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Let $A$ be ${\mathbb C}^2$ with the pointvise multiplication and the involution $(x,y)^*=(\overline{x},\overline{y})$, i.e., continuous functions on two points. But the norm let be $\|(x,y)\|=|x|+|y|$.

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To avoid confusing yourself with sequences of sequences, I recommend thinking of the elements of $C_0$ as functions $f:\mathbb{N} \to\mathbb{R}$ such that $\lim_{n\to\infty}f(n)=0$. Then the claim becomes: if $(f_k)_{k=1}^\infty$ is a Cauchy sequence of functions with respect to the uniform norm, there is $f\in C_0$ such that $f_k\to f$ uniformly. The ...

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I assume you know that a function $F$ of bounded variation on $[a,b]$ can be written as the difference of monotone functions $F=F_{+}-F_{-}$. And I'll assume you also know that $F_{\pm}$ have derivatives a.e., and these derivatives are non-negative where they exist. One pair of functions $F_{\pm}$ can be defined using the variation function $V_{a}^{x}(F)$: ...

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The post by Nick Thompson answer you question. Let me just add that the point here is not that $Tx$ is not in $L^2$. As a matter of fact, the operator maps $L^2$ to itself. The point is that the norm of the operator is not bounded. So when you say "I tried all the obvious candidates for counter-examples and the outcomes live in $L^2$", you are not addressing ...

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Let $\varepsilon > 0$, and $x(s):= s^{\varepsilon-1/2}$. Then $\left\| x\right\|_{2}^{2} =\frac{1}{2\varepsilon}$, and $\left\| T[x]\right\|_{2}^{2} = \frac{1}{2\varepsilon^3}$. So \begin{align} \left\|T[x]\right\|_{2} = \frac{1}{\varepsilon} \left\|x\right\|_{2}, \end{align} i.e., $\left\|T\right\|_{2} \ge \frac{1}{\varepsilon}$, for every $\varepsilon ... 1 "...from our results so far we cannot conclude that the integral is positive." As Martin pointed out in a comment, this does not mean that there are examples to the contrary, only that no proof is yet at hand. 0 I think a way to define$H^{s}$for negative$s$is as follows: $$\mu\in H^{s}\leftrightarrow (1+|\xi|^{2})^{s}(F\mu)(\xi)\in L^{2}$$ and you define the norm on the domain$\Omega$by restriction map. For detail see the Wikipedia article. But it does not seem to me that this norm is explicitly computable in most cases. For example, the Fourier transform ... 0 Hint$X$is compact and statements about the empty set are vacuously true. 0 Hint: Use$K=X$. Then think what$K^c$is. 0 Another place where you can find this is the book of Nigel Higson and John Roe, "Analytic K-homology". It gives a complete account of BDF theory including the Weyl-von Neumann result. 3 1) Take$T:l^p\to l^p,(x_1,x_2,x_3,\ldots)\mapsto(0,x_1,x_2,x_3,\ldots)$(which is your idea for (4)). Then$T$is one to one, because if$T(x)=T(y)$, then$(0,x_1,x_2,x_3,\ldots)=(0,y_1,y_2,y_3\ldots)$and hence$x_i=y_i$for each$i\in\mathbb N$. But$T$is not onto, since there is no sequence$x\in l^p$such that$T(x)=(1,0,0,0,0,\ldots)$. 2) Take ... 0 In finite (fixed) dimension all the norms are equivalent. The function $$x\longmapsto\frac{x}{\|x\|_2}$$ is continuous and bijective form$S_1$(compact by Heine-Borel) to$S_2$, so the inverse is continuous. 1 Note that on$\Omega=\mathbb{R}^2$we find that$H^1_0(\Omega)=H^1(\Omega)$. To see that such a constant$C$can not exists think of the following example. Consider the functions$u_k:\mathbb{R} \to \mathbb{R}\$, defined by $$u_k(x) := \chi_{\{[-k,k]\}}(x) + (k+1 - |x|)\chi_{\{[-(k+1),-k]\cup[k,k+1]\}}(x)$$ and simply make them rotational symmetric to get ...

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Your equation is only of first-order, only first derivatives with respect t only one variable appear there. There is no classification of such equations into elliptic, parabolic, hyperbolic equations. If you talk about systems of first-order equations, they can be classified, if they can be written as second-order equations. E.g. the system $$... 1 It may require too much background for your class presentation, but you might be interested in an application of this to statistical mechanics. See my paper with R.R. Phelps: "Some convexity questions arising in statistical mechanics," Math. Scand. 54(1984), 133-156. The "pressure" is a real-valued convex function on a Banach space of "interactions"; under ... 0 let x_n+M be a cauchy sequence in X/M then x_n is a Cauchy sequence in X and as X is a Banach Space and hence complete.Thus x_n converges to some x\in X.Hence x_n+M converges x+M \in X/M 0 Here is an elementary proof, with your notations. If one among f,g is 0, then choose \lambda = 0. If f\not = 0 and g\not = 0, then you have in fact equality of the kernels as both are hyperplanes, note H this hyperplane. Let e\in H, and D the line generated by e, which is a supplementary subspace of H in V as f and g are non zero ... 1 The heat equation is parabolic:$$ \frac{\partial f}{\partial t} = \frac{\partial^{2}f}{\partial^{2}x}+\frac{\partial^{2}f}{\partial y^{2}}+\frac{\partial^{2}f}{\partial z^{2}} $$Laplace's equation is elliptic:$$ \frac{\partial^{2}f}{\partial^{2}x}+\frac{\partial^{2}f}{\partial y^{2}}+\frac{\partial^{2}f}{\partial z^{2}} = 0 $$The ... 2 Yes. Note that f is a function not just of dimension but of the volume of the domain. 1 Let e_n denote the element of \ell_1 whose m'th coordinate is 0 if m\ne n and 1 if m=n. To see more directly why the bounded linear map, referenced in my comment to the main post, T:\ell_1\rightarrow\ell_1 defined by$$Te_1=e_1;\ Te_n=e_1+e_n, n>1$$is not an adjoint operator, consider the sequence (e_n) in \ell_1. Note for any ... 3 The first condition you will want to impose is that if N is a null-set, then so is g^{-1}(N). To see this, note that in general \chi_{M} \circ g = \chi_{g^{-1}(M)}, where \chi_{M} is the characteristic function/indicator function of the set M. Hence, if there was a null-set N \subset [a,b] such that g^{-1}(N) is not a null-set, then you would ... 1 HINT: Use the change of variable formula to compare$$\int_I |f( s)|^2 d s= \int_I |f( \phi(t))|^2 | \phi'(t)|\, d t \ \text{and}\\ \int_I |f( \phi(t))|^2 d t \\ $$0 Indeed, each continuous bijection from a compact topological space into a Hausdorff space is homeomorphism(See topological books). Similarly for open functions. 0 The function \psi:\>X\to{\mathbb R} defined by$$\psi(f):=\int_0^1\left|{\rm Im}\bigl(f(x)\bigr)\right|\>dx$$is continuous on X:$$|\psi(f)-\psi(g)|\leq\int_0^1|f(x)-g(x)|\>dx=\|f-g\|\ .$$Therefore S:=\psi^{-1}(0) is closed in X. 0 No, you can't change the order of differentiation in the term$$ \frac{d}{dx} \frac{\partial J}{\partial u'}. $$The idea here is that u and u' are functions of x, and the notation (writing \frac{d}{dx} as a total derivative instead of a partial derivative) is supposed to suggest that the \tfrac{d}{dx} sees that x-dependence. For instance, if ... 1 Take any point y\in (x-\delta(x),x+\delta(x)) and consider r=min\{y-(x-\delta(x)),x+\delta(x)-y\} Then r>0 and then consider the open interval \left(y-\frac{r}{2},y+\frac{r}{2}\right) and try to prove that this lies entirely in your given set. 1 Hint: Show the complement is open. Show that the nearest element of X to f is \mathrm{Re}\,f. 0 Hints: (a) Pe_j=e_j for all j implies Pf=f for f\in F. (b) If (f-g) \perp F, then (f-g,e_j)=0 and P(f-g)=0 by definition. Also use (a) (c) f = Pf+(f-Pf). Let u=Pf and v=f-Pf. If u,u'\in F, v,v' \perp F with u+v=u'+v' then (u-u')=(v'-v) is orthogonal to itself. (d) Use (c) to decompose f=u+v. Show u \perp F^{\perp} so f-u = ... 1 DIFFERENTIATING AT THE ORIGIN. Take the vector space to be the ordinary (x,y) plane. In polar coordinates, restricting to r \geq 0, let$$ f(r, \theta) = r \sin 3 \theta . $$A line through the origin is given by fixing a value of \theta; doing so, the Gateaux ratio is just the constant \sin 3 \theta , so that is also the limit. The Gateaux ratio ... 1 Let A be an element of F\setminus F_n (i.e., A has at least n+1 elements and may even be infinite). Pick n+1 distinct points a_0,\ldots, a_{n}\in A and let r=\min_{0\le i<j\le n}p(a_i,a_j)>0. Now let B be any element of F_n. Then for any b\in B there exits at most one i with p(a_i,b)<\frac r2, hence by pigeon-hole there ... 0 To prove that F_n is closed. Take an element K not in F_n, then we can take n+1 points x_1,x_2,...,x_{n+1} in K. Let 2r be the minimum distance between any two of these n+1 points. Then$$B(K,r)$$is disjoint from F_n. The reason is that any element J of B(K,r) ought to have a point in the n+1 balls (in X) with centers ... 1 Since$$ p=p^2=(p_1+\cdots+p_n)^2=\sum_{i=1}^{n}\sum_{j=1}^{n}p_i p_j= =p_{1}^{2}+\cdots+p_{n}^{2}+\sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)=p+\sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)$$we have$$ \sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)=0. \tag5 $$Without loss of generality we may assume that A is a C^*-subalgebra of B(H), where H is a ... 0 let f(x) = \dfrac{a_1}{x - \lambda_1} + \dfrac{a_2}{x - \lambda_2} + \dfrac{a_3}{x - \lambda_3}. you can verify that f is continuous everywhere except at \lambda_1, \lambda_2 and \lambda_3. at these points f has a vertical asymptote. then$$\lim_{x \to \lambda_1 + }f(x) = \infty, \lim_{x \to \lambda_2 - }f(x) = -\infty.$$therefore by ... 0 Suppose that x\in V\cap\operatorname{cl}Y, and let U be an open nbhd of x. V is an open nbhd of x, so without loss of generality we may assume that U\subseteq V. Then U\cap Y\ne\varnothing, since x\in\operatorname{cl}Y, so$$U\cap(V\cap Y)=U\cap Y\ne\varnothing\;,$$and x\in\operatorname{cl}(V\cap Y). 2 Just as an example, consider the operator \frac{d^{2}}{dx^{2}} on the set \mathscr{D} of all twice absolutely continuous functions f \in L^{2}[0,2\pi] with f'' \in L^{2}[0,2\pi]. Then T_{\alpha,\beta}=\frac{d^{2}}{dx^{2}} is selfadjoint on the domain \mathcal{D}(T_{\alpha,\beta}), 0 \le \alpha,\beta < \pi consisting of all f \in ... 1 No, it is not. The reason why is that, if \mathbb{K} is complex, then the inner product is conjugate linear and therefore only induces an anti-isomorphism, not an isomorphism, between H and its dual. We find that y is identified with the conjugate of the gradient in this case. To be more concrete, pick H = \mathbb{C} (over \mathbb{C}). Then pick ... 1 So does the author just mean that the basis is not a Hamel basis? Yes, precisely that; "a basis, in the sense of algebra" is a Hamel basis. I think that a total orthonormal sequence must be a Schauder basis Yes, that is correct, but a total orthonormal family need not be countable, in contrast to a sequence, and if you have an uncountable ... 1 The standard definition requires A to be bounded. There are many text on nonlinear analysis, like this one. 0 In you post, I read C is bounded in BV norm, i.e., the function of bounded variation. If you know Sobolev space, you could just think BV is the relaxation of W^{1,1} and hence similar to W^{1,1} space, bounded sequence in BV will compact in L^1, and even in L^q for any q<N/(N-1) where N is the dimension of space. For reference, ... 0 For some reasons I read that you wanted to have T self-adjoint on one domain and not-selfadjoint on the other. If T is densely defined (its domain is dense in H), then its adjoint exists. If moreover the adjoint is densely defined then T is closable (there exists a closed operator which graph contains the graph of T). It may happen that its ... 2 As a first approach you can do this component wise. The trouble with this is that usually you want to have a definition which does not depend on the choice of a basis, so this is what you would have to check in this case. A (very) general definition can be found in several places, e.g. in Rudin's functional analysis, chapter 3, section 'Vector valued ... 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\}$$for k \in \mathbb{N}^n. It is straightforward to show that \mathcal{D}_K is then a Fréchet ... 2 The problem here is that \delta isn't really a function--rigorously, it's a functional. The idea that \int_{-\infty}^\infty f(t) \delta(t) dt=f(0) for any real function is perfectly fine, but the notion that \delta is some bizarre function that assumes an infinitely tall value in an infinitesimally small region to have unit area can't be made rigorous. ... 0 If I set x_i=|T(e_i)|^{q-1} when T(e_i)>0 and x_i=-|T(e_i)|^{q-1} when T(e_i)<0 then x_iT(e_i)=|T(e_i)|^q and \frac{|T(\mathbf{x})|}{\|\mathbf{x}\|_p}=(\sum_{i=1}^{k}|T(e_i)|^q)^\frac{1}{q}, for \mathbf{x}=(x_1,...,x_k). 2 The result perhaps can be proved inductively showing that \text{index}\,( T- \delta) = \text{index}\,T for \delta of rank 1. A useful particular case is showing by hand that \text{index}( I - \delta) = 0, starting with rank of \delta=1. The different cases presented suggests that there could be a uniform proof. We sketch one below. Break it into ... 1 First note that given any x,n\in L^2(-1,1)  with n\neq 0 there is a unique scalar \lambda and y \in L^2(-1,1) such that x = \lambda n + y and y \bot n. For existence, let \lambda = { 1\over \|n\|^2}\langle n , x \rangle  and y = x-\lambda n. It is easy to check that y \bot n. For uniqueness, suppose \lambda_1 n + y_1 = \lambda_2 n + ... 2 It is a consequence of Poisson summation formula. You just have to prove that, if$$ f(x) = \exp\left(-\pi a x^2+2\pi i b x\right), $$then its Fourier transform is:$$ \widehat{f}(s) = \frac{1}{\sqrt{2\pi a}}\exp\left(-\frac{(2b\pi+s)^2}{4a\pi}\right).$$3 Try something like f_n(x) = ne^{-nx}. Then$$\|f_n\|_1 = \int_0^1 ne^{-nx} \, dx = 1 - e^{-n}$$for all n but \|f_n\|_\infty = n. 6 This is not true. Simply take f = \chi_{[-1,1]} (the indicator function of the interval [-1,1]). Then f \in L^p for all p \in (0,\infty], but if \widehat{f} \in L^1 was true, then Fourier inversion would imply that$$ f = \mathcal{F}^{-1} \widehat{f} \in C_0 $$would be (almost everywhere equal to) a continuous function. This is clearly not the ... 1 Your professor is wrong and Umberto P. is right. If we take$$ f(x)= K_0(|x|) $$where K_0 is a modified Bessel function of the second kind, we have f\geq 0 and:$$ \int_{-\infty}^{+\infty}K_0(|x|)\,dx = \pi,\qquad \int_{0}^{+\infty}K_0(|x|)^2\,dx = \frac{\pi^2}{2},$$so f\in L^1\cap L^2(\mathbb{R}), but:$$ \widehat{f}(s) = ...

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