\int_{\mathbb R}(f_n\,')^2 = n\int_{\mathbb R}(f\,')^2 \to \infty.$$

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Yes, because $A(e_k) = f(c_k) e_k$ determines $A$ on the linear span $V$ of the orthonormal basis, and if $A$ is bounded that determines it on the closure of $V$, which is $H$.

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This inequality follows by integration of $$\langle A(t\,v) - A(0), v \rangle \ge A(\frac12 \, v) - A(0), v \rangle$$ for all $t \ge \frac12$. Note that this inequality is equivalent to $$\langle A(t \, v) - A(\frac12 \, v) , v \rangle = \frac1{t - \frac12} \, \langle A(t\,v) - A(\frac12\, v), t \, v - \frac12 \, v\rangle\ge 0$$ for all $t \ge \frac12$ ...

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Here I discuss in more detail the case mentioned in the section "second update". This special distribution was produced with a seed $\alpha$ built as follows: its $n$-th binary digit is $1$ if $\mbox{Rand}(n)< p$, and $0$ otherwise, using a pseudo random number generator. I used $p=0.75$ in my example. Now $x_n$ (introduced in the first paragraph in my ...

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Assuming that $u_1$, $u_2$, and $u_3$ are continuous function than the supremum norm can be written as $$\|\vec{u}\|_{\infty}=\sup\{\sup |u_i|\text{ with }i=1,2,3\}.$$ If the functions $u_i$ are continuous and defined on a compact set, then the $\sup$ can be replaced by $\max$. For the more general cases, I would refer to the Wikipedia page https://en....

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Your notation is fine, but incomplete in two ways. Firstly you have not specified a domain and a codomain for the operator $\mathscr{F}$. For the domain we can take the smooth functions $\mathcal{C}^{\infty}(X,Y)$, where $X$, $Y \subseteq \mathbb{R}^d$ are open. However, for negative $n$ this will not work for the codomain, as the constants you introduce do ...

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This proof is fine, however there are two points I would like to make: 1) Expand upon your first sentence. 2) After your first sentence, the result follows from a nice corollary to the Hahn-Banach theorem: If $X$ is a locally convex space (e.g. a Hilbert space) and $C\subset X$ is convex, then $C$ is closed in the original topology iff it is closed ...

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Define $$Av = v(a),\;\;\; Bv=v(b).$$ These are continuous linear functionals on $H^1(I)$ because, for example, $$v(a) = -\left.\frac{b-t}{b-a}v(t)\right|_{t=a}^{b}\\= -\int_{a}^{b}\frac{d}{dt}\left(\frac{b-t}{b-a}v(t)\right)dt \\ = \int_{a}^{b}\frac{1}{b-a}v(t)-\frac{b-t}{b-a}v'(t)dt,$$ and $$|Av|=|v(a)| \le C(\|v\|_{L^2}+\|v'\|_{... 2 This is false in general. For instance, let B:H_0\to H_0 be any operator that is injective but whose image is not dense (say, B could be an isometry to a proper subspace of H_0). Let H=H_0\oplus\mathbb{R} and let A:H\to H be B on the first coordinate and 0 on the second. Then \lambda=0 is an eigenvalue of A, and A_0 is just B. But ... 0 Approximate f(x) uniformly by a trigonometric polynomial h(x). By Stone-Weierstrass this can be done. Then explicitly find a solution u(x,t) such that u(x,1) = h(x). 2 A space is called 2nd-countable iff it has a countable base (basis). A separable metrizable space is 2nd-countable but a non-metrizable separable space need not be 2nd-countable. (1). Suppose the topology T on the set X is generated by the metric d. Let D be a dense subset of X. Then B=\{B_d(y,q):y\in D\land q\in \Bbb Q^+\} is a base for T. ... 2 Realcompactness is a not very well-known property of topological spaces; it does not come up in elementary topology courses, usually. I'll quote some of the results in paragraph 3.11 of Engelking's (IMHO very good) standard work General Topology (revised and completed edition). Gilman and Jerrison also cover it in Rings of Continuous Functions. The "... 1 Consider A and B with rank 1 such that A = u u^T and B = v v^T for som u,v. For arbitrary C = w w^T you have Trace(AC) = trace(u u^T w w^T ) = u^t w \,\, trace(u w^T ) and Trace(BC) = v^T w \,\, trace(v w^T ). Therefore any w that is orthogonal to u but not to v will suffice. Of course if u and v are parallel this does not exist. 4 Not an answer, just some thoughts. Let's try applying Fourier transform:$$f(x)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty e^{i k x} g(k) dk$$The equation becomes:$$\int_{-\infty}^\infty \left(2 e^{i k x}-e^{i k x/2} (1+e^{i k /2}) \right) g(k) dk=0$$Making a simple substitution:$$\int_{-\infty}^\infty e^{i k x} \left(g(k)- (1+e^{i k }) g(2k) \...

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I think you need to use the Szego-Jacobi parameters or something. I remember this from a long time ago. http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.64.1023

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A little too long for a comment : there is no generic method to find a closed form for orthogonal polynomials wrt to a particular inner-product. It is the opposite, to obtain Hermite,Laguerre,Legendre they started from a closed-form two variable analytic function $$f(x,t) = \sum_n g_n(x) t^n$$ satisfying $$\int_a^b f(x,t) f(x,u) w(x) dx = h(ut)$$ for some $... 0 All known orthogonal polynomials, including those you have mentioned, are solutions of differential equations that arise in various physical phenomena. Legendre developed his poylnomials as coefficients of expansion of a Newtonian potential; Hermite polynomials are eigenstates of the quantum harmonic oscillator; The Legendre polynomials arise naturally when ... 0 To show that$K$is independent of the choice of extension$\tilde{g}$, consider another extension$\hat{g}\in H^1(\Omega)\cap C(\overline{\Omega})$with$\hat{g}=g$on$\partial\Omega$. Put $$\hat{K}:=\{v \in H^1(\Omega)\ |\ v-\hat{g}\in H^1_0(\Omega) \}.$$ We have to verify that$K=\hat{K}$. To this end consider$v\in K$. Then$v-\tilde{g}\in H^1_0(\...

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For the functional $\int_0^1\left(\frac{x'^2}{2}-ax\right)dt$ and the Euler-Lagrange equation you need to calculate $$\frac{\partial{\cal L} }{\partial x}=-a,$$ and $$\frac{\partial{\cal L} }{\partial x'}=x'.$$ Then the equation is $$x''+a=0$$ which upon integration you will get $$x(t)=-\frac{at^2}{2}+Bt+C$$ for some other constants $B,C$.

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For $p \in [1,2]$ $$\|f 1_{|f| > 1}\|_p^p \le \|f \|_2^2, \qquad \qquad \|f 1_{|f| \le 1}\|_p^p \le \|f \|_1$$ With Cauchy-Schwarz $$\|f \|_1\le \|f\|_2\|1_M\|_2$$ Thus $$\|f\|_p^p=\|f 1_{|f| > 1}\|_p^p+\|f 1_{|f| \le 1}\|_p^p \le \|f\|_2^2+ \|1_M \|_2 \|f\|_2$$ Let $U_{n,s} = \sum_{k=s}^\infty a^n_k e_k$. You have said $\sup_n \|U_{n,1}\|_2 < \... 0 Because$\{y_n\}_{n=1}^\infty$is a countable dense subset, for any$y\in Y$we can find a subsequence$\{x_n\}$of$\{y_n\}$such that$x_n\to y$. Suppose$\epsilon > 0$and$ |y\| =d > 0$. We can find some$N$such that if$n\geq N$, then$\|x_n - y\| < \epsilon d$. Since$x_n\to y$, we have that$|\|x_k\| - d| \to 0and therefore $$|\epsilon\|... 1 Let's take a 1D example. We can combine the integrals and simplify the notation a bit:$$F\left(u,u'\right)=\int_{\Omega}\frac{(u-v)^2+(u')^2}{2}\,dx.Then L=\dfrac{(u-v)^2+(u')^2}{2} and the EL equation says \begin{align*} \frac{\partial L}{\partial u}-\frac{d}{dx}\,\frac{\partial L}{\partial u'}&=0 \\ u-v-\frac{d}{dx}u'&=0 \\ u-v-u''&=0. \... 1 Let x_n=0 for all n and y_n=\frac 1 n for N_1 <n<N_2, y_n=0 for all other n. Then \|x-y\|_2 \to 0 as N_1 \to \infty but \|y\|_1 \to \infty as N_2 \to \infty. Hence no open ball around x in \ell^{2} norm is contained in the open unit ball w.r.t. \ell^{1}. 1 The proof definitely works when f is real-valued. For complex-valued f \in C_K(X), choose \alpha \in \mathbb{C} with |\alpha| = 1 so that \lambda(f) = \alpha |\lambda(f)| holds. (I.e., decompose \lambda(f) in polar form.) Then|\lambda(f)| = |\lambda(\alpha^{-1}f)| = |\lambda(\operatorname{Re}(\alpha^{-1}f))| \leq \|\operatorname{Re}(\alpha^{-... 1 Okay so I am adding a new partial answer as the one I had before was wrong. Turns out that it's not always true thatA_\lambda$maps$\mathcal{N}(\lambda-A)^{\perp}$into itself. Here's an example. Consider$L$, the left shift on$\ell^{2}(\Bbb{Z}_{>0})$seen as a real Hilbert space. That is, $$L(x_1, x_2, \ldots):=(x_2, x_3, \ldots)$$ Then, any$\...

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I believe yes. Let $X$ be Banach and $A: X\to X$ finite rank. We will define sub-spaces $\mathcal{I, K, C}$ of $X$. $\mathcal I$ is the image of $A$, it is finite dimensional. $\mathcal K$ is the kernel of $A$ and it is closed, we choose a complement $\mathcal C$ of $\mathcal K$, this complement is finite dimensional. Now $A$ mainly lives on the space $\... 1 First, it's important to precise that$\ell^1(\mathbb R)\subset \ell^2(\mathbb R)$(otherwise, the question has no sense). A counter example is $$y_n=\frac{x_n}{\|x_n\|_{\ell^1}},$$ where$x_n=(\underbrace{n,...,n}_{n\text{ times}},0,...,0,...)$. 4 It is true that the question here only says there is an inclusion, but the answer actually writes down explicitly an inequality which on one side has the$L_p$norm and on the other side has a quantity involving the semi-norms in${\cal S}(\mathbb{R})$, which proves that the inclusion is indeed continuous with respect to the usual topologies on$L_p$and${\...

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(I wanted to add this as a comment, but I don't have enough reputation) Not really an answer to your question, but your visualization method looks quite similar to the diagrams in Counterexamples in Topology, see for example pages 16 and 21. What makes that approach very useful is that the authors provide examples for each area in the diagram ("landmarks" ...

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$$y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$$ So $$y^{'}(x)=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$ import these power series to main differential equation $y^{'}(x)=1+xy(x)$to receive to $$\sum_{n=0}^{\infty}na_{n}x^{n-1}=1+\sum_{n=0}^{\infty}a_{n}x^{n+1}$$ by simplifying this equation: $$a_{1}=1 , a_{3}=\frac{1}{3}, a_{5}=\frac{1}{15}, \cdot\cdot\cdot,a_{2n-1}=... 1 If the conditions of the Picard–Lindelöf theorem are satisfied then the function sequence (y_n) defined iteratively by$$ y_0(x) = 0 \, , \\ y_{n+1}(x) = Ty(x) = x + \int_0^x t y_n(t) \, dt $$converge to a solution of the initial value problem. This is called Picard-iteration. The first iterates are$$ \begin{align} y_0(x) &= 0 \\ y_1(x) &= x + ...

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Let us define a $T: X \to X$ by the following rule of assignment: let $x \in X$, then for all $n \in \mathbb{N}$ the $(n+1)$-th term of $T(x)$ is given by $\frac{x_n}{2}$, and the first term of $T(x)$ is 1. That is: $$x \mapsto \bigg(1, \frac{x_1}{2}, \frac{x_2}{2}, \ldots \bigg).$$ Note that for all $x,y \in X$, $$\|T(x) - T(y) \| < \gamma \|x-y\|$$ ...

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In general $f$ is far from unique. But for example it's unique if $E$ is a Hilbert space.

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If you are already familiar with the supremum norm, you could introduce weighted variants hereof. For example for any $\alpha<0$ you can define \begin{align*} ||u||_\alpha := \sup_{x\in\mathbb{R}}\, (1+|x|)^\alpha\,|u(x)|. \end{align*} These norms will generate different topologies for different $\alpha$. Consider for example the sequence $u_n(x):=\chi_{[... 2 For 1, take$\mathcal H=\ell^2(\mathbb Z)\oplus\mathbb C$, and define$U_n$by$U_n=U^n\oplus 1$, where$U\in B(\ell^2(\mathbb Z))$is the unilateral shift. Then$(U_n)$converges weakly to$0\oplus 1$, which is a rank-$1$projection. For 2, this cannot happen. For if$P$is a non-trivial projection, there exists a non-zero$x\in\ker P$. Thus$\|Px\|=0$, ... 6 Your profile says you are a PhD Candidate, so perhaps you are interested in some more details. Also maybe this answer is a bit off-topic and a bit overly advertising! But I found the things below extremely helpful for my own understanding how mathematics can be structured (digitally). I would like to elaborate on user87690's answer. They are correct that ... 1 The only bounded operator such that and$||Ax||_H=\langle Ax, x \rangle_H$for all$x$is the zero operator. Proof: replace$x$by$n$and divide by$n^{2}$. You get$\frac 1 n ||Ax||_H=\langle Ax, x \rangle_H$. Let$n \to \infty$to get$\langle Ax, x \rangle_H=0$. Go back to the given equation to get$Ax=0$. This is true for every$x$so$A=0$. Answer ... 9 Ad the issue with inner product Banach space vs. Hilbert space: Every inner product space induces a norm and every norm induces a metric. A Banach space is a normed vector space such that the induced metric is complete. A Hilbert space is an inner product space such that the induced metric is complete. So in your diagram, Hilbert spaces should really be the ... 2 What does the Hahn-Banach theorem guarantee in terms of uniqueness? What exactly do we know about the functional$f$? The answer is "very little". We know the value it takes at$x_0$, and hence on the linear subspace spanned by$x_0$. The next natural question "what are possible candidates for the kernel of$f$?" There are probably a lot of these! Any ... 11 My advice is to place a lot more landmarks like$\mathbb R^n$. Ideally, every area should have at least one point in it, which will serve to prove that the area really belongs there. It will also clarify what the relationships really mean. For example, all manifolds are metrizable, but not uniquely. So if you want "manifolds" to extend outside of "metric ... 2 We use the max norm $$|| T(y) - T(z) ||_{\infty} = \max_{x \in [-1, 1]} | T(y)(x) - T(z)(x) |$$ For any$x \in [-1, 1]$, we write $$| T(y)(x) - T(z)(x) | = \left| \int_{0}^{x} t( y(t) - z(t) )dt \right|$$ and $$| T(y)(x) - T(z)(x) | \leq \left| \int_{0}^{x} t dt \right| ||y-z ||_{\infty} \leq \frac{1}{2} ||y-z ||_{\infty}$$ 1 If$(y_n)$is an orthonormal set in a Hilbert space then there are several ways of characterizing completeness of this orthonormal set. One of the equivalent properties is vectors of the form$\sum\limits_{k=1}^{n} a_ny_n$form a dense subset of the Hilbert space. In our case this property is obvious that$x_n \in K(H)$for all$n$,$(x_n)$is orthonormal ... 1 I dont't know how much theory background you have, so I will try to make it as elementary as possible. Let$K$be a nonzero (for$K=0$it is trivial) compact self-adjoint operator on a Hilbert space$H$. Then by the Spectral Theorem, there is an orthonormal system$(x_n)_n$and a system of nonzero numbers$(\lambda_n)_n$(the eigenvalues of$K$possibly ... 1 What you're essentially asking is whether or not the$*$-homomorphism$A\to B_A(X)$is always isometric, or equivalently, injective. As stated in the comments, this is not true for every Hilbert$C^*$-module. I don't know if this property has a name, and it isn't given a name in any of the references on Hilbert$C^*$-modules I have found. Note that if$...

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Yes. The product space $\prod_{[0,1]}[0,1]$ has the product topology (aka the topology of pointwise convergence) and a basic non-empty set has the form $\prod_{x \in [0,1]} U_x$ where all $U_x=[0,1]$ except for a $x$ in some finite set $F \subseteq [0,1]$ when $U_x$ is some non-empty open subset of $[0,1]$. Pick $y_x \in U_x$ for those $x$ and find a ...

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