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6

It looks like it requires countable choice, but you can certainly modify it so that it no longer requires choice. Namely, define $a_1=r_1$ and $b_1=b$. Then, for each $n \in \mathbb{N}$, given $a_n,b_n$, there are two possibilities: $r_{n+1} \le a_n$ or $r_{n+1} \ge b_n$. In this case, let $a_{n+1} = \frac{2a_n+b_n}{3}$ and $b_{n+1} = \frac{a_n+2b_n}{3}$; ...


5

$\binom{k}{2}\leq n$ is equivalent to $(2k-1)^2 \leq 8n+1$, hence the largest triangular number $\leq n$ is given by $\binom{k}{2}$ with: $$ k = \left\lfloor \frac{1+\sqrt{8n+1}}{2}\right\rfloor.$$


5

All of them. Define $b_1 = a_1$, $c_1 = 0$ and $$b_{n + 1} = \begin{cases}b_n & a_{n + 1} \le a_{n} \\ b_n + a_{n + 1} - a_n & a_{n + 1} > a_n\end{cases}$$ $$c_{n + 1} = \begin{cases}c_n + a_n - a_{n + 1} & a_{n + 1} \le a_{n} \\ c_n & a_{n + 1} > a_n\end{cases}$$ for $n > 0$. It is obvious that $b_n$ and $c_n$ are increasing and ...


4

If $a\in\mathbb{N}$, factor $1+t^a$ then exploit: $$ \int \frac{\log(1-\alpha t)}{1+t}\,dt = \log(1-\alpha t)\log\left(\frac{\alpha+\alpha t}{1+\alpha}\right)+\text{Li}_2\left(\frac{1-\alpha t}{1+\alpha}\right). $$ The last line can be easily checked through differentiation.


3

This is not correct. $f_n\to f$ a.e. iff $$\mu(\{x :\lim_{n\to\infty} |f_n(x)-f(x)|\ne0\}) = 0. $$ Convergence in measure has the limit outside the measure - one must take care when interchanging limit operations! Now, convergence in measure implies that there is a subsequence that converges a.e., and in a finite measure space (i.e. $\mu(X)<\infty$), ...


3

You can prove the result at once by writing $$\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \ge 2\sqrt{x^2\cdot \frac{1}{x^2}} + 2 = 4,$$ then taking square roots.


2

The proof is presented in a way that supposedly use dependent choice, which is in fact stronger than countable choice. However this can be avoided in one of several ways. As Clive suggested, using $(a,b)$ as a bootstrap, you can generate algorithmically smaller and smaller intervals. As Carl suggested, you can use the fact that the rational numbers are ...


2

You can use $f(x)=x^{2}-2$ then $f(3)=7>0$ and $f(1)=-1<0$, now use Weierstras theorem for continuous functions. for the method that you want, let $\alpha$ be the Lub for then $b=2-\alpha^2>0$ there is $n_0$ such that $b=2-\alpha^2>\dfrac{1}{n_0}$. now it is may choose $n>n_0$ such that $(\alpha+\dfrac{1}{n})^2<2$ contrary that $\alpha$ is ...


2

This is not exactly an answer to the question as formulated, but it may clarify the precise formulation of the distinction. Convergence a.e. can be written using countable intersections and unions. First define $$A_{n,m} = \{ x : |f_n(x) - f(x)| > 1/m \}.$$ Now, for the convergence to fail at $x$, there must be some $m$ such that $x \in A_{n,m}$ for ...


2

Another Proof : Let , $f(c)=0$. We want to show $f\equiv 0$ in $\mathbb R=(-\infty,c]\cup[c,\infty)=A\cup B\text{ (say) }$. $\bullet$ In $A$ , $f'(x)\ge 0$ implies $f$ is monotone increasing in $A$. So , $x\le c\implies f(x)\le f(c)=0$. Also , $f(x)\ge 0$ (given). Hence , $f(x)=0$ in $A$. $\bullet$ In $B$ , define $F(x)=e^{-x}f(x)$. Then ...


1

I'm not an expert in this subject, so pardon if I've made any mistakes, but I believe the answer is yes. $\phi^t(ax)$ is the integral curve which passes through $ax$ at $t = 0$ with velocity $X(ax)$. To check that $a \phi^t(x)$ is equal to this integral curve, it suffices to check that it is an integral curve which passes through $ax$ at $t = 0$ (which it ...


1

Yes, your re-characterization is wrong, becuase it is very well known that convergence in measure does not imply convergence almost everywhere. Here is an example. Consider the intervals $E_{i,n}=[{i-1\over n},{i\over n}]$ and their characteristic functions $\chi_{E_{i,n}}$, where for each $n$ $i$ runs between $1$ and $n$. So the few first intervals are ...


1

You really only need to consider your first equation in order to obtain a contradiction. Suppose that there exists $w \in L^{1}_{\mbox{loc}}(\mathbb{R})$ such that $$ \int w\varphi dx = \varphi(0),\;\;\;\varphi \in \mathcal{C}_{c}^{\infty}(\mathbb{R}). $$ That equation is inconsistent. To see why, find ...


1

Each $\ln_a^b f_n$ is simply a number; so if each number is defined but the sequence of numbers doesn't converge, then extending from Riemann integrals to Lebesgue integrals (which will all have the same value) doesn't change anything. If you want an example where $\int_a^b (\lim f_n)$ doesn't exist as a Riemann integral but does exist as a Lebesgue ...


1

For illustration let us consider the case where the ambient space is $\mathbb{R}^3$ and $M$ is the $(x,y)$-plane (which is a surface of 0 Gaussian curvature). Let $g$ be a function of compact support in $\mathbb{R}^3$ whose support include the origin. Let $g_{\nu}(x,y,z) = g(x, y, z/\nu)$. Its inverse Fourier transform scales like $$ ...


1

Since $f(x)^2 \le M(f^2)$ for all $x\in A$, then $|f(x)| \le \sqrt{M(f^2)}$ for all $x\in A$. Hence $M(|f|) \le \sqrt{M(f^2)}$, or $M(|f|)^2 \le M(f^2)$. By a similar argument, $m(|f|)^2 \ge m(f^2)$.


1

Yes, the intended meaning is what you're thinking. Now, an element $x$ belongs to $f^{-1}(B)$ if and only if $f(x)\in B$. Take $a\in A$: can you say that $f(a)\in f(A)$?


1

Here's an argument that should work in $\mathbb{R}^n$ for $n \geq 3$. I guarantee that it is not the most elegant approach. Assume $\gamma$ contains no segment along a line through the origin. (If $\gamma$ contains one or more such segments, shift it by a translation $\tau$ along a vector not in one of those lines. Then carry out the following procedure, ...


1

For example, put $A:=\{1,\frac{1}{2},\frac{1}{4},...\}$. Then the set $B:=\{0\}\cup\bigcup_n \left(\frac{1}{2^n} + \frac{1}{2^n} A\right)$ is compact (closed and bounded) and has derived set $B'=\{0\}\cup \{\frac{1}{2},\frac{1}{4},...\}$.


1

Let $\varepsilon > 0$. Since $\lim\limits_{x\to \infty} f(x) = a$, there exists $M > 0$ such that $|f(x) - a| < \varepsilon$ for all $x \ge M$. So for $x > M$, $$\left|\frac{1}{x}\int_0^x f(t)\, dt - a\right| \le \frac{1}{x}\int_0^M |f(t) - a|\, dt + \frac{1}{x}\int_M^x |f(t) - a|\, dt < \frac{C}{x} + \frac{\epsilon(x - M)}{x},$$ where $C = ...



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