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This is not an answer but it is too long for a comment. mjqxxxx's answer contains all the required steps. Concerning the approximation made for $B$, consider the definition $$B=\frac{(1-c)r}{1-(1-c)r}\exp\left[ \frac{(1-c)r}{1-(1-c)r} \right]$$ and let us define $a=(1-c)r$ which makes $$B=\frac a{1-a}\exp\left[ \frac{a}{1-a} \right]$$ and develop as a ...

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Let $B > 0$ such that $|f(x)| \le B$ for all $x$. Then, $$\left| \frac{f(x)g(x) - f(c)g(c)}{x - c} \right| = \cdots \le B \left| \frac{g(x) - g(c)}{x -c} \right| = B \left| \frac{g(x) - g(c)}{x -c} - g'(c) \right| < B \epsilon$$

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I don't think I understand your proof, but here is one For $\epsilon > 0$, there is $N$ such that $n > N$ gives $|x_n| < \epsilon$. Let $A = \{n \in \Bbb N: \sigma(n) \le N\}$. Let $n_0 = \max A$. Then, for all $n > n_0$, $n \notin A$, so $\sigma(n) > N$, hence $|x_{\sigma(n)}| < \epsilon$, i.e. $|y_n| < \epsilon$.

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The basic idea is correct. Let me propose a more rigorous operationalization. Fix $\varepsilon>0$. Since $x_n\to 0$, there must exist some $N\in\mathbb N$ such that for any integer $n> N$, one has $|x_n|<\varepsilon$. For each $i\in\{1,\ldots,N\}$, one can find a unique integer $m_i$ such that $\sigma(m_i)=i$. Let $M\equiv\max_{i\in\{1,\ldots,N\}}... 0 Fix$\epsilon > 0$. Consider the subset of the image of$\sigma$such that for all$c$in this subset,$x_c$is not within$\epsilon$of$0$. This is finite. In particular, the preimage of this subset has a maximal element, say$M$since$\sigma$is a bijection. Hence for$n > M$,$|y_n|<\epsilon$, QED. 1 You need more than the fact that$A\cap\left\{ \left(p-r,p+r\right)-\left\{ p \right\} \right\} \neq\emptyset$. This only says that the sequence is eventually within the punctured neighborhoods of$p$. You need to show that is eventually entirely within said neighborhoods. In symbols, you need to show:$\exists N\in\mathbb{N}$s.t.$\forall n>N$you have ... 0 Well i work in this: For first Consider the open neighborhood$B(1,r)$. By the archimedean property, there exists some$r ∈ N$so that$r>\frac{1}{n}$. Then$B(1,\frac{1}{r})\subset B(1,r)$and$A\cap\left\{ \left(p-r,p+r\right)-\left\{ p\right\} \right\} \neq\phi$Then p it's limit point. I'm working in the two. 5 Alternatively, show that$Q(x):=\int_0^1\,f(t)\,P(t-x)\,\text{d}t$satisfies$\frac{\text{d}^k}{\text{d}x^k}\,Q(x)=0$for some integer$k>0$. As$P$is a polynomial,$\frac{\partial^k}{\partial x^k}\,P(t-x)=0$if$k:=1+\deg(P). Thus, the Leibniz Integral Rule gives $$\frac{\text{d}^k}{\text{d}x^k}\,Q(x)=\int_0^1\,f(t)\,\frac{\partial^k}{\partial x^k}\,... 1 For A: Let n be such that {1\over n} < r. Then$$ A\cap\left\{ \left(p-r,p+r\right)-\left\{ p \right\} \right\} =A\cap\left\{ \left(1-r,1+r\right)-\left\{ 1 \right\} \right\} $$contains 1-{1\over n} and is therefore not empty. Since r was arbitrary it follows that p=1 is a limit point of A. Try to do the same proof for B. 6 P(X)=a_0+a_1X+\cdots+a_nX^n. P(t-X)=a_0+a_1(t-X)+\cdots+a_n(t-X)^n=a_0+a_1t-a_1X+a_2t^2+2a_2tX+a_2X^2+\cdots \implies f(t)P(t-X)=f(t)a_0+tf(t)-a_1X+t^2f(t)a_2+2a_2tf(t)X+f(t)a_2X^2+\cdots The above is meant for you to visualize what is happening. When you integrate with respect to t, everything alongside the X's will turn to constants. 1 @Wolfy , From Theorem 2.15 we know$$\int \sum_{1}^{\infty}|f_j| = \sum_{1}^{\infty}\int |f_j| $$So, of course, we also know$$\int \sum_{1}^{\infty}|f_j| \leq \sum_{1}^{\infty}\int |f_j| $$but we don't need this inequality since we know the equality holds. Here is the proof with some adjusments / clarifications Theorem 2.25 - Suppose that \{f_j\}... 0 the set F=\{ (x,y) ; f(x)\geq y\} is the complaimentary of your set \{ (x,y) ; f(x)< y\} let (x_n, y_n) a sequence of F whitch converge to (x,y) so we have :$$ f(x_n)\geq y_n \qquad \textrm{for all } n\\ \implies \lim_{n\to \infty}f(x_n)\geq \lim_{n\to \infty} y_n\\ \implies f(x)\geq y $$where for the last line we use the continuity of ... 0 The function g(x,y) = y - f(x) is continuous ,thus g^{-1}(-\infty,0) must be open, proving the statement. 0 Hint: Consider g(x,y)=f(x)-y, g^{-1}((0,+\infty))=\{(x,y):f(x)-y>0\} is open since it is the preimage of an open subset. 1 1) First we show by contradiction that the statement is true for \displaystyle a\ge\frac{e}{e-1}: \hspace{.2 in}Let d\ge ac\ln c, and assume that d<c\ln d. \hspace{.2 in}Then c\ln d>d\ge ac\ln c\implies\ln d>a\ln c=\ln c^a\implies d>c^a, \hspace{.2 in}so c\ln d>d>c^a\implies \ln d>c^{a-1}>\left(\frac{d}{\ln d}\... 2 For a\ne -1, we have using integration by parts$$\begin{align} I(a)&=\lim_{\epsilon \to 0^+}\int_\epsilon^1 x^a \log(x)\,dx\\\\ &=\lim_{\epsilon \to 0^+}\left(\left.\frac{x^{a+1}\log(x)}{a+1}\right|_{\epsilon}^1-\frac{1}{a+1}\int_\epsilon^1 x^{a}\,dx\right)\\\\ &=\lim_{\epsilon \to 0^+}\left(-\frac{\epsilon^{a+1}\log(\epsilon)}{a+1}-\frac{1}{(... 2 Since\|f - f_n \|_p \to 0$, we can extract a subsequence$f_{n_j}$so that$\| f - f_{n_j} \|_p \le \frac{1}{2^j}$. Put $$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$ Then$g \in L^p$and by the triangle inequality we have $$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$ From this subsequence, you can extract a ... 1 Problem 2.3.19 - Suppose$\{f_n\}\subset L^1(\mu)$and$f_n\rightarrow f$uniformly. a.) If$\mu(X) < \infty$, then$f\in L^1(\mu)$and$\int f_n \rightarrow \int f$. b.) If$\mu(X) = \infty$, the conclusions of (a) can fail. (Find examples on$\mathbb{R}$with Lebesgue measure). Proof Initial remark: We know from the definition of integral ... 1 Something like$f(x) = 10 e^{-100 x^2}$on$[-1,1]$The idea is that the Simpson's rule weight heavily the middle point, so if your function is nearly constant but has a very narrow and high spike at the middle point, it will be worse than just taking the endpoints 0 This is how I would tackle this, maybe you can draw some conclusions from that. Let$S$be the set of discontinuities of$f$and assume$\mu(S)>0$. For$\epsilon>0$let$S_\epsilon$be the set of points$x$such that there are sequences$\xi_i\to x$and$\eta_i\to x$such that$\lim f(\xi_i)$and$\lim f(\eta_i)$exist and differ by more than$\...

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The cone $C=\{y\colon Ax=y,x\ge 0\}$ is finitely generated (by finitely many columns of $A$) and convex. By Minkowsky-Weyl theorem (en easy proof via Fourier-Motzkin eliminations can be found here, Theorem 1) it is a polyhedral cone, that is, $C=\{y\colon By\le 0\}$. From the last representation it is clear that $C$ is closed as an intersection of closed ...

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Let us write this expression under the form: $$f(x,y) = -5(x^2+y^2)^2 + 4(x^2 +y^2) + 4 x^2y^2$$ Let us convert this expression into polar coordinates. One gets: $$F(r,\theta) = -5r^4 + 4r^2 + r^4 \sin ^2(2 \theta)$$ (using relationship $\sin (2 \theta) = 2 \sin \theta \cos \theta$). Therefore: $$F(r,\theta) = r^2 (r^2 (\sin ^2(2 \theta) - 5) + 4)$$ ...

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First note that as $c\rightarrow 1$, $(1-c)\rightarrow 0$, and so $$B \sim (1-c)r e^{(1-c)r} \sim (1-c) r.$$ Moreover, for small $x$, $W_0(x) \sim x$. So $$W_0\left[B\left(1+\frac{x}{rc}\right)\right]-W_0[B]\sim\frac{B x}{rc}\sim(1-c)x.$$ Hence $$A=\lim_{c\rightarrow 1}\exp\left[-\left(\frac{1}{1-c}\right)\left((1-c)x\right)\right]=e^{-x}.$$

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Carothers' book that Bungo mentioned seems like it would be a good fit for you, but given what you wrote I think the following might be an even better fit: Introduction to Topology and Modern Analysis by George F. Simmons

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I have made the required adjustments /corrections in your proof. Proposition 2.21 - The set of integrable real-valued functions on $X$ is a real vector space, and the integral is a linear functional on it. Proof - To prove that the set of integrable real-valued functions on $X$ is a real vector space, it is enough to prove that for all $a,b\in \... 1 The old classic General Topology by John Kelley has a strong emphasis on analysis. I prefer it to Munkres for being better structured and more concise. That said, it is still a topology book, and I am uncertain if it is really the case that every young analyst should know all it covers (as suggested by Kelley himself sixty years ago). A few do deem it to be ... 0 If the sequence of non-negative real numbers has the following definition: $$x_{n+m}\le \frac{x_n+x_{n+1}+\cdots+x_{n+m-1}}{m}$$ for$n\in\mathbb{N} \quad m\in \mathbb{N^{>1}}$then it is straightforward to check the two properties. Consider$n=1$$$x_{m+1}\leq\sum_{i=0}^{m-1} \frac{x_{1+i}}{m}$$ Since$x_{m+1}$is no more than the average of the first ... 1 @Wolfy, It is true: since we can deduce Fatou's lemma from the Monotone Convergence Theorem, and the Dominate Convergence Theorem is a consequence of Fatou's lemma, we can prove the Dominate Convergence Theorem from the Monotone Convergence Theorem. However, such path would make us to prove again (inside the proof of the Dominate Convergence Theorem) ... 1 For (a), the direction ($\Leftarrow$) can be shown by differentiate$f(tx)=t^{a}f(x)$with respect to$t$and put$t=1$. For the opposite direction, put$g(x)=|x|^{-a}f(x)$where$|x|=\sqrt{x_{1}^{2}+\cdots+x_{n}^{2}}$is standard euclidean norm. By using product rule and given relation$\langle \nabla f(x), x\rangle=af(x), we can prove \begin{align} \nabla ... 1 Real analysis : authors: Folland, Rudin, Royden (hist newest edition), Frank jones (especially for gamma functions) Point set topology: authors: Munkres, Hocking & Young (this is not standard, but I like it personally) In my very opinion, one should learn multivariable calculus before studying real analysis. I prefer texts written in abstract spirit ... 2 You might enjoy an analysis book which has strong emphasis on function spaces as an application of general metric space theory. A good one along these lines is Carothers' Real Analysis. The book is divided into three parts: metric spaces, function spaces, and Lebesgue measure/integration. A typical quote which illustrates the spirit of the book: In the ... 1 This is a nice example to try, stop, and make your intuition rigorous. What your intuition apparently materialized is the following, written rigorously: Givenf,g: \mathbb{R}\to \mathbb{R}$such that$f$is continuous,$g$is uniformly continuous and$\lim\limits_{x \to \infty}f(x)-g(x)=A \in \mathbb{R}$and$\lim\limits_{x \to - \infty}f(x)-g(x)=B \in ...

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What you really want is to understand the difference between continuity and uniform continuity, so let's start with the definition of continuity. (I'll stick to functions of a single variable because for understanding this difference you don't have to introduce anything more complicated.) A function $f(x)$ is continuous at point $x_0$ iff for any given $\... 1 Uniform continuous means you can, given$\varepsilon$, in the$\varepsilon, \delta$definintion of continuity, the value of$\delta$uniformly for all$x$. In case you have a pole (a zero of a polynomial in the denominator without a zero in the nominator) this is never true near the pole, but these are not the only examples. (I don't prove this but only ... 1 But with you attempt, you do not reach the conclusion. Let$K>0$, then we have $$E[|X|1_{F}] = E[|X|1_F1_{|X|>K}+|X|1_F1_{|X|\le K}]\le E[|X|1_{|X|>K}]+E[K1_F]= E[|X|1_{|X|>K}]+KP(F).$$ 2$\mathcal{C}$is UI$\Rightarrow(i)$Let$\epsilon>0$. Then, since$\mathcal{C}$UI there exists$K>0$s.t. $$E[|X|] = E[|X|1_{|X|>K}+|X|1_{|X|\le K}]\le \epsilon + K$$ for all$X\in\mathcal{C}$. Thus$(i)$.$\mathcal{C}$is UI$\Rightarrow(ii)$You want to find for a given$\epsilon>0$a$\delta>0$such ... 1 Heine Borel Theorem was discovered much later than other forms of completeness of real number system and it is a very powerful and non-intuitive result about closed intervals. Therefore it is reasonable to expect that it does not have an obvious or trivial proof. Your approach fails primarily because it is not guaranteed that the numbers$a,r_{1}, r_{2}, \...

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Yes, we can choose such a $K$, or just notice that $$\lim_{K\to +\infty}K^{1-p}A=0$$ and that the bound of $\mathbb E\left(\left|X\right|;\left|X\right|\gt K\right)$ is uniform with respect to $X\in\mathcal C$. The follows from the inclusion $$\left\{\left|X\right|\gt K+k^{-1}\right\}\subset \left\{\left|X-X_n\right|\gt k^{-1}\right\}.$$ To see this, it ...

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Simon Stevin used unending decimals to represent all numbers (whether rational or not) already at the end of the 16th century. In the 17th century, Descartes seems to have been the first to use the term real to describe ordinary numbers.

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I finally solve it. Assume $\sum_{n=1}^{\infty}a_nf_n(x)<\infty$ almost everywhere in $[0,1]$. Let $g_N(x)=\sum_{n=1}^Na_nf_n(x)$. $g_N$ is measurable. Let $g(x)=\sum_{n=1}^{\infty}a_nf_n(x)$. $g(x)$ is also measurable, and $g(x)<\infty$ a.e. Hence, $g_N(x)\to g(x)$ a.e. as $N\to\infty$. By Egoroff's theorem, for any $\epsilon>0$ exists $E\subset[0,... 1 Yes, midpoint quasi-concavity and continuity imply quasi-concavity. Consider an upper level set$E(t) = \{x : f(x)\ge t\}$of a midpoint quasi-concave and continuous function. For any two points$a,b\in E(t)$, their midpoint$(a+b)/2$is also in$E(t)$. But then the points at$1/4$and$3/4$of the distance from$a$to$b$are also in$E(t)$, namely $$\... 1 Say f\in C_c(X). Standard construction: For n=1,2\dots and j\in\Bbb Z define$$E_{n,j}=\{x:j/n<f(x)\le(j+1)/n\}$$and set$$\phi_n=\sum_{j\in\Bbb Z}\frac jn\chi_{E_{n,j}}.$$Then \phi_n is \sigma(\mathcal E)-measurable and \phi_n\to f (uniformly). 1 Of course this is immediate from basic uniqueness results for differential equations (f''+f=0, c''+c=0, f(0)=c(0) and f'(0)=c'(0), so f-c.) A cute direct proof: Define$$\phi(x)=(f(x)-c(x))^2+(g(x)-s(x))^2.$$Show that \phi'(x)=0 and deduce that \phi(x)=\phi(0)=0. 1 The idea is simple. If f(c) < M then by continuity we have an interval around c where the values of f are less than M. In fact a stronger statement is true and that is what we need. We can find an interval of [c - h, c + h] such that all values of f are less than a specific number M' which is itself less than M. Next it should be easy to ... 1 We have:$$ S_N = \sum_{n=0}^{N}\frac{3^{n+1}}{1+2^{n+1}} = \sum_{n=0}^{N}\left(\frac{3}{2}\right)^{n+1}\frac{1}{1+\frac{1}{2^{n+1}}}=\sum_{n=0}^{N}\left(\frac{3}{2}\right)^{n+1}\sum_{m\geq 0}\frac{(-1)^m}{2^{(n+1)m}}$$hence by exchanging the sum on m and the sum on n,$$ S_N = 3\sum_{m\geq 0}(-1)^{m}\frac{2^{(m+1)(N+1)}-3^{N+1}}{(2^{m+1}-3)2^{(m+1)(N+1)... 1 Let$x=\tan{u} \rightarrow dx=\sec^2{u}du$Then integral becomes$\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}} \frac{du}{\sqrt{\tan^2{u}-1}}=-\coth{\tan{u}}|^{\frac{\pi}{2}}_{\frac{\pi}{4}}=-1+\coth{1}=0.313035$1 You can see that$(a \bmod{n})\bmod{n}$must be equivalent to$a\bmod{n}$. This is obvious because$a \bmod{n} \in [0,n-1]$and so the second$\operatorname{mod}$cannot have an effect. Moreover, if we consider what the$\operatorname{mod}$operation does, then it makes sense that if$b\mod{n} = 0$then we would expect $$a+b\equiv a\pmod{n}$$ 0 The best part about the inequality $$\sum_{n = 1}^{N}x_{i}^{\beta} \leq \left(\sum_{i = 1}^{N}x_{i}\right)^{\beta}\tag{1}$$ is that it is homogeneous in variables$x_{i}$and hence we can assume without any loss of generality that$\sum x_{i} = 1$. Since each each$x_{i}$is positive and they sum to$1$it follows that each$x_{i}$lies between$0$and$1$... 0 Yes,$g$is smooth. We first prove the following proposition: If$g$is$k$times differentiable, then the$k$th derivative of$g^2$is of the form $$a_0 \ gg^{(k)} + a_1 \ g^{(1)}g^{(k-1)} + a_2 \ g^{(2)}g^{(k-2)} + \cdots$$ where the$a_i$are positive integers. Proof by induction: For$k=0$, the$k$th derivative of$g^2$is just$g^2 = 1 \ g g^...

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EDIT. It turns out that my proof contains an error: $f'(x)$ may take zero infinitely often as $x \to 0$. Then the L'Hospital argument breaks down. Also, the reference in @levap's comment claims that $g$ need not be smooth, citing  f(t) = \begin{cases} \mathrm{e}^{-1/|t|} (\sin^2 (\pi/|t|) + \mathrm{e}^{-1/t^2}), & t \neq 0 \\ 0, & t = 0 \end{...

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