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Too long to be a comment. Spectrum of a non-closed operator is not defined. If A is not closed, then $A−z, z\in\mathbb C$ is never closed and $(A−z)D(A)$ cannot be equal to $H$. So far you have shown that $z\notin \overline{N(A)}$ implies that $(A-z)^{-1}$ is bounded on the range of $(z-A)$ which is closed. But the $Ran(z-A)$ need not to be the whole $H.$ ...

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(b) is easy enough and you should be able to do it. For (a), do you know the identity $$f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}?$$

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Some hints. Hint 1: Since $f$ is continuous, for any $\epsilon\gt0$, there is a $\delta\gt0$ so that $$|x-\bar{x}|\le\delta\implies|f(x)-f(\bar{x})|\le\epsilon$$ Hint 2: $x\in\partial B_a(\bar{x})\implies|x-\bar{x}|=a$. Hint 3: $$\left|\int_{\partial B_a(\bar{x})}f(x)\,\mathrm{d}x-\int_{\partial ... 0 I won't provide you with a finished proof, but I'll provide you with an outline of such a proof. First, look at the geometric situation. The left-hand side integrates f over the surface of a ball with radius x around \bar{x}, and scales the result by the area of that surface. Now, assume for for a second that f is constant within B_a(\bar{x}). ... 0 Let y=f² y'=2ff' f*(1-f'²)=2xf' (4f²-4f²f'²)=8xff' 4y-y'²=4xy' y'²+4xy'=4y y=ax+4b a²+4ax=4ax+4b a²=4b y=ax+(a²/4) f(x)=(+or-)sqrt(ax+(a²/4)) a=constant 2 The statement is false. Let Q=\mathbb{Q}\cap [0,1]. Since Q is countably infinite, we can find a bijection \phi:\mathbb{Z}\to Q. Define f:\mathbb{R}\to Q to be the function with the value \phi(z) on the half-open interval (z,z+1]. It is not hard to show that f is measurable and it is clearly bounded. Let s:\mathbb{R}\to\mathbb{R} be a ... 0 I posted this same kind of analysis for someone studying the Poisson kernel a couple of days ago. For vectors x,y \in \mathbb{R}^{d},$$ |x|^{2}=|x+y-y|^{2}=|x+y|^{2}-2(x+y)\cdot y+|y|^{2} \le |x+y|^{2}+2|x+y||y|+|y|^{2}. $$Because 2ab \le a^{2}+b^{2} for any real numbers a, b, then$$ |x|^{2} \le ...

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We start with $\displaystyle \zeta(s,z) = \sum_{n=0}^{\infty} \frac{1}{(z+n)^{s}}$. First, substitute in $mz$ for $z$: \begin{align} \zeta(s,mz) &= \sum_{n=0}^{\infty} \frac{1}{(mz+n)^{s}} \\ &= \frac{1}{m^s}\sum_{n=0}^{\infty} \frac{1}{(z+\frac{n}{m})^{s}}. \end{align} Now write $n = n'm + n''$, where now $n'$ will range from $0$ to $\infty$ ...

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Alternate proof if you know that compactness is equivalent to sequential compactness: suppose $Y$ is compact, and $x$ is a limit point of $Y$. Let $(x_n) \subset Y$ with $x_n \to x \in X$ and for all $n$, $x_n \neq x$. $(x_n)$ has a convergent subsequence converging to some $y \in Y$ because $Y$ is compact. But since the whole sequence $(x_n)$ is ...

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Let $F$ be our compact set. Then take $x\in X\setminus F$. Because {$x$} and $F$ are compact ,then they have a distance,say $d$.And thus the open disk $D(x,\frac {d}{2})\subset X\setminus F$. So $X\setminus F$ is open and thus $F$ is closed.

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The problem with your reasoning is that $K$ may have an open cover $\mathscr{U}$ that does not arise from an open cover of $Y$, so that compactness of $Y$ doesn’t tell you anything about $\mathscr{U}$. It’s easiest, I think, to prove the contrapositive: if $A\subseteq X$ is not closed, then $A$ is not compact. Let $\langle X,d\rangle$ be a metric space, and ...

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If you accept that both $e^x$ and $lnx$ are continuous, then, working with $x$ in $(0, \infty)$ you can do this: $x^{1/5}=e^{(1/5)lnx}$ , is the composition of the two continuous functions $e^x$ and $\frac{1}{5}lnx$ , so, as the composition of two continuous functions, it is continuous.

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It may help if you unfold the definition of limit in pointwise convergence. Then pointwise convergence means that for each $x$ and $\epsilon$ you can find an $N$ such that (bla bla bla). Here the $N$ is allowed to depend both on $x$ and $\epsilon$. In uniform convergence the requirement is strengthened. Here for each $\epsilon$ you need to be able to find ...

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$f_n\to f$ pointwise on $(a,b)$ if for each fixed $x\in(a,b)$, $|f_n(x)-f(x)|\to 0$ as $n\to\infty$. Notice this is a pointwise (local) criterion. On the other hand, $f_n\to f$ uniformly on $(a,b)$ if $\sup_{a< x<b}|f_n(x)-f(x)|\to 0$ as $n\to\infty$. This is a "global" criterion in that is requires the maximum of all the pointwise errors on $[a,b]$ ...

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The binomial coefficient is asymptotic $n^{-1/2},$ which means that the series converges absolutely when $|x| < |1+x|.$

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It is not true in general that $|x|^{\frac{1}{5}}<|x|$ (this is true iff $|x|>1$). Here's how I would go about a proof. First treat continuity at $0$ as a separate case ($\delta=\varepsilon^{5}$). Now let $x\in\mathbb{R}$ and $\varepsilon>0$ be arbitrary. Set $\delta=\min\{\varepsilon|x|^{\frac{4}{5}},|x|\}$. We need to do a bit before we ...

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First note that $f(y) > 0$(Why?). Now let $g(x) = \ln(f(a^x))$, where $a>0$. We then have $$g(x+y) = \ln(f(a^{x+y})) = \ln(f(a^xa^y)) = \ln(f(a^x) f(a^y)) = g(x) + g(y)$$ This is the Cauchy function equation and if $g(x)$ is continuous, the solution is $$g(x) = cx$$ Hence, we have $$f(a^x) = e^{cx} \implies f(x) = x^t$$ Note that in the above proof, we ...

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In a convergent alternating series, the sum of the series is between any two consecutive partial sums. Using this idea and the Leibiniz formula, you can get arbitrarily precise upper and lower bounds on the value of $\pi$. You can then use these bounds to bound the error from 3.141. Note that if you just need any upper bound, you can use the fact the ...

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Suppose not. Then there exists a $\varepsilon>0$, such that for every $n\in\mathbb N$, there exist $x_n,y_n\in K\cup A$, with $d(x_n,y_n)<1/n$ and $|f(x_n)-f(y_n)|\ge\varepsilon$. If $K$ contains infinitely many terms of the sequence $\{x_n\}_{n\in\mathbb N}$, then compactness of $K$ guarantees the existence of a convergent subsequence $x_{n_k}\to ... 0 I think I solved the problem (I print screened my solution). If anyone has comments, I would greatly appreciate them. This took me a while. 0 In 2 and 3 dimensions we can observe from example that the derivative satisfies what is required of a tangent vector. If you understand why is that it works in the case$f:R \rightarrow R$, then you will get why it works in the case for 2 and 3 dimensions. For higher dimensions this is just a generalization as you cannot visualize 4d space or higher (I can ... 3 Simply follow the line... For every$(x,y,z)$in$D$, consider$g:t\mapsto g(t)=f(x+t,y+t,z+t)$, then $$g'(t)=(\partial_xf+\partial_yf+\partial_zf)(x+t,y+t,z+t)=f(x+t,y+t,z+t)=g(t)$$ hence$g(t)=g(0)\mathrm e^t$for every$t$such that$(x+s,y+s,z+s)$is in$D$for every$s$between$0$and$t$. There exists some finite$T$depending on$(x,y,z)$such ... 0 Hint: The fact that your function$s(x)$is measurable means that for any measurable set$A\subseteq\mathbb{R}$,$s^{-1}(A)$is a measurable set. What does your function$slook like? We have $$s(x)=\begin{cases}7 & \text{if }x\in E\cap F\\ 5 & \text{if }x\in E\setminus F\\ 2 & \text{if }x\in F\setminus E\\ 0 & \text{if }x\notin E\cup ... 1 Your argument seems fine to me. There's just one ambiguity about one chart 'getting' the whole manifold. By your work it seems as if the chart map is surjective, but really if we know the manifold is compact couldn't a chart miss a few points but still give us enough information to determine the whole manifold. So in that sense we'd have a chart that gets ... 2 HINT:$$\displaystyle \int\frac{1+x^2}{1+x^4}dx=\int\frac{\frac1{x^2}+1}{x^2+\frac1{x^2}}dx$$As \displaystyle\int\left(\frac1{x^2}+1\right)dx=x-\frac1x write \displaystyle x^2+\frac1{x^2}=\left(x-\frac1x\right)^2+2 and set \displaystyle x-\frac1x=u Then use Trigonometric substitution 0 Proof: "\to" Since f(x) \rightarrow l \ as \ x \rightarrow a,for any \epsilon>0,there is \delta>0 so that if |x-a|<\delta,we have$$|f(x)-l|<\epsilon.$$Then for any sequence x_n\to{a},it is cauchy,i.e. for any \epsilon>0 there is N>0 such that |x_n-x_m|<\epsilon for all n,m>N. Let \epsilon=\delta,we will ... 2 As stated, the claim is false as it is equivalent to \;f\; being continuous at \;x=a\; . For example, take$$f(x)=\sin\frac1x\;,\;\;a=0\;,\;\;x_n:=\frac1{2n\pi }$$then$$\lim_{x\to 0}f(x)\;\;\text{doesn't exist, yet}\;\;\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}\sin2n\pi=0$$3$$ \begin{align} \int\frac{1+x^2}{1+x^4}\,\mathrm{d}x &=\frac12\int\frac{\mathrm{d}x}{1-\sqrt2x+x^2}+\frac12\int\frac{\mathrm{d}x}{1+\sqrt2x+x^2}\tag{1}\\ &=\int\frac{\mathrm{d}x}{\left(\sqrt2x-1\right)^2+1}+\int\frac{\mathrm{d}x}{\left(\sqrt2x+1\right)^2+1}\tag{2}\\ ... 3 1) Without loss of generality we can consider the region D as following: $$D = \{(x,y): (x-x_0)^2+(y-y_0)^2 \leqslant R^2 \},$$ where R > 0. It does not change anything dramatically, one can easily prove for a random domain D. It does not really affect the proof. Let us now consider the calculation of A, the integral becomes simple if we use the polar ... 1 it may help you to get a handle on this if you observe that the triangle inequality for a metric implies $$\forall y.\mid f_x(y) - f_z(y) \mid \le d(x,z)$$ 1 If\overline x$is not a strict local minimum, then every neighborhood of$\overline x$contains points$z\neq x$where the value of$f$is$\leq$its value at$\overline x$. Apply that repeatedly, taking as your neighborhoods each of the balls around$\overline x$of radius$1/k$. That gives you, for each$k$a point, called$z$above, but which I'll now ... 0 In general, Fredholm integral equation of first kind arise typically with a compact operator in applications and hence can be notoriously conditioned when solving numerically. Also, sometimes the kernel$K$can in fact be a finite rank operator, in which case you cannot even solve it analytically, let alone numerically. This is one reason why people work ... 1 You did not prove that the "pasted" function (call it, say,$h$) is continuous. You have to prove that$f(x_n)$converges for any$x_n$. Here you proved it for two particular sequences. If you want to use sequences, your proof must start with "Let$x_n$be a sequence in$[0,2]$converging to 1." For instance, you may split this sequence in two ... 2 Concerning the first point, your mistake is I only naively (erroneously) the word in parentheses. You are right,$w$lies in the plane$x_n = 0$. What the author probably thought of was the straight line through$y_0 = h^{-1}(x_0)$with direction$w$, which lies in the plane$x_n = c_0$, and thus its image under$h$is a curve in$S$whose tangent ... 1 Since you can't take sums in metric spaces, I think that the most natural space to ask this question is a Banach Space. If I understood you correctly, you want to answer the following: Question: Suppose$\sum_{n=1}^\infty a_n$is a series in a Banach Space$X$such that every subseries$\sum_{k=1}^\infty a_{n(k)}$converges in$X$. Does ... 0 If f(0) = 0 we are done. If the former is not true then, since f: [0,1] -> [0,1], it must be that f(0) > 0. Applying the same reasoning we conclude that f(1) = 1 (and we are done) or f(1) < 1. Lets asume then that f(0)>0 and f(1)<1 . Take g: [0,1] -> [0,1] such that g(x) = f(x) - x. If 0 belongs to Image(g) then we are done. If it does not, then ... 1 Set$f_n(x) = \frac{1}{n} \sin(nx)$. It converges to zero in$(C^1([0,1]),d_{\infty})$but it doesn't converge to zero in$(C^1([0,1]),d)$, because there is always a point$x$such that$f'_n(x) = 1$. 6 Suppose we know that$\lim_{h \to 0} F(h) = 0$. This means that for every$\epsilon > 0$, there exists$\delta > 0$such that $$|h| < \delta \implies |F(h)| < \epsilon.$$ Fix$x \in \mathbb{R}^n$. If$|t| < \delta/|x|$, then$|tx| = |t||x| < \delta$, so$|F(tx)| < \epsilon$. This says exactly that$\lim_{t \to 0} F(tx) = 0$. In ... 3 The second chain of inequality shows that $$\lim _{h\to 0} \frac{|\lambda (h) - \mu (h)|}{|h|}=0.$$ A necessary condition for the convergence is that$\phi (t)\to 0$if$t\to 0$, where$\phi$is the quotient calculated in$tx$. It's something like “$a_n \to a$implies$a_{n_k}\to a$for each subsequence$a_{n_k}$”. 5 You fix$x \neq 0$and$h=tx$for$t \to 0$. You have proved that $$\lim_{h \to 0} \frac{|\lambda (h)-\mu (h)|}{|h|}=0,$$ so $$\lim_{t \to 0} \frac{|\lambda (tx)-\mu (tx)|}{|tx|}=0$$ as well. But$\lambda (tx)=t \lambda (x)$and$\mu (tx)=t \mu (x)$by linearity, and you conclude. 4 The last inequality is wrong. Just take$n=m+1$and$a_n = -a_m$. Then$|\sum_{k=m}^na_k \alpha| = 0$, but it certainly doesn't follow that$|\sum_{k=m}^na_kb_k| = 0$for all possible choices of$b_m, b_n$. 2 Let $$a_n=\frac{1}{n\,(\log n)^2}$$$\{a_n\}$is summable; use the integral test and substitute$u=\log x$. But$\{a_n\log n\}$is not; again use the integral test with the same substitution. As JessicaK notes, Cauchy's condensation works too: $$2^na_{2^n}=2^n\frac{1}{2^n(\log 2^n)^\varepsilon}=\frac{1}{n^\varepsilon(\log 2)^\varepsilon}$$ where ... 1 For real$x$,$y$, and$t$, $$x^{2}+y^{2} = ((x-t)+t)^{2}+y^{2} = (x-t)^{2}+2(x-t)t+t^{2}+y^{2}.$$ For real$a$and$b$, one has$ab \le |ab| \le \frac{1}{2}(a^{2}+b^{2})$. For$\rho > 0$, one also has$2ab \le \rho a^{2}+\frac{1}{\rho}b^{2}$. Hence, if$\alpha > 0$,$y \ge 0$, and$|t| \le \alpha y$, $$x^{2}+y^{2} \le ... 2 I think the statement should be understood from the perspective of Possion kernel. The classical Possion kernel is given by$$P_{y}(x)=\frac{y}{x^{2}+y^{2}},u(x+iy)=\frac{1}{\pi}\int^{\infty}_{-\infty}P_{y}(x-t)f(t)dt$$(see the wikipedia article http://en.wikipedia.org/wiki/Poisson_kernel) In other words at here both x,y are fixed and only t is allowed ... 2 Take a Cantor set and rotate it so that each of its points becomes a circle. That's K. 2 Construct a Kantor set in the real line, and join each of its points to i and to -i. Alternatively, consider the union of all circles of center 1/n and radius 1/n. 2 Now that you know the answer is \text{ln}(2), perhaps you want to try some kind of estimate using Taylor's Theorem... Note that \xi>0 and so 1+\xi > 1.$$|\sum_{k=1}^{n} \frac{(-x)^k}{k}-\text{ln}(1+x)| \leq \frac{\frac{n!}{(1+\xi)^n}\cdot x^{n+1}}{(n+1)!} < \frac{x^{n+1}}{n+1}$$So in our case (it works in all possible values of x, since ... 4 Your series does not converge to zero - the sequence it sums does, though. (You don't need to prove this, but your series sums to -\text{ln}(2).) Hint: Try grouping successive terms together. In particular, this becomes$$\sum_{n=0}^\infty \left(-\frac 1 {2n+1} + \frac 1 {2n+2}\right)$$See if you can work with that. 1 The formula(s) you want to prove are examples of "the Poisson summation formula", eminently googleable... There are at least two proofs. The more elementary one is to observe that, given Schwartz function$f$, the function$F(x)=\sum_{\lambda\in\Lambda} f(x+\lambda)$for any lattice$\Lambda$is a$\Lambda$-periodic function, so descends to$\mathbb ...

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My understanding of a k-submanifold $C$ of $\mathbb R^n$ is that there must be an open set $U$ in $\mathbb R^n$ , and some chart map $\gamma$ , so that $\gamma (C\cap U)=(x_1,x_2,..,x_k, 0,0,..,0)$, i.e., C is embedded in $\mathbb R^n$ in the same way as the "standard embedding" of $\mathbb R^k$ in $\mathbb R^n$. In the case of a curve $C$, you need an ...

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