# Tag Info

2

Let: $B: C[a,b] \times C[a,b] \to \Bbb R$ be the (check that it is) bilinear form given by: $$B(x,y) = \int_a^b x(t)y(t) dt$$ We have that: $$|B(x,y)| \le \int_a^b |x(t)||y(t)| dt \le (b-a)\|x\|_{\infty} \|y\|_{\infty}$$ Hence, $B$ is continuous and in particular $\|B\| \le b-a$. Now, $f(x) = B(x,x)$ is the composition of two continuous functions, ...

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Without the requirement $\|B\|=1$ the exact value of the infimum is known: $$\inf \{\|A-B\| \colon \operatorname{rank} B = p\} = \sigma_{p+1}$$ where $\sigma_1\ge \dots \ge \sigma_n$ are the singular values of $A$, in nonincreasing order. This is the min-max principle for singular values. If you require $\|B\|=1$, the singular value $\sigma_{p+1}$ still ...

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Yes, it does. Indeed, let $g_n(x) : = \inf_{y\in B_{1/n}(x)}f(y)$. Then it is easy to see that $g_n(x)\le g_{n+1}(x)$ and hence $$g(x) = \sup_{n \in \mathbb{N}} g_n(x) = \lim_{n \to \infty} g_n(x) = h(x).$$

2

If $X = \{p\}$ then $X$ is connected. If $X \neq \{p\}$ then $X$ is not connected. This follows from the fact that in a metric space singletons are closed, together with the fact that in a connected space the only sets that are both open and closed are the empty set and the set itself.

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None of the cross terms survive in the following: \begin{align} \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt & = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{n=\infty}^{\infty}A_n e^{int}\sum_{n=-\infty}^{\infty}\overline{a_n}e^{-int}dt \\ & =\sum_{n=-\infty}^{\infty}A_n\overline{a_n}\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{int}e^{-int}dt \\ ...

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The Spectral Theorem for $A$ is given in terms of a Borel Spectral measure $E$ $$Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x,$$ and $x \in \mathcal{D}(A)$ iff $$\int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty.$$ The operator $e^{iA^2}$ is defined through the functional calculus as $$e^{iA^2}x = ... 0 How about a bounded quasinilpotent operator T \in \mathscr{L}(L^2[0,1]) defined by$$ (Tf)(x) = \int_{0}^{x}f(t)dt. $$This operator has \sigma(T)=\{0\}. The resolvent cannot have a pole at 0 because T is not nilpotent of any order. So the resolvent has an essential singularity. You can solve for the resolvent by solving ... 0 A way to interpret the delta function in this context is through an integral. Start with the example of the Fourier transform composed with its inverse.$$ f = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(t)e^{-ist}dt\right)e^{isx}ds. $$This may be correctly written as$$ ...

1

Sure, you can do it this way. If $y \neq 0$ is killed by all of the $f_i$, then if $x$ is in a basic weak neighborhood corresponding to the $f_i$, then $x+ty$ is also in this neighborhood for all $t$, since $f_i(x) = f_i(x+ty)$. In particular, the weak neighborhood will be unbounded. Ultimately, I think the point is that the kernel of a linear functional ...

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This is in fact never true; suppose you have a matrix $A$ of rank $k$. Then, we can take linearly independent $v_1, \dots, v_k \in \text{Ran}(A)$, so there exist $u_1, \dots, u_k$ such that $Au_i = v_i$ for each $i$. Suppose for a contradiction the statement is true for some $p < k$, so there is a sequence $(B_n)$ of $n \times n$ norm 1 matrices of rank ...

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Orthornormality refers to the basis $e_i$. When a basis is orthonormal it means the inner product between any two elements of the basis $e_i,e_j$ is $\langle e_i, e_j \rangle = \delta_{ij}$ (see kronecker delta). More generally, two vectors $u,v$ are orthogonal if $\langle u, v \rangle = 0$. The normality part comes from elements of the basis having norm ...

2

The indices are different because the sums are independent from one another. If one intends to multiply the sums, then this distinction is a critical one. Suppose that we were to multiply the summations $\sum_{n=1}^2a_n$ and $\sum_{n=1}^2b_n$ and naively failed to make this distinction. Then, we would have incorrectly \begin{align} ... 1 That's because it is notationally bad and depending on how you read it, it may even give different results. For example, consider simple vectors in \mathbb{R}^{3} given by v=a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}=\sum_{n=1}^{3}a_{n}e_{n}. Then if we use same index, \begin{aligned} ... 0 No. The determinant is a continuous map on \mathbb{R}^{n\times n}. If A is a matrix with nonzero determinant (full rank), then any matrix B sufficiently close to A will also have nonzero determinant (and thus will also be full rank), by continuity. 0 You have already basically proved what you want when you wrote |T_n(x)| \le \left(\sum_{k=1}^{n}|a_k|^2\right)^{1/2}\|x\|. All you have to note is that \{ a_k \} \in \ell^2, which gives |T_n(x)| \le \|a\|_2\|x\|_2 for all n and fixed x. That's enough to apply the uniform boundedness principle. The uniform boundedness principle is the simplest way ... 3 An alternate approach is induction on n=\dim(W). The base case n=0 is clear, so the hard part is the induction step. For this, it's enough to prove the following result: if M is a closed subspace of V and x\in V,x\not\in M, then M+\mathbb{C}x is also a closed subspace. Indeed, by the Hahn-Banach theorem there is a continuous linear functional ... 0 This is not true. Let A=C[0,1]. Consider the increasing sequence of closed sets closed sets F_0=\varnothing, F_i=[0,2^{-i}], and set I_i=\left\{f\in A:f|_{F_i}=0\right\}. Then we are in the setting you described. We can identify A/I_i with C(F_i), via the map A/I_i\to C(F_i), f+I_i\mapsto f|_{F_i}. With this A/I_{i+1}\to A/I_i is ... 1 HINT (I don't know if it works): Take \phi\in \mathcal S' (a continuous linear functional) with the property that\tag{1}\langle \phi, G\rangle = 0\qquad \forall G(x)=Ae^{\frac{|x-x_0|^2}{2}}, $$where x_0\in\mathbb{R}^n, A\in\mathbb{R} are arbitrary. Claim (to be proved): \phi=0. Once this is proved, the exercise is done by standard duality ... 4 I’d attack it much more directly. HINT: Suppose that a=\langle a_n:n\in\Bbb Z^+\rangle\notin\ell_\infty. Then a has a subsequence \langle a_{n_k}:k\in\Bbb Z^+\rangle such that |a_{n_k}|\ge k for each k\in\Bbb Z^+. For each k\in\Bbb Z^+ let$$x_{n_k}=\frac1{ka_{n_k}}\;,$$and let all the other terms of x be 0. Show that x\in\ell_1, ... 4 Notice that for each vector x, one has$$\|T^* T^2(x)\|^2 = \langle T^* T^2(x),T^* T^2(x) \rangle = \langle TT^*T^2(x), T^2(x)\rangle = \langle T^*T^3(x),T^2(x) \rangle = \langle T^3(x),T^3(x) \rangle = \|T^3(x)\|^2.$$Thus$$\|T^3\| = \sup_{\|x\|=1} \|T^3(x)\| = \sup_{\|x\|=1}\|T^*T^2(x)\| = \|T^*T^2\|.$$0 In polar coordinates, since the spherical surface form is R\sin \theta d\theta d\phi , in order for the integration to be one, the dirac delta should be: 1/(R\sin \theta_0) \delta(\theta-\theta_0,\phi -\phi_0) 1 Diagonalization will help here. (When in doubt and working with normal matrices, try utilizing diagonalization!) Write T = UDU^*, then T^* = UD^* U^*, giving that T^*T^2 = UD^*D^2U^*. However T^3 = UD^3 U^*. Since unitary conjugation does not change the operator norm, this boils down to considering D^*D^2 and D^3. Here Dg(x) = f(x)g(x) for ... 2 Yes, it is correct. The operators T_N are of finite rank (hence compact) and converge to T. Thus, T is compact as well. 3 This is studied in potential theory: the function u is the Newtonian potential of f,$$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$where K(x)=c_n|x|^{2-n} for n\ne 2 and K(x)=c_2\log|x| for n=2. In dimensions n\ge 3 the kernel K decays at infinity, so u(x)\to 0 as |x|\to\infty in this case, provided f is reasonable (integrable and ... 1 Apart from a few typos, your derivation looks correct. There should be e^{-\alpha^2 t} in the final equation and I get x + \tan(x) = 0 (which is \cot(x) = -\frac{1}{x}), as the defining equation for the \alpha_n's. This follows from$$X(x) = \sin(\alpha x) \implies 0 = X(1) + X'(1) = \sin(\alpha) + \alpha \cos(\alpha) \implies \tan(\alpha) + \alpha ...

1

Let $C=\max\{\|L\|_{\infty}^2,1\}$. Assuming that $\alpha<\frac12$, we may compute \begin{align} \int_0^1\int_0^1|k(x,y)|^2\ \mathsf dy\ \mathsf dx &= \int_0^1\int_0^1 L(x,y)^2|x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\\\ &\leqslant C \int_0^1\int_0^1 |x-y|^{-2\alpha}\ \mathsf dy\ \mathsf dx\\ &= 2C\int_0^1\int_0^x (x-y)^{-2\alpha}\ \mathsf ...

0

always with the same problem, i want to calculate the solution with separate variable methode. That what i try: we put $$u(x,t)= X(x) T(t) \neq 0$$ then we have $$\dfrac{T'(t)}{T(t)}= \dfrac{X"(x)}{X(x)}$$ So we obtain twe equations $$T'(t) + \lambda T(t)= 0$$ and $$X''(x)+ \lambda X(x)=0, X(0)=0, X(1)+ X'(1)=0$$ this second equation admits eigenvalues ...

0

Your question seems to deal with convergence of sequences of functions. In your case, you have the sequence $(f_n)_{n\geq 0}$ where $f_n(x)=|x|^n+(-1)^n$ and you seem to be interested in the function this sequence converges to and in what sense (uniform, pointwise). Hence, the literature is convergence of function sequences.

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Take any nontrivial, nonnegative, symmetric function $g \in C_c^\infty (\Bbb{R}^d)$. If we let $h := \mathcal{F}^{-1}(g)$, then $h$ is real-valued (why?) and $$h(0) = \int g(x) \, dx > 0,$$ since $g \geq 0$ and $g \not \equiv 0$. By continuity of $h$ and by rescaling (i.e., replace $g$ by $Cg$ for some large $C>1$), we get $h \geq 1$ on $B_{2\delta} ... 0 You have the following typos: the display after "yields" should read $$\|f_n(x)-f(x)\|<\epsilon$$ because you've taken the limit for$m\to\infty$with$n$fixed and you've used the continuity of the norm in$X$. You're missing a$\|$bracket before$f(x)\|$elsewhere. Now using the inequality in 1. and the triangle inequality (in the form ... 2 Define$\phi:(B+I)/I\to B/B\cap I$by $$\phi(b+j+I)=b+B\cap I,\ \ \ \ \ b\in B,\ j\in I.$$ Of course we need to check that this is well-defined. If$b_1+j_1=b_2+j_2$, then $$b_1-b_2=j_2-j_1\in B\cap I,$$ so$b_1+B\cap I=b_2+B\cap I$. The map is obviously linear, multiplicative,$*$-preserving, and onto. As for injectivity, if$b_1+B\cap I=b_2+B\cap I$, ... 1 Take a positive, nonzero$u \in C^{\infty}_c(\mathbb{R}^d)$and consider$\{u_{\epsilon}\}$, where$u_{\epsilon}(x) = u(x\epsilon^{-1})$. For simplicity, let$s = 1and notice that (by a change of variables) $$\|u_{\epsilon}\|_{L^{2^*}} = \Big(\int |u_{\epsilon}|^{2^*}\Big)^{\frac{1}{2^*}} = \epsilon^{\frac{d}{2^*}}\|u\|_{L^{2^*}}$$ $$\|\nabla u\|_{L^2} = ... 2 Given f and \epsilon, choose a polynomial p with \Vert f-p\Vert_{\infty,X}<\epsilon (where \Vert\cdot\Vert_{\infty,X} is the supremum norm oin X). Now see the corresponding polynomial function in \mathcal{A}, p:\mathcal{A}\to\mathcal{A}. (Remember: the functional calculus respects this notation, i.e., p(a), in the functional calculus, is ... 1 If you have y \in H^1(\Omega) \cap C(\bar\Omega) you can use a truncation argument to obtain x \in H^1(\Omega) \cap C_c(\bar\Omega) such that \|x-y\|_{H^1} is arbitrarily small. 2 Defining \phi_\lambda(x)=\phi(\lambda x) for smooth \phi, the requirement is$$ u(\phi_{\lambda})=\lambda^{-m-d}u(\phi)\quad\forall\phi\in C_0^{\infty}. $$If u happens to be a continuous function (and hence u(\phi)=\int u(x)\phi(x)), this is equivalent to what you wrote. 0 Note: There is nothing about completeness of \mathcal{H} needed to carry out of the following steps. Because P_n is monotone, then (P_nx,x) is monotone in n for each fixed x, and is bounded above by (x,x), which forces convergence of \lim_n(P_n x,x) for all x. Then, using polarization, the following expression must also have a limit in n ... 0 I haven't worked out the details - also I suspect we still haven't been given all the relevant definitions. But in case it helps, here's how the argument "must" go in outline, if it's by R-N: Somehow we reduce to the case \mu(\Omega)<\infty. Define a complex measure \nu by$$\nu(E)=x^*(\chi_E).$$Detail: Something shows somehow that \nu is in fact ... 0 As suggested by daw, take u(x) = 1. If you have some regularity of \Omega, then$$\int_{\mathbb R^n} u \nabla v \, \mathrm{d}x = \int_{\partial\Omega} \frac\partial{\partial n} v \, \mathrm{d}sfor v \in C_c^\infty(\mathbb R^n). The right hand side cannot be written as \int_{\mathbb{R}^n} w \, v \, \mathrm{d}x for some w \in ... 2 You want to show: \int_a^b \frac{1}{L} e^{-2 \pi inx/L}e^{2\pi imx/L}dx=0 \begin{align} <a_n|a_m> & = \int_a^b \frac{1}{L} e^{-2 \pi inx/L}e^{2\pi imx/L}dx \\ & = \frac{1}{L} \int_a^b e^{2 \pi i x(m-n)/L} dx\\ & = \frac{1}{L} \frac{L}{2 \pi i(m-n)} \left( e^{2 \pi i (m-n) b/L}-e^{2 \pi i(m-n)a/L}\right) \\ & = \frac{1}{2 \pi ... 2 Example, A = \frac{1}{i}\frac{d}{dx} $$on the domain \mathcal{D}(A) of absolutely continuous functions f \in L^2[0,1] for which f' \in L^2[0,1] and f(0)=0. Then A^* is the same as A except that the condition f(0)=0 is replaced by f(1)=0. Then A^{\star\star}=A because A is closed and densely-defined. However, ... 1 Setting v=u_1-u_2, define I(t)=\frac{1}{2}\int_0^1v(x,t)^2dx. Then$$\frac{dI}{dt}=\int_0^1v(x,t)v_t(x,t)dx=\int_0^1vv_{xx}dx$$Integrating by parts, this becomes:$$\frac{dI}{dt}=\left[vv_x\right]_0^1-\int_0^1v_x^2dx$$Now, note: vv_x\vert_{x=0}=0,vv_x\vert_{x=1}=-v(1,t)^2 by the conditions given, so: ... 1 If \Omega is bounded, it suffices to take u(x)\equiv 1. Then \hat u cannot be in W^{1,p}(\mathbb R^n) for p>n since it is discontinuous. (Sobolev embedding) 0 Consider the two cases of V_n being monotone increasing and monotone decreasing in n. In the first case you have P_n P_{n+1}=P_n and in the second you have P_n P_{n+1}=P_{n+1}. At any rate you have either P_n(P_{n+1}- P_{n})=0 or P_{n+1}(P_{n+1}-P_{n})=0. First consider V_n is increasing. So (P_{n+1}-P_n)(z) is in V_n^\perp. This means ... 1 Assume V_n \subseteq V_{n+1} for all n. Then P_{n+1}P_n=P_n. Using adjoint, P_nP_{n+1}=P_n must also hold because P_k^*=P_k for all k. Therefore, for all x,$$ (P_nx,x) = (P_nx,P_nx)=\|P_nx\|^2=\|P_nP_{n+1}x\|^2 \le \|P_{n+1}x\|^2=(P_{n+1}x,x). $$0 In every metric space (X,d) a sequence x_n converges to x if and only if every subsequence x_{n_k} has a further subsequence converging to x (easy proof by contradiction). All you need to now is thus that convergence in measure is convergence in a metric space (e.g. d(f,g)=\int \min\lbrace 1,|f(x)-g(x)|\rbrace \, d\mu(x) is a suitable metric on ... 2 A linear operator is continuous if and only if it is bounded. By one of the comments above, it is possible to show that the sequence a_k must be bounded (for otherwise, Tx for x\in\ell^1 would not itself be an element of \ell^1). Therefore, |a_k|\leq C for some C\geq 0. Next, recall that a linear operator T:\ell^1\rightarrow\ell^1 is ... 2 By the monotone convergence theorem, you always will have$$\int f^p = \lim_{k \rightarrow \infty} \int f_k^p$$Taking pth roots (and using continuity of the function x \rightarrow x^{1 \over p}), one therefore has$$||f||_p = \lim_{k \rightarrow \infty} ||f_k||_p \tag 1$Since the$f_k$increase to$f$, the limit in$(1)$is an increasing limit. So one ... -1 Oh, I just read your updated version of the question. If fk->f in Lp, we know that f is in Lp. Considering fk is a non-negative increasing sequence of functions, to converge at f it must be that sup(k) ||fk||p=||f||p. f is in Lp, so ||f||p is finite. -1 Well, fk is a sequence of increasing functions to f, so with sufficient work you should be able to show that sup(k) ||fk||p = ||f||p. 5 You need to show the three vectors are linearly independent. In this case I would use this trick; so that you don't need to worry about them being functions and the equality to hold for every value of$x$. If you consider$D: \mathcal{F} \rightarrow \mathcal{F}$, the derivative operator, is an endomorphism in$\mathcal F\$ (i.e. a linear map from ...

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