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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.


1

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 $$\log(n)^2 - \log(n+1)^2 + 1 > 0$$ which 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 $$ \begin{align} a_n &=\left(1+\frac1n\right)^n\frac{n^n}{n!}\\ &=\left(1+\frac1n\right)^n\frac{n^{n-1}}{(n-1)!}\\ &=\left(1+\frac1n\right)^n\,a_{n-1}\\[9pt] &\le ea_{n-1} \end{align} $$ Since $a_1=2\lt e$, 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!} ...


0

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 ...


0

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 ...


1

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


0

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.


1

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 ...


1

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.


1

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 ...


2

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.


1

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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