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0

The given property of $X$ (separated because of $d(x,y)=0 \iff x=y$ sometimes called axiom of separation since ensuring the uniqueness of limits) implies that all sequence contains a convergent subsequence so this space $X$ is sequentially compact then compact (and separated because metric). NOTE.- For french mathematicians, compact is always separated. ...


1

1) Injectivity: Let $x_1,x_2\in X$ and $x_1+g(x_1)=x_2+g(x_2) \Leftrightarrow x_1-x_2=g(x_2)-g(x_1) \Rightarrow L\|x_2-x_1\|\ge\|g(x_2)-g(x_1)\|=\|x_2-x_1\|$. Therefore $x_1=x_2$, because $L<1$. 2) Surjectivity Let $c\in X$ be an arbitrary element. We want to find $x\in X$ such that $f(x)=c$, i.e $x+g(x)=c \Leftrightarrow x=c-g(x):= r(x)$. For the ...


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Fix $\delta > 0$. Choose $x_1 \in X $. Choose $x_2$ such that $d(x_1,x_2)>\delta$. Now choose $x_3$ such that $d(x_1,x_3)>\delta$ and $d(x_2,x_3)>\delta$. Continue this process... Basically, we want to cover $X$ with disjoint $\delta$ balls. Since, {$x_1, x_2, \dots$} forms and infinite set and clearly does not have a limit $\implies$ ...


3

Using the axioms of $\mathbb R$ (ordered field where each bounded set has a supremum) you can show that there is a square root function $s:\mathbb R^+_0\to \mathbb R^+_0$. So you want from $s$ that it satisfies $s(x)^2=x$ for all $x\in\mathbb R^+_0$. With that condition it is uniquely determined. To see that $\sqrt{1+\sqrt{2}}$ exists, you only have to ...


1

Use the fact that an open subset of a complete metric space is homeomorphic to a complete metric space. Let $U$ be a non-empty open subset of $\Omega$. Then the subspace $U$ has a complete metric $d$ that generates the subspace topology. For $n\in\Bbb N$ let $H_n=G_n\cap U$; each $H_n$ is a dense, open subset of $U$, so by your argument $\bigcap_{n\in\Bbb ...


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You can use the definition of convergence: For $\epsilon >0$ you can pick a positive integer $N<\frac{1}{\epsilon}$ then it's easy to show $|s_n-0|<\epsilon$ whenever $n\geq N$


2

Note that $(1/k)$ for $k$ odd as a subsequence of $(1/n)$ is convergent with limit $0$. The constant sequence $0$ obviously converge to $0$. So in this case $(s_n)$ converges to $0$.


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Answering this question so that it no longer remains unanswered. Maybe limit "from above" and limit "from below" respectively? – AdLibitum


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I re-worked my proof but I am not sure it is entirely correct, please provide comments or suggestions if needed Suppose $E\in M_\mu$, and $\mu(E) < \infty$, by lemma 1.17 $$\mu(E) = \inf\left\lbrace \sum_{1}^{\infty}\mu((a_j,b_j)): E\subset \bigcup_{1}^{\infty} (a_j,b_j)\right\rbrace$$ Since, $E\subset \bigcup_{j}(a_j,b_j) \Rightarrow E\subset ...


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$\sum_k a_k=\sum_k \dfrac{1}{2k+1}>\sum_k \dfrac{1}{2k+2}=\dfrac{1}{2}\sum \dfrac{1}{k+1}.$ If you mean sequence note that $0‎\leq‎ s_n<\dfrac{1}{n}$ thus use $0=\lim 0<\lim s_n<\lim\dfrac{1}{n}=0$


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If we consider $$f(x)=\frac{1+\sin(1/x)}{2},$$ we get a differentiable function that is not uniformly continuous.


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This is just an elaboration of (my interpretation of) André's comment: If $f:(0,1) \to \mathbb{R}$ is continuous, and $L_0=\lim_{x \downarrow 0} f(x)$ and $L_\infty=\lim_{x \to \infty} f(x)$ exist (and are finite), then $f$ is bounded. To see this, note that there is some $\delta>0$ such that if $x < \delta$, then $|f(x)| \le L_0 +1$, and similarly ...


0

$$\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n+1} + \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}= \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n+1} - \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}= \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n+1} - (-1)^{n}\frac{1}{n}= \sum_{n=1}^\infty a_{k+1}-a_{k}$$ now use telescopi rule


1

\begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \, (2n+1)}{n \, (n+1)} &= \sum_{n=1}^{\infty} (-1)^{n+1} \, \left( \frac{1}{n} + \frac{1}{n+1} \right) \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{(-1)^{n}}{n} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \left( \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} + 1 \right) ...


2

$$ \sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}=\sum_{n=0}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}+1\\ =\sum_{n=1}^{\infty}{\frac{{(-1)}^{n}}{n}}+1 $$ So summing the two summation yields: $$ \sum_{n=1}^{\infty} {{(-1)}^{n+1}\frac{2n+1}{n(n+1)}}=\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n+1}}+\sum_{n=1}^{\infty}{\frac{{(-1)}^{n+1}}{n}}\\ ...


2

You have to be careful in performing rearrangements since your series is conditionally convergent but not absolutely convergent (see, for instance, the Riemann series theorem). ...


0

Yes that's the limit. You can easily show that the expression is correct by induction. The series is a geometric series, which has a well known formula for the sum to infinity. The limit of $a^n$ as $n\rightarrow\infty$ is 0.


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As you showed the general formula for the $n$th term, the sum in the bracket is a geometric series and will also converge provided $|a| < 1$. Oussama Boussif's answer then gives the correct limit to $\frac b {(1-a)}$. Interestingly independent of $x_0$. Sorry, I can't comment due to low reputation. Here is a good link for some standard formulae: ...


6

You are extremely close to the answer, you just have to recall that for any $a\in (0,1) $ we have: $$ \sum_{i\geq 0}a^i = \lim_{n\to +\infty}\sum_{i=0}^{n-1}a^i = \frac{1}{1-a}.$$ Given that, you may also notice that: $$ \left(x_{n+1}-\frac{b}{1-a}\right) = a\cdot\left(x_{n}-\frac{b}{1-a}\right)$$ holds, so you know how fast the convergence is.


5

This only answers the "to what does this converge?" part of the question and does not prove the sequence actually converges. A neat trick for these sorts of problems is to assume the sequence converges and then use simple algebra to find the limit. If $x_n$ converges, then $\lim_{n\to\infty} x_n = \lim_{n\to\infty}x_{n+1} = x$. Then, we just solve the ...


0

You are proving that if $E$ is measurable, then for every $\epsilon > 0$ there exists a closed set $F \subset E$ satisfying $m^*(E) - m^*(F) < \epsilon$. Your first step is to choose $F_k \subset E$ satisfying $m^*(E\setminus F_k) < \epsilon/2^k$. How do you know this can be done? If it can, take $k=0$ to find $F_0$ satisfying $m^*(E) - m^*(F_0) ...


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If $f(x,y)=|x|$, then $\frac{\partial f}{\partial x}(0,y)$ does not exist for any $y$. Therefore certainly $\frac{\partial^2 f}{\partial y\partial x}(0,y)$ does not exist, because it does not make sense to take the derivative of a function which is not even defined. On the other hand, $\frac{\partial f}{\partial y}(x,y)=0$ for all $(x,y)$, and thus ...


0

As you pointed out to Felipeh, it's important to remember that the bounded open (with respect to Euclidean topology on $\mathbb{R}^{n}$) sets $B(x,\delta)$ are a priori "balls", not actual balls define by a metric or some weaker quasidistance function. Perhaps, you have already shown this, but I don't think it is that obvious that $\mu$ is finite on the ...


4

Just exploiting the arctangent Taylor series, $$\begin{eqnarray*} \int_{0}^{\frac{\pi}{2}} \arctan(\sin^3 x)\,dx &=& \sum_{m\geq 0}\frac{(-1)^m}{2m+1}\int_{0}^{\frac{\pi}{2}}\sin^{6m+3}(x)\,dx\\ &=& \frac{3\sqrt{\pi}}{4}\sum_{m\geq 0}\frac{(-1)^m\,\Gamma\left(3m+2\right)}{(3m+\frac{3}{2})\Gamma\left(3m+\frac{5}{2}\right)}\end{eqnarray*} $$ ...


0

As said in a comment above: $L^2$ spaces are Hilbert spaces (for the inner product corresponding to the measure), and therefore complete: Cauchy sequences converge.


6

Note that $d'(x,y)\leq d(x,y)$ for any $x,y$ in your space, and $d(x,y)\leq 2d'(x,y)$ if $d(x,y)\leq 1$, so a sequence is Cauchy with respect to one metric if and only if it is Cauchy in the other metric.


0

Consider the following lemma: If $\nu$ is a signed measure and $\lambda$, $\mu$ are positive measures such that $\nu=\lambda-\mu$, then $\lambda\geqslant \nu^+$ and $\mu\geqslant \nu^-$. To see this, let $(P,N)$ be a Hahn decomposition for $\nu$, then for $E\subset P$, $$\nu(E) = \nu^+(E) = \lambda(E) - \mu(E)\implies \lambda(E) = ...


3

$f$ is trivially positive on $\mathbb{R}^+$, and the (convexity) inequality $e^y\geq 1+y$ implies $\log(1+x)\leq x$ for any $x>0$, hence your function is always between $0$ and $1$.


0

Hint:- Notice that $e^x\ge x+1$ for all $x>0$.


4

The map $x \mapsto \frac{x}{1+|x|}$ has derivative $\frac{1}{(1 + |x|)^2} > 0$ and is therefore monotone on the whole real line. The assertion follows directly from this.


1

The uniform limit of a sequence of continuous functions is always continuous, so this will only work if the characteristic function of the compact set is continuous itself. However, this is in general not the case. I assume you try to find such a sequence in order to prove something related to measure theory. It is probably not necessary to have uniform ...


0

If $f_n$ converge uniformly, then the limit (in this case $\chi_K$) will be continuous. In most cases, this is false (for example for $K \neq \emptyset$ and $X = \Bbb{R}^n$). To see this, note that (by the intermediate value theorem) $\chi_K$ would have to attain the value $1/2$ if it was continuous.


0

To be more precise than the comments, use the usual approximation by convolution. This will give you a $C^\infty$ function which is bounded by the $L^\infty$ norm of $f$ and is arbitrarily close to $f$ in $L^p$. Now truncate by a suitable $C_c^\infty$ bump function $\varphi$ which fulfills $0 \leq \varphi \leq 1$.


0

There are (at least) two ways to interpret your question. The first is that you are asking if the value of the integral is actually given by a Riemann sum. This is not the case, even for Riemann integrable functions. Consider for example the function $$ f:\left[0,1\right]\to\mathbb{R},x\mapsto\begin{cases} 1, & x\in\left[0,\frac{1}{\sqrt{2}}\right],\\ 0, ...


1

(Converted and fixed from a comment) Let $F(t,x) = \rho(t) \exp(itx)$, then the solution to the PDE $$ \partial_t u = F $$ with $0$ initial data is exactly $$ u(t,x) = \int_0^t \rho(s) \exp(isx) \mathrm{d}s. $$


1

Take $a_n =n ,b_n =n , c_n =n^2 , d_n =1$, $f(x) =\ln x$ then we have $$\frac{f(a_n ) b_n }{f(c_n ) d_n } =\frac{n\ln n}{2\ln n} =\frac{n}{2}\to \infty$$


2

Let we assume $x,y\in(0,1)$ and $|x-y|=h$. We want to prove: $$ \sup_{\substack{x,y\in (0,1)\\ |x-y|=h}}\left|x \log x-y\log y\right|\leq -h\log h \tag{1}$$ but since $f(x)=x\log x$ is a convex function on $I=(0,1)$, it is enough to prove $(1)$ for $\min(x,y)\to 0^+$ and $\max(x,y)\to 1^-$. The first case is trivial, since $\lim_{x\to 0^+}x\log x = 0$. So ...


1

I assume that you intend $a_n$ as a sequence satisfying $a_{n+1} = \arctan a_n$. Your 'last' theorem is not correct. Consider $b_n = 1-1/n$. You can prove that $|b_{n+1}/b_n|<1$ for all $n$ but $b_n\to 1$ as $n\to \infty$. Despite of this, you can prove that the limit of $a_n$ is zero. From the fact $\arctan x < x$ for positive $x$, we get $\langle ...


1

Another Approach: let $(X,d)$ be a metric space and let $\{p_n\}$ be a cauchy sequence with a convergent subsequence, say convergent to $L \in X$. Now consider the completion $\overline{X}$ of $X$: by definition every Cauchy sequence in $\overline{X}$ converges, so our sequence $\{p_n\}$ converges in $\overline{X}$, say to $M$. But then every subsequence ...


0

Denseness of the periodic points implies an underlying structure. If, in addition, there's a dense orbit (topological mixing) we can say that every open set of the space contains points with very different behavior. So not only is regular (periodic) behavior possible, it is very close to very 'random looking' behavior... so that not only do we have ...


2

Yes, that's exactly right - you just break up the absolute value into parts: $$f(x)=\frac{1}{|x-2|}-x=\begin{cases}\frac{1}{x-2}-x&\text{if }x>2\\\frac{1}{2-x}-x&\text{if }x<2\end{cases}$$ From here, you can proceed as normal with the derivative arguments - that is breaking up the intervals into pieces where the derivative is positive and ...


4

Hint: Notice that $$f(x)=\begin{cases} -\frac{1}{x-2}-x&x<2\\ \frac{1}{x-2}-x&x>2 \end{cases}$$


2

I claim that $f=0$ almost everywhere. To prove this, we first need a lemma. Lemma: Let $A\subset [0,1]$ be the set of numbers $x$ such that every finite string of $0$s and $1$s appears somewhere in the binary expansion of $x$. Then $A$ has measure $1$. Proof: There are only countably many such finite strings, so it suffices to show that for any finite ...


2

First let's convince ourselves that the equality cannot hold: let $\nu_1$ be a non-trivial measure, i.e. $|\nu_1|(A) > 0$ for some $A$, and let $\nu_2 = -\nu_1$. Then clearly $$0 = |\nu_1 + \nu_2|(A) < 2|\nu_1|(A).$$ You make a mistake when looking at $$(\nu_1^+(A)+\nu_2^+(A))-(\nu_1^-(A)+\nu_2^-(A))$$ you conclude that $$(\nu_1^+(A)+\nu_2^+(A)) = ...


1

The concept is okay, but there are a number of issues with how you've written it up: Let $ϵ_2>ϵ_1>0$. Let $ϵ_2$ be such that, for all $n,N\in \Bbb N, |\sup\limits_n \inf\limits_{n≥N} a_n−a|>ϵ_2$. First, as stated you are introducing $\epsilon_2$ twice. You should say "let $\epsilon_2$ be such that ..., and let $ϵ_2>ϵ_1>0$", the latter ...


1

As a means of trying to prove things more then one way Good idea; I would recommend trying to prove it by translating the $\epsilon$-$N$ definitions alone, without considering subsequences. Try not to rely on proof by contradiction. Let $\epsilon_2> \epsilon_1 > 0$. Let $\epsilon_2$ be such that... That's a confusing way of putting it. At ...


2

$\phi$ is actually defined on $\mathbb R^2$; the author could have written $$ \phi: \mathbb R^2 \to \mathbb R: (a, b) \mapsto a^2 + a b^2 $$ The problem is that our notation for derivatives depends on the choice of "dummy variables" that we use in denoting a function (sigh...on the good side, it often makes the chain rule work out in really nifty ways). ...


2

Let $n$ be such that $m^*(V)>1/n$, choose $n$ distinct rationals $q_1,\dots,q_n\in\mathbb{Q}/\mathbb{Z}$, and write $V_k=\bigcup_{i=1}^k V\oplus q_i$. Then if $V\oplus q_k$ is measurable, taking $A=V_k$ we find that $$m^*(V_k)=m^*(V\oplus q_k)+m^*(V_{k-1}).$$ Thus if $V$ were measurable, we could conclude by translation-invariance of $m^*$ and induction ...


3

Your first line should have a "$\leq$" instead of "$=$", since you are using the triangle inequality: $$ |x_ky_k-ab|=|x_ky_k-ay_k+ay_k-ab|\leq|x_k-a||y_k|+|a||y_k-b|. $$ Your next idea is totally fine, but you are making things more complicated than they are :) First, fix $\epsilon>0$. Since $(x_k)$ converges to $a$ and since $(y_k)$ converges to $b$, we ...


5

The desired result drops right out if you apply the fundamental theorem of calculus to the function $$ g(t) = f\bigl(x + t(y - x)\bigr), $$ since $$ g'(t) = \langle \nabla f\bigl(x + t(y - x)\bigr), y - x\rangle $$ by the chain rule.



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