# Tag Info

1

Use contraposition: if $X'$ is finite dimensional, then $X''$ is as well. Since $X$ can be embedded isometrically in $X''$, it must be finite dimensional.

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Again, Hahn-Banach to the rescue : Choose an infinite basis $\mathcal{B}$ of $X$. Start with $v_1\in \mathcal{B}$, and choose $f_1 \in X'$ such that $\|f_1\| = 1$ and $f_1(v_1) = 1$. Now choose $v_2 \in \mathcal{B}$ and use Hahn-Banach to produce $f_2 \in X'$ such that $f_2(v_1) = 0$ and $f_2(v_2) = 1$. Thus proceeding, construct $\mathcal{D}_n := \{f_1, ... 1 As far as I can see, you are trying to find and upper bound for $$\frac{f(x)}{g(x)}=\left(\frac{1+|x|^2}{1+|x|}\right)^{\frac s2}$$ and a lower positive bound for its inverse: $$\frac{g(x)}{f(x)}=\left(\frac{1+|x|}{1+|x|^2}\right)^{\frac s2}$$ If$s>0$, the first tends to infinity and the second to$0$when$|x|\rightarrow \infty$, so there is no such ... 0 Use Hahn-Banach : Suppose$\|x_0\| > 3$, then$\exists f \in X'$such that$\|f\| = 1$and$f(x_0) = \|x_0\|$. Hence,$\|2f\| \leq 2$, but $$|2f(x_0)| = 2\|x_0\| > 6$$ 0 I think you are getting confused in how to use the definitions provided : Suppose$K$is compact and$f$is continuous - you want to show that$G_f$is compact. So choose a sequence$(x_n, f(x_n)) \subset G_f$. So far so good - how do you choose a convergent subsequence? Well, you work on the first component first : Since$(x_n) \subset K$, choose a ... 1 I agree with points$1$to$4$. For the last point, let's define for$k \in \mathbb N-\{1\}$$$g_k(x) = \left\{ \begin{array}{l l} -k^4x+k^2-k^4 & \quad x \in [-1,-1+\frac{1}{k^2}] \\ 0 & \quad x \in [-1+\frac{1}{k^2},1] \end{array} \right.$$ It looks a lot like your picture. Note that$\forall k, \int_{-1}^{1}g_k=\frac{1}{2}$... 0 If$n=0$then the result holds. Now, differentiate both sides. The LHS gives$\frac12 x^{-\frac12}$while the RHS gives$x^{-1}$. Thus, the rate of change of$\ln x$is less than that of$\sqrt{x}$. As both rates of change are decreasing,$\ln x$is always less than$\sqrt{x}$, as required. 1 If some$Q_k$is finite, then write$Q_k = \{x_1,x_2,\ldots, x_n\}$. Suppose$\cap Q_k = \emptyset$, then for each$1\leq i\leq n, \exists j = j_i$such that$x_i \notin Q_{j_i}$. Since$Q_{j_i}$contains$Q_n$for all$n\geq j_i$, we may assume that$j_i\geq k$. This is true for each$i$, hence if we choose$m = \max\{j_1, j_2, \ldots , j_n\}, then $$... 1 If Q_k has only finitely many points, then we have that |Q_k|\geq|Q_{k+1}|\geq\ldots, this is a decreasing sequence of positive integers, so it must stabilize. This means that for some n, every m>n satisfies Q_m=Q_n, so clearly the intersection is non-empty, since it is equal to Q_n as well. The problem when each Q_k is infinite is that ... 1 let \sqrt n > \ln n, then$$\sqrt{n+1} > \sqrt{(\log n)^2-( \log(n+1))^2+( \log(n+1))^2+1}> \log (n+1) $$hence proved by induction, that is if we show that$$\log(n)^2 - \log(n+1)^2 + 1 > 0which you can find the maximum using calculus which is less than 1. Also you can try this, \begin{align*} \log(n+1)^2 - \log(n)^2 &= (\log n + ... 1 Notice that\frac{\ln(n)}{\sqrt{n}} = 2 \frac{\ln(\sqrt{n})}{\sqrt{n}},$$so it is sufficient to show that$$\frac{\ln(x)}{x}< \frac{1}{2} \ \text{for all} \ x >0 \hspace{1cm} (1)$$If f(x)= \frac{\ln(x)}{x} then f'(x)= \frac{1-\ln(x)}{x^2}, so f is nondecreasing on (0,e] and nonincreasing on [e,+ \infty). Therefore, f has a global ... 7 The most important inequality about the exponential is$$\tag1e^x\ge 1+x\qquad\text{for all }x\in\mathbb R\text{ with equality iff }x=0.$$From this we find$$\tag2 e^{x}=(e^{x/2})^2\stackrel{(a)}\ge (1+x/2)^2=(1-x/2)^2+2x\stackrel{(b)}\ge 2x$$where (a) holds for x\ge-2 and is strict for x\ne0, and (b) is strict for x\ne 2. Hence e^{x}>2x at ... 0 By AM-GM inequality (\frac{a+b}{2} \geq \sqrt{ab} ), notice with a = x^2 and b = y^4 which are obviously positive$$ \frac{x^2+y^4}{2} \geq xy^2 \implies 1 \geq \frac{2xy^2}{x^2 + y^4} $$Hence, we have shown that f \leq 1 . Furthermore, to show f \geq -1 , notice$$ (x + y^2)^2 \geq 0 \iff x^2 + y^4 + 2xy^2 \geq 0 \iff 2xy^2 \geq -1(x^2+y^4) ... 1\dfrac{1}{n^2-1}\geqslant\dfrac{1}{n^2+\cos\pi n}$So, if$\sum\dfrac{1}{n^2-1}$then$\sum\dfrac{1}{n^2+\cos\pi n}$converges. Now, by Cauchy Condensations Test, if$\sum\dfrac{2^n}{2^{2n}-1}$converges, the sum converges.$\dfrac{2^n}{2^{2n}-1}=\dfrac{2^n}{(2^n+1)(2^n-1)}<\dfrac{1}{2^n-1}$We know that$\sum\dfrac{1}{2^n-1}$converges. ... 1$n^2 + \cos n\pi \geq n^2 - 1 \geq 0.5n^2$for$n \geq 2. So: $$\sum_{n=2}^\infty \dfrac{1}{n^2 + \cos n\pi} < \displaystyle \sum_{n=2}^\infty \dfrac{2}{n^2}.$$ By comparison test, the series converges. 2 Let $$a_n=\frac{(n+1)^n}{n!}$$ then \begin{align} a_n &=\left(1+\frac1n\right)^n\frac{n^n}{n!}\\ &=\left(1+\frac1n\right)^n\frac{n^{n-1}}{(n-1)!}\\ &=\left(1+\frac1n\right)^n\,a_{n-1}\\[9pt] &\le ea_{n-1} \end{align} Sincea_1=2\lt e$, inductively, we have that$a_n\lt e^n$. 1 Define a sequence$a_1 = 3$, and$a_{n+1} = a_n - \dfrac{a_n}{n}$for$n \geq 1$then$\dfrac{a_{n+1}}{a_n} = 1 -\dfrac{1}{n}$, and$\lim_{n \to \infty} |\dfrac{a_{n+1}}{a_n}| = 1$. This example shows that the above convergence test can not be used since the limit of the ratio could be$1$while the ratio test requires that it be less than$1$for the series ... 5 Consider the function$f(x) = \sqrt x - log(x)$and differentiate.$f'(x) = \frac{1}{2 \sqrt x} - \frac{1}{x} = \frac{\sqrt x - 2}{2x} \ge 0$for all$x \ge 4$. So the function is monotone increasing for all$x \ge 4$. Also$f(4) = \sqrt 4 - \log(4) > 0$Thus$f(x) > 0$for all$x \ge 4$and$\sqrt x - \log(x) > 0$. Now put$x = n \ge 4$and get ... -1 It doesn't matter that you have used the "intermediary variable"$c_n$. You have shown that $$|\frac{a_{n+1}}{a_n}| < 1$$ and if this is true also for$n \to \infty$then, by the ratio test, the series converges absolutely. 0 Stirling approximation write $$n! \simeq \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ Then the inequality is satisfied if $$n+1 \lt (2 \pi )^{\frac{1}{2 n}} n^{\frac{1}{2 n}+1} \simeq (2 \pi )^{\frac{1}{2 n}}[n+\frac{1}{2} \log \left(n\right)]$$ which is satisfied for any$n \geq 1$1 To prove that inequality, it means that we need to prove $${(n+1)^n\over n!}<e^n$$ Because${(1+{1\over n})^n}<e$, so${(n+1)^n\over n!}$=${(1+{1\over n})^n}$${(1+{1\over n-1})^{n-1}}2^1\over1$$1\over1!$$<e^n 9 Use the binomial theorem on (1 + n)^n and try.$$(1 + n)^n = 1 + \frac{n!}{1!(n - 1)!} n + \frac{n!}{2!(n - 2)!} n^2 + \frac{n!}{3!(n - 3)!} n^3 + \dots + \frac{n!}{n!(n - n)!} n^n \\ = n! \{\frac{1}{0! n!} + \frac{n}{1! (n-1)!} + \frac{n^2}{2! (n-2)!} + \frac{n^3}{3! (n-3)!} + \dots + \frac{n^n}{n! (n-n)!}\} \\ \le n! \{1 + \frac{n}{1!} + \frac{n^2}{2!} ...

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You're right. But in order to prove that you need to show that if $X\subseteq\Bbb Q$ contains two distinct $x<y$ then we can write $X$ as the union of two disjoint relatively open sets. That is to say, there are two open sets in $\Bbb R$, $U$ and $V$ which are non-empty, disjoint and $X=(X\cap U)\cup(X\cap V)$. HINT: Pick an irrational number ...

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If $\displaystyle \lim_{n \to \infty} S_n = a$, we show $\displaystyle \lim_{n \to \infty} S_{3n} = a$. Let $\epsilon > 0$ be given, there exists $N$ such that if $n > N$ then $|S_n - a| < \epsilon$. Now choose $M$ such that $M > \dfrac{N}{3}$, then if $n > M$, then $3n > 3M > N$, so $|S_{3n} - a| < \epsilon$ If $\displaystyle ... 4 The identity you give is quite pretty! Yes they are equal. Let $$\phi(t) = \sum_{n \ge 0}\frac{(b-a)^{n+1}}{(n+1)!} \left( f^{(n)}(a) \cdot t^{n+1} + f^{(n)}(b) \cdot (-1)^n (1-t)^{n+1}\right).$$ The first series in question is$\frac12(\phi(1)+\phi(0))$and the second is$\phi(\frac12)$. The "why" is that these are equal because, in fact,$\phi$is ... 1 Since$f$is uniformly continuous, you can find a$\delta$such that whenever we have$|x-y|<\delta$, then$|f(x)-f(y)|<1$. Take$M$big enough so that$D\subset [-M,M]$. By the archimedean property, we can cover this interval with a finite number of intervals of length$\delta$. Take those who intersect$D$and call them$I_1,\ldots,I_n$. For each ... 1 I am assuming$f:D\to \mathbb{R}$is uniformly continuous and$D$is totally bounded. Prove that$\overline{D}$is bounded : Since$D$is totally bounded, so is$\overline{D}$. Hence$\overline{D}$is compact Extend$f$to a continuous function$g:\overline{D} \to \mathbb{R}$as follows: a) For a$x\in \overline{D}$, choose a sequence$x_n \in D$such ... 0 (Open) subsets$U$of a metric space (or more generally a topological space)$X$for which the equality$\mathrm{Int} ( \mathrm{cl} ( U ) ) = U$are called regular open set (or sometimes open domains). Not all open sets have this property; one simple example, in the real line, is $$U = ( -1 , 0 ) \cup ( 0 , 1 ).$$ For an open set to be regular open, you ... 0 For contradiction, suppose there exists$x \in D $such that$f(x) > M $for$M \in \mathbb{R}^>$. Since$f$is continuous at$x$, we can obtain a$\delta > 0 $such that if$|y-x| < \delta$, then$|f(y) - f(x)| < \epsilon $Take$\epsilon = M$, then $$f(x) < M + f(y) = K$$ Contradiction. 0 Hint Prove that there is a monotonic subsequence$\{a_{n_k}\}$and let be$n_0$the limit of that subsequence. 4 Intuition: this multiplier evolves$f$by the Schrodinger equation by time one (after making Planck's constant one). And this maps the Dirac delta function to something whose absolute value is constant, which is a counterexample for$p=1$. So approximate the Dirac delta function by a narrow bell curve. This will work for$p<2$. For$p>2$, do the ... 1 Think about what the Ratio Test says. Given a power series $$\sum_{n=0}^\infty a_n x^n$$ the ratio test says that the series converges absolutely if $$L = \lim_{n\to\infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| |x| < 1$$ (it may or may not converge absolutely if that is$1$, but this won't affect ... 0 I think you wrote the radius down incorrectly. The ratio test says that the series will converge absolutely if $$\lim_{n\to\infty}{\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|}=|x|\lim_{n\to\infty}{\left|{a_{n+1}\over a_n}\right|}<1$$ which happens when $$|x|<\lim{\left|a_n\over a_{n+1}\right|}$$ so the radius is that. I'm not sure what level of rigour ... 1 Yes. For every countable ordinal we can find a set of real numbers whose Cantor-Bendixson rank is that ordinal. To see this, first note that every countable ordinal embeds (order-wise, and thus topologically wise) into the rational numbers, and so into the real numbers. So it suffices to find an ordinal whose rank is$\omega$. What does that mean? It means ... 0 you can use the zero point theorem! if$a_0>0$and$\dfrac{a_1}{2}+\ldots+\dfrac{a_n}{n+1}<0$then the condition you write will turn out to be a critical condition:$a_0+\dfrac{a_1}{2}+\ldots+\dfrac{a_n}{n+1}=0$since$f(0)=a_0>0$and ... 0 You haven't specified a nature of$A$. If$A\subset \mathbb R$, the equivalence of 1 and 2 holds. It also holds if the domain of$f$is a metric space (or a first-countable topological space). Anyway, the comment by David Mitra is essentially the solution. I give a slightly different version of "(1) implies (2)". First, take any sequence$(x_n)$in ... -1 Because the dual space of$l^2$is$l^2$. 1 The inverse tangent function, for example, maps$\mathbb R\to(-\pi/2,\pi/2)$. You will not find an example of a continuous function that sends a closed interval into a non-closed set, because continuous functions map compact sets to compact sets. 0 Consider$e^{-x}$. Then$[0,\infty)$goes to? 0 The "let$h=hv$" seems to be inconsistent notation so we will let$h=mv$where$v$is a unit vector. Now:$DF(x+mv)-F(x)+r=F(mv)+r\Rightarrow DF(x+mv)=F(mv)+F(x)=F(x+mv)$now take$m\to 0$to get:$DF(x)=F(x)$0 The statement is trivial for the zero solution, so we'll ignore it for the rest of the answer. Note that$ -\int_0^T e^{-x} y'y''dx = \int_0^T yy'dx = \frac{1}{2}\big(y(T)\big)^2 - \frac{1}{2}\big(y(0)\big)^2$and$ -\int_0^t y'y''dx = \int_0^t e^x yy'dx. $Proposition 1. Let$q \colon [0,\infty) \to \mathbb R$be positive, nondecreasing, and$C^1$. Let ... 1 First of all,$f_n\to f$in$L^2$implies$f_n\to f$in measure, so the second assumption is redundant. There is a standard counterexample to show that convergence in$L^p$,$1\le p<\infty$, does not imply convergence a.e., much less "almost uniformly". Namely, enumerate dyadic subintervals of$[0,1]$as$I_1,I_2,\dots$(order does not matter), and let ... 1 What is the norm of the linear functional$L(x)$? It is the smallest constant$M$such that $$|L(x)| \leq M ||x||_2$$ holds for every$x \in \ell^2$. Okay, so now note that your linear functional is of the form$\langle a, x \rangle$, so apply the Cauchy inequality to get $$|L(x)| = |\langle a, x \rangle| \leq ||a||_2 ||x||_2.$$ This means$M \leq ...

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The composition of linear functions is bilinear: $$R(S+T)=RS+RT,$$ $$(S+T)R=SR+TR,$$ $$\cdots$$ See Derivative Bilinear map.

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Since $$\int_{a}^{b}dt\int_{a}^{b}ds|K(t,s)|^{2}<\infty$$ $K(t,s)$ is a Hilbert-Schmidt operator and such operators are compact.

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In principle, this could be almost anything and $$B=\begin{cases}219&\text{if }A=0\\ 224&\text{if }A=50\\ 231&\text{if }A=100\\ 246&\text{if }A=200\\ 255&\text{if }A=255\\ 42&\text{otherwise}\\ \end{cases}$$ would fit. Of course, one expects something "smoother", and there are still many options, for example a polynomial of degree ...

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For $p<1$, we say that $f_n\to f$ in $L^p$ if $\lim_{n\to\infty}\int |f_n-f|^p=0$. Define $g_n:=|f_n-f|^p$: it converges to $0$ in $\mathbb L^1$ by assumption and $g_n\to |f-g|^p$ almost everywhere. We deduce from the case $p=1$ that $|f-g|^p=0$ a.e. hence $f=g$ a.e.

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If $x\in\mathbb{Q}$ then $x=\frac{a}{b}$ for some $a,b\in\mathbb{Z}$ so we have for $$n>b$$ $n!x\in\mathbb{Z}$ and therefore $$\lim_{m\to \infty}{\cos^m{(n!2\pi x)}}=\lim_{m\to \infty}{1^m}=1$$ Thus $\forall \epsilon>0$ we can take $n>N=b$ to get $$\left|\lim_{m\to \infty}{\cos^m{(n!2\pi x)}}-1\right|=|1-1|=0<\epsilon$$ so the limit is $1$. If ...

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We quite obviously have: $$\lim_{m\to\infty}\cos^m(n!2\pi x)=\cases{1&if n!x\in\mathbb{Z}\\0&if 2n!x\notin\mathbb{Z}\\\mathrm{doesn't\ exist}&else}$$ And thus: $$\lim_{\mathbb{N}\ni n\to\infty}\left(\lim_{m\to\infty}\cos^m(n!2\pi x)\right)=\cases{1&if x\in\mathbb{Q}\\0&else}$$ I'll leave it to you to check the details.

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The only outer measures for which your condition holds are the so-called regular outer measures $^{[1]}$, that is, the ones for which every set $A\subset X$ admits a measurable set $E$ such that $A\subset E$ and $m^\star(A)=m(E)$. (In Munroe's book Introduction to measure and integration, such a set $E$ is called a measurable cover for $A$.) To prove this ...

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