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Found a counterexample (finally!). Let $A = \left(\begin{array}{rr} 1 & 0 \\ 4 & 1\end{array}\right)$ and $B = \left(\begin{array}{rr} 1 & -1 \\ 0 & 1\end{array}\right)$, then $AB = \left(\begin{array}{rr} 1 & -1 \\ 4 & -3\end{array}\right)$. Then we have that $\sigma(A) = \{1\}$, $\sigma(B) = \{1\}$ but $\sigma(AB) = \{-1\}$. This ...

0

Since $\int_0^1 f_n$ is monotone decreasing in $n$, argue by contradiction, we assume that $$\int_0^1 f_n \geq 2 \epsilon \quad \forall n.$$ By the definition of Riemann integral $\int_0^1 f_n = \sup_{g\leq f_n, \text{ piecewise constant }} \int_0^1 g$, there exists a piecewise constant function $g$ for each $f_n$ such that $$\int_0^1 g \geq ... 1 Look at the complement of K:=\mathrm{supp}(\varphi). This is an open set on which \varphi=0. What is the derivative of \varphi at a point in this set? What does this tell you about the support of \varphi^\prime? 1 Such a function will be trivially impossible unless you relax the requirement that an analytic function be equal to its taylor series in a neighborhood of any point. Once that restriction is relaxed, the following is an example of a piecewise function with all derivatives continuous: f(0) = 0 f(x) = g(x) = e^{(\tfrac{-1}{x^2})} for x < 0 f(x) = ... 0 Let's look at f. In what follows, C_i are functions which are bounded away from both 0 and \infty. The prefactor may be written as$$C_1(x) (1-x)^{-m/2+1/4}$$The integrand may be written as$$C_2(t) (1-t)^{m-1}$$Thus for m \neq 0, the integral may be written as$$C_3(x) (1-x)^m$$while for m=0 we have$$C_4(x) \ln(1-x)$$Thus overall ... 0 Hint: in both cases, show that the involved sequence is bounded in L^2 and that$$\int_{[0,1]} u_k(x) g(x)\mathrm dx\to \int_{[0,1]} u(x) g(x)\mathrm dx$$for each smooth function g, where u is the wanted weak limit. 1 Weak convergence in L^2(0,T;W^{1,2}(\Omega)) means by definition that we have u^*(u_k) \to u^*(u) for any functional u^*\in L^2(0,T; W^{1,2}(\Omega))^* = L^2(0,T;W^{1,2}(\Omega)^*). To see that this implies weak convergence in L^2(0,T;L^2(\Omega)), note that W^{1,2}(\Omega) \to L^2(\Omega) is a continuous injection. Hence, its dual, the ... 4 No, this cannot be done. If \phi\in C_c^\infty(R), then its Fourier transform extends to be an entire analytic function (cf. Paley--Wiener theorem). Should it be constant for (real) x in some neighborhood of x_0, it would be identically constant by the uniqueness theorem for analytic functions, and hence zero, because the Fourier transform of a smooth ... 2 For sequences, there is the following simple characterization: Proposition. Suppose f_n, f \in C_0(X). The following are equivalent: f_n \to f weakly; f_n \to f pointwise and \sup_n \|f_n\|_\infty < \infty. Proof. 1 \implies 2: The evaluation maps f \mapsto f(x) are continuous linear functionals, so f_n \to f pointwise, ... 1 Assume g has support in [-M,M]. Then, for x\in[-N,N],$$ h(x)=\int f(x-y)g(y)dy=\int_{[-M,M]} f(x-y)g(y)dy=\int \tilde f(x-y)g(y)dy $$where \tilde f=1_{[-N-M,N+M]}f, 1_K denoting the indicator function of the set K. Now show that \tilde f\in L^1 and conclude that h\in C^k. 2 Here is an approach. Assume we have the ode$$ a(x)y''+b(x)y'+c(x)y = 0 \longrightarrow (*)$$and we want to have the Sturm Liouville Form. Multiply (*) by the function \mu(x) as$$ \mu\,a y''+\mu b y'+\mu c y = 0 .$$To determine \mu(x) we have$$(\mu a y' )' +\mu c y = 0  \implies (\mu a)' = \mu b  \implies \mu(x) = e^{\int ...

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I believe the correct approach is to divide both sides by $-(1-x^2)$ then use the approach of integrating factor. In this particular case, you need to multiply the equation by $$\exp\big(\int-\frac{x}{1-x^2}dx\bigg)$$ then you should be able to do it. You should replace $x$ by $2x$ in the integrating factor (as in Lengendre polynomial) to see what happens. ...

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You don't want to show that $Y$ is not closed, you want to show that $Y$ is not compact. That is, you don't want a sequence $(f_n)$ of members of $Y$ that converges to a limit $f$ such that $f \notin Y$ - that would prove that $Y$ is not closed, but $Y$ is closed (why?). You want a sequence $(f_n)$ that doesn't have any convergent subsequence and $f_n(x) = ... 0 This was resolved in comments:$\|T^*\|=\|T\|$for every bounded linear operator between normed spaces. In fact, we only need the easier part here:$\|T^*\|\le \|T\|$, which comes directly from the fact that$T^*$acts on linear functionals by composing them with$T$. 0 The region of integration for$\int_{a}^{s}\int_{a}^{u}\cdots\,dt\,du$is a triangle; if you draw the region then you can see how to write the integral as one where the orders of integration of swapped. However, you can approach this purely algebraically so that you don't have to rely on visualizing the region. Let$H(t)$be the Heaviside step function where ... 1 This is true for any measure. See Rudin's Real and Complex Analysis Example 4.5(b) and the proofs referred to there. For any measure space$(X,d\mu)$,$L^2(X,d\mu)$will always be a Hilbert space. Depending on the measure, it may not be separable, or it may be finite dimensional, but it will always be a Hilbert space. (Also, the reference works with ... 1 Lemma. If$g\colon\mathbf R^n\to\mathbf R$is continuous and$g(x)\leqslant 0$for each$x\in A$, then$g(x)\leqslant 0$for each$x\in \overline A$. We use this with$g(x):=|f(x)|-K$and$A:=\Omega\setminus N$. To show the lemma, take$x$in the closure of$A$and$(x_n)_{n\geqslant 1}$a sequence of elements of$A$converging to$x$. Then ... 1 Let$\{ e_{j} \}$be an orthonormal basis of$L^{2}[0,1]$consisting of real functions. Define$f_{i,j}(x,y)=e_{i}(x)e_{j}(y)$. Then$\{ f_{i,j} \}_{i,j}$is a complete orthonormal basis of$L^{2}([0,1]\times[0,1])$; this can be verified by showing that$(f,f_{i,j})=0$for all$i$,$j$for a continuous$f(x,y)$implies$f=0. Notice that \begin{align} ... 2 This is more of a comment, but too long: I think part of the problem is, that you do not empose restrictions on the "measure". It should be straightforward to convince yourself that a positive measure does not suffice in general, so you need signed measures. In your first example, the measure will be something like\mu(A) = \int \phi(x) \chi_{A}(x,x) ... 1 Can we accept as fact the following representation of the delta function? \begin{align} \delta(x-y) = \sum_{j\in \mathbb{Z}} e_{j}(x)\overline{e_{j}(y)} \end{align} If so, then \begin{align} \mathrm{tr}(K) &= \sum_{j\in \mathbb{Z}} \int_{0}^{1}Ke_{j}(x)\overline{e_{j}(x)} \, \mathrm{d}x \\ &= \sum_{j\in \mathbb{Z}} ... 1 Well the biggest problem is that it is a classical result of analysis/topology that the space of continuous functions is closed under theL^\infty$topology i.e. the topology of a.e. uniform convergence (in the case of continuous functions we take a continuous representative, and a convergent sequence in uniformly Cauchy, thus converges uniformly to a ... 1 Taking for simplicity$\Omega$compact, if$C^\infty_c(\Omega)$were to be dense in$(L^\infty,\|\cdot\|_\infty)$then having that$C^\infty_c(\Omega)$is separable because every differentiable function can be approximated by rational polynomials, so would have to be$L^\infty(\Omega)$. 1 The compactly supported part is most important. Try to$L^\infty$-approximate the constant function$g \equiv 1$by a compactly supported function$f$. There will be a set of positive measure where$f$is zero, so$\| f - g \|_\infty$will always be at least$1$no matter what$f$we pick. 0 First we shall assume that the basis are orthonormal, i.e. $$\left\langle {{x_i},{x_j}} \right\rangle = \left\{ \begin{array}{l}1,i = j\\0,i \ne j\end{array} \right.$$ now, merely expanding the arbitrary vector in terms of these basis yields $$x = \sum\limits_{i = 1}^\infty {\left\langle {x,{x_i}} \right\rangle {x_i}}$$ Now, the "energy" of the vector ... 1 Your answer is correct: you need the boundedness of$S_j$. Assuming that, here is a shorter proof. Let$P\subset C^k[0,1]$be the space of polynomials of degree at most$k-1$. The functionals of the form$\int f^{(k)}\,d\nu$vanish on$P$and therefore define functionals on the quotient space$C^k/P$. Also, every functional on$C^k/P$is of this form, by the ... 0 I think I got it now. First of all, the functionals$(S_0,\ldots,S_{k-1})$must be bounded (with respect to the$C^k$-norm), otherwise the conjecture may not hold. Indeed, if$k=1$, then one can construct an unbounded linear functional$S:C^1([0,1])\to\mathbb C$(using Zorn's lemma and the fact that$C^1([0,1])$is infinite-dimensional) that separates the ... 1 The duality$l^p(X)^*=l^q(X^*)$for$1<p<\infty$holds for every Banach space. Indeed,$c_{00}(X)$, the space of finitely supported sequences, is dense in$l^p(X)$. Therefore, every linear functional on$l^p(X)$is determined by its values on sequences with one nonzero element. This identifies such a functional with an$X^*$-valued sequence$(x_n^*)$. ... 0 It's more natural to think of these functions as defined on the unit circle in the complex plane. Then we look at the completion of the linear span of the functions$z^n$,$n=0,1,2,3\dots$. Of course the completion depends on the norm we use. If the uniform norm is used, we get the Disk Algebra. If$L^p$norm is used, we get the Hardy space$H^p$. If by ... 2 The most natural way (to me) is to interpret the space$AC^1$as the space of functions whose first derivative is in$AC$. Remark: the class$AC$, as defined, is precisely the class of absolutely continuous functions. Googling "AC1" together with "absolutely continuous" brought up, for example, this paper in which$AC^{1}$notation is explained ... 0 Hint. We have by Fubini $$\int_a^s \int_a^u y(t) \,dt\,du = \int_a^s\int_t^a y(t) \,du \, dt = \int_a^s (t-a)y(t)\,dt.$$ 0 You have the right idea. Let's introduce$g=(-\Delta)^{-1}f$, so that the claim becomes more readable: $$\int_{\Omega} |\nabla g|^2=\langle -\Delta g,g \rangle_{H^{-1} \times H^1_0}\tag1$$ Recall the general definition of$-\Delta g$: it is the distribution$T$that acts on test functions$\varphi$by $$T(\varphi) = -\int g\Delta \varphi\tag2$$ When ... 3 If$u\in\mathbb L^p$for some$p>1$, then take$u_n:=u\chi_{\{|u|\leqslant n\}}$. If$u$does not belong to any$\mathbb L^p$space for any$p>1$, then it is not possible: if$\lVert u_n-u\rVert_1\to 0$and$(u_n)_n$is bounded in$\mathbb L^p$, then extract a subsequence$(u_{n_k})_{k\geqslant 1}$which converges almost everywhere to$u$. Then using ... 0 What would it mean to say ‘the solution of$u$depends on$F$continuously‘ It means that for every$F$and every$\epsilon>0$there exists$\delta>0$such that for every functional$G$with$\|F-G\|<\delta$we have$|\tilde u-u\|<\epsilon$, where$\tilde u$is the solution of the variational inequality with$G$instead of$F$. how is ... 0 I don't know how the inequality$|(u(t)-v(t))^+| \leq |(u_0-v_0)^+|$could be called, but it certainly implies the comparison principle. Indeed, if$u_0\le v_0$, then the right hand side is zero, hence the left hand side is zero, hence$u(t)\le v(t)$. Maybe it could be called the comparison-contraction principle. 4 Take$X=c_{00}$---the space of all sequences which are almost everywhere$0$and as$x_n$---the sequence having$\frac{1}{2^n}$on$n$-th place and$0$elsewhere. 1 If$E$is reflexive, then is a dual space, namely$E=(E^*)^*$Converse is false:$\ell^\infty$is not reflexive, but is a dual space:$\ell^\infty= (\ell^1)^*$0 By the lemma you cited, you can lift$W_3$to$X_{n+1}$(maintaining convexity, balancedness), and then to$X_{n+2}$, to$X_{n+3}$, etc... If you let$\widetilde{W}^\infty_3$be the union of all those ''lifts'', then$\widetilde{W}^\infty_3\in\mathcal{U}$, so also$\widetilde{W}_3:=\widetilde{W}^\infty_3\cap W_1\cap W_2\in\mathcal{U}$, and$\widetilde{W}_3$... 0 Starting with step 3 in the question above, we need to modify things slightly. First, change the definition of$Z$slightly from the one above. $$Z = \{\alpha x + \beta y : x \in O_1, y \in V, |\alpha| + |\beta| \le 1\}$$ So we are using the condition$|\alpha| + |\beta| \le 1$whereas, above we use$|\alpha| + |\beta| = 1.$It's easy to see: $$Z = ... 2 1) These seminorms are not the same (supremum is not the same as sum), but (for the same n and K) they are equivalent. 2) Accordingly, all three ways of defining convergence in C^\infty(O) are equivalent. Thanks to @Daniel Fisher for the seminorm remark ;) 3 The Fourier transform$$ \hat{f}(s)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ist}\,dt $$is built from functions e^{-ist} which are not in L^{2} on the real line \mathbb{R}. This is an import example physically and mathematically. It tells you have "wave packets" that are in L^{2} even though the components of the pack are not! ... 4 For U \subset V, we have a natural (continuous) injection$$\iota^U_V \colon \mathscr{D}(U) \hookrightarrow \mathscr{D}(V).$$Its transpose,$$\rho^V_U \colon \mathscr{D}'(V) \to \mathscr{D}'(U)$$is called the restriction of the distributions on V to distributions on U. For regular distributions, that corresponds to the restriction of the locally ... 0 Yes it is. Sketch of proof: 1) induction on nonnegative integer k,s (trivial for k=0 and for s=0; after that, induction k\to k+1 using commutation of x with \Delta, and likewise for s\to s+1). 2) For negative k,s by duality. (Omit step 2 if you only need nonnegative k and s.) 3) Now for arbitrary (not necessarily integer) k,s by ... 3 Hint: \Omega=(0,1), f(x)=\log x. 3 One way your identity 1_{A}B^{s}f = B^{s}(1_{A}f) can fail: 1_{A}f may not be in the domain of B^{s} even if f is differentiable. For example, let f be piecewise linear with f(-2)=0 and derivative$$ f'=\chi_{[-2,-1]}-\chi_{[1,2]}. $$Then f \in \mathcal{D}(B^{p}) for 0 \le \rho \le 1. However, ... 6 These are some reasons that I can see at the moment (In the case I recall something else it shall be added to this list): 1) In fact any signal in reality is a function in L^2(I) where I is a time interval, since its energy or power is finite, i.e.$$ \int_I |x(t)|^2 {\rm d}t < \infty $$2) The Fourier series intrinsically means that any periodic ... 0 Same approach as previous answers, using limited operators: If T\colon X \to X is a compact operator and S \colon X \to S is a bounded operator, then both TS and ST are compact. Let S: C^1[0,1] \to C^1[0,1] such that u(t) \mapsto u(\sqrt{t}). Readily we see that \|S\| = 1, so S is bounded. But TS = Id. If T is compact so is Id, but ... 2 A simple example of why this is headed in a wrong direction: If X=\mathbb [0,1] and Y_1=[0,1], then the integral might be defined as normal. If Y_2=[-1/2,1/2], then the integral defined as normal also exists. But, as metric spaces without knowing their "real number" structure, Y_1 and Y_2 are essentially the same metric space, but the integral ... 3 Let \phi, \psi be non-zero solutions of L\phi = 0, L\psi=0, chosen to satisfy \phi'(0)=0, \psi(1)=0. Let w be the Wronskian w=W(\phi,\psi)=\phi\psi'-\psi\phi'. I believe the solution u of Lu=f with u'(0)=u(1)=0 is$$ u = \psi(x)\int_{0}^{x}\frac{\phi(t)f(t)}{w(t)(t-2)}\,dt ... 0 Since my attempt to close the question as a duplicate failed, I'll post an answer: the statement follows from a general theorem on the continuity of Nemytskii operator, which is stated and proved here. 1 Yes, this works for any autonomous first-order equation$u_t = A(u)$, for the reason you stated. A more formal way to express this is to introduce the flow maps$\Phi_t$which are defined so that$\Phi_t(u_0)=u_t$. These maps form a semigroup:$\Phi_{t+s}=\Phi_t\circ \Phi_s\$. So, if you are interested in a property that is preserved under composition (such ...

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