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0

It is bounded since: $$f(0)\leq f(x)\leq f(1)$$ Denote for shorthand: $$\mathcal{S}:=\{x:\Delta f(x)>0\}$$ So countably many jumps: $$\sum_{x\in\mathcal{S}}\Delta f(x)\leq f(1)-f(0)<\infty\implies\#\mathcal{S}\leq\#\mathbb{N}$$ Thus it is integrable.


1

$$ f(\alpha(x))=f(0)+f'(0)\alpha(x)+o(\alpha(x))=f(0)+f'(0)\alpha(x)+o(x) \quad\Rightarrow $$ $$ \quad\Rightarrow\frac{\alpha(x)}{x}=\frac{1}{f'(0)}\frac{f(\alpha(x))-f(0)}{x}+o(1)\quad\Rightarrow $$ $$ \lim_{x\to 0^+}\frac{\alpha(x)}{x}=\frac{1}{f'(0)}\lim_{x\to 0^+}\frac{f(\alpha(x))-f(0)}{x}=\frac{1}{f'(0)}\lim_{x\to 0^+} ...


2

Let $X$ be a non-empty set and $\mathscr A$ an algebra on it. A premeasure on a $\mathscr A$ is a function $\lambda:\mathscr A\to[0,\infty]$ such that $\lambda(\varnothing)=0$; and if $A_1,A_2,\ldots$ is a countable collection of disjoint sets in $\mathscr A$ and if their union is contained in $\mathscr A$, then $$\lambda\left(\bigcup_{n=1}^{\infty} ...


0

Let $g :[0,8] \rightarrow [0,8]$ be continuous. The function $g \circ f$, for $f(x)=4x-4$ is a continuous function from $4(1)-4$ to $4(3)-4=8$, i.e. it is a continuous function from the closed ball $[0,8]$ into itself, then use Brower's fixed point theorem for compact sets :https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem . Note that we only need ...


2

Let $h(x) = f(x) - 4x + 4$. $h$ is continuous on $[1,3]$. Moreover: $h(1) = f(1) \ge 0$ and $h(3) = f(3) - 8 \le 0$ since $\operatorname{Range}(f) = [0,8]$. Thus, by IVT..


5

If you let $f(x)=0$ for $x\le 0$ and $f(x)=1$ for $x>0$ you get a counterexample. The result you want requires that you assume that $f$ is continuous at $x_0$. Assuming that, there's probably a more elementary proof than the following simple but not quite so elementary proof: Say $x_0=0$ for convenience. Subtract a constant so $f(-1)=0$. Now since $f$ ...


0

Prove it by induction on $m$. For $m = 1$, we have $$-C + \phi_n \leq \phi_{n} \leq C + \phi_n$$ as expected. Now assume that it holds for all values $\leq m$. Then for $m + 1$, we have \begin{align} \phi_{n(m+1)} &= \phi_{n + nm} \\ &\leq C + \phi_n + \phi_{nm}\circ f^n &\text{ by almost additivity} \\ &\leq C + \phi_n + Cm + ...


1

Let's to use polar coordinates $x=\rho\cos\phi$, $y=\rho\sin\phi$, and your limit is $$ \lim_{\rho\to0} \frac{\rho^4 \cos^2\phi \sin^2\phi}{\rho^2\cos^2\phi + \rho^4\sin^4\phi} = \lim_{\rho\to0} \frac{\rho^2 \cos^2\phi \sin^2\phi}{\cos^2\phi + \rho^2\sin^4\phi} $$ But $\lim_{\rho\to0}\rho^2 \cos^2\phi \sin^2\phi = 0$. If $\cos\phi\ne0$, then $$ ...


0

For example if X= {0, 1, 2} with d(x, y) defined as d(x, y)= |x- y| then there are only 2^3= 8 subsets so at most 8 "open balls".


0

It is not clear to me what you are asking exactly. Given a ball $B=B(x;r)$ in $\Bbb R^n$ all the balls $B(x;\rho)$ with $\rho<r$ are inside $B$ and so in whatever open set $A$ the ball $B$ is a subset of. This is entirely trivial. A bit less trivial question would be if you can find infinitely many disjoint balls inside an open set $A\subseteq\Bbb R^n$, ...


2

It may be the case that these open balls are actually equal to each-other despite having different radii. For instance, if we equip $\mathbb{Z}$ with the usual metric, then $B_{1/2}(0) = B_{1/3}(0)$, for instance.


6

In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the "discrete metric" $d(x,y)=0$ if $x=y$ and $1$ otherwise, all balls $B_r(x)$ for $r \leq 1$ are the same (they are just $\{ x \}$) while all balls $B_r(x)$ for $r>1$ are also the same ...


2

We have: $$ \int_{0}^{\pi}\frac{dx}{\left(\sin^2 x\right)^{\frac{1}{3}}}=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}$$ hence: $$\begin{eqnarray*} \left|\int_{\pi}^{+\infty}\frac{dx}{x^2\left(\sin^2 x\right)^{\frac{1}{3}}}\,dx\right|&\leq& ...


1

Assuming we are using the usual topology of $\mathbb{R}$. a. $A_1 = \left\{ \frac{1}{n\pi} \,|\, n \in \mathbb{Z} -\{0\}\right\}$. So clearly $A_1$ is NOT open , since it has isolate points (actually all of its points are isolated points). It is also clear that $A_1$ is NOT closed, since $0$ is in its closure and $0\notin A_1$. In other words, there is a ...


0

Hint. You have an Euler-type integral, you may then rewrite your initial integral in terms of the Appel hypergeometric function $$ F_1(a,b_1,b_2,c;x,y)=\dfrac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1\frac{v^{a-1}(1-v)^{c-a-1}}{(1-xv)^{-b_1}(1-yv)^{-b_2}}~dv $$ where $\Re c>\Re a>0$ with the change of variable $$ v=\frac{u}{1-e^{-\lambda x}}. $$


0

I think a simpler and more intuitive way to approach this problem is the following: A (subset of a) metric space is compact iff it is complete and totally bounded. A closed subset of a complete metric space is complete. A closed interval is a closed set, and $\mathbb R$ is complete, so a closed interval is complete. A closed interval is bounded, and any ...


0

The definition of derivatives on the boundary of a domain is not completely agreed upon. My preferred definition of differentiability of a function $f\colon A \to \mathbb R$ with $A\subset \mathbb R^n$ in any point $x_0\in A$ is the following. We say that $f$ is differentiable in $x_0$ if there exists a linear map $L$ such that $$ \lim_{x\to x_0} ...


1

The claim is not true for merely continuous and bounded $g$. For instance, take $$ g(v):=\min( \sqrt{\|v\|},1). $$ Fix $x$ such that $\Pi x \ne \hat\Pi x$. Then for $s>0$ such that $\|s\hat\Pi x\|\le1$ and $\|s\Pi x\|\le1$ $$ \|g(s\hat\Pi x)-g(s\Pi x) \|= \sqrt s\left|\sqrt{\|\hat\Pi x\|} - \sqrt{\Pi x}\right|, $$ hence for $s\searrow 0$ the quantity $$ ...


2

$\int_0^1\vert f_n(x)-1\vert dx=\int_0^{1/n}\vert f_n(x)-1\vert dx+\int_{1/n}^1\vert f_n(x)-1\vert dx$ Since $f_n(x)=1$ for $x\geq\frac{1}{n}$, the second term is equal to $0$. You can also remove the absolute value in the first term since for $0\leq x\leq\frac{1}{n}$, $f_n(x)=nx\leq 1$.


1

The continuity of $F$ is proved by zhw. As you probably know, the key point is to show that for any compact set $K\subseteq \mathbb H$, one can find an integrable function $g_K$ such that $\vert f(x,y,t)\vert\leq g_K (t)$ when $(x,y)\in K$. Then you can apply the "standrad" theorem on continuity for integrals depending on a parameter; or reprove this theorem ...


1

The difference between the formulae you have given lies in the definition of $m_j$ and $M_j$, that is: $$M_j:=\sup_{x\in [x_{i-1}, x_i]} \{f(x)\} \\ m_j:=\inf_{x\in [x_{i-1}, x_i]} \{f(x)\}.$$ Here the definition of sumpremum and infimum of a set. The upper and lower Riemann sums are defined in terms of $M_j$ and $m_j$ (see, for example, Rudin's ...


1

We have to prove that $\lim_{x\rightarrow a} f(x)$ exists. It is enough to prove that for any two sequences $\{x_n\}$ and $\{y_n\}$ tending to $a$ the sequences $\{f(x_n)\}$ and $\{f(y_n)\}$ converge to the the same limit. If $\{x_n\}$ and $\{y_n\}$ converge to $a$ then for any $\delta>0$ there exist $N_x$ and $N_y$ such that $|x_n-a|<\delta$ for any ...


4

I'd say so. A sum of non-negative reals is 0 iff every term was itself zero. This is easy to prove independently.


1

I find doing proofs this way (especially in analysis) very helpful in understanding the details of what is happening. Here is a complete proof for the forward direction; of course I would shorten in considerably if I were to tun it in as an assignment or just thinking through it for myself. $f: (X, d) \rightarrow (Y, \rho)$ is uniformly continuous $E ...


3

For any function $f : \mathbb R \rightarrow \mathbb R$, the set $f^{-1}( a, \infty)$ always exists, and it is defined by $$ \{ x: \in \mathbb R : f(x) > a \}, $$ although it is not always measurable.


10

$f^{-1}$ here does not mean the inverse of $f$. It is a horrifically bad overuse of notation and can be very misleading. $f^{-1}(A)$ where $A$ is a set is the pre-image, i.e. $$f^{-1}(A) = \{x\in X: f(x) \in A\}.$$ The pre-image always exists. If $f$ were invertible, then this would coincide with what you think; however the pre-image takes care of the ...


18

$f^{-1}(A)$ is the preimage of the set $A$, it exists even if $f$ is not invertible. For $f: X\to Y$, it is defined as $$f^{-1}(A) = \{x\in X: f(x)\in A\}.$$ The notation is slightly confusing, I would say, but one gets used to it. It isn't really that bad of a notation, since, if $f$ is actually invertible and its inverse is $g$, then $g(A) = f^{-1}(A)$ ...


0

There's more than logical inference to be done in this step. I think that, in effect, you need the following lemma: For all real numbers $r$, if for every $p,q \in E$ we have $d(p,q) \le r$, then $\operatorname{diam}(E) \le r$. The proof of this lemma essentially involves looking at the definition of $\operatorname{diam}(E)$, which is defined as a ...


0

Start with this visualization of Lipschitz continuity by A. di M.: As you slide the focus of the green overlay along the graph of the function, the rest of the graph stays within the green area. The Lipschitz condition says that the modulus of continuity is bounded by a linear function, so you can use a green area with straight sides. The Hölder ...


0

The idea is that neither gives a perfect answer for the area "under" $f$ (technically the "signed area" under $f$, meaning the area between $f$ and the $x$-axis, where the area counts as negative when $f$ is negative). But if $f$ is a Riemann integrable function, then if you take $P$ to be a finer and finer partition, the upper and lower sums $U(P,f)$ and ...


8

For a non-Taylor series way of evaluating, \begin{align*} \lim_{h \to 0} \frac{f(x+2015h) - f(x+2014h)}{h} &= \lim_{h \to 0} \left( \frac{f(x+2015h) - f(x)}{h} - \frac{f(x+2014h) - f(x)}{h} \right) \\ &= \lim_{h \to 0} \frac{f(x+2015h) - f(x)}{h} - \, \lim_{h \to 0} \frac{f(x+2014h) - f(x)}{h} \\ &= 2015 \lim_{h \to 0} \frac{f(x+2015h) - ...


1

Use Taylor expansion \begin{align*} & \frac{f(x + 2015h) - f(x + 2014h)}{h} \\ = &\frac{[f(x) + f'(x)(2015h) + o(h)] - [f(x) + f'(x)(2014h) + o(h)]}{h} \\ = & \frac{f'(x)h + o(h)}{h} \\ = & f'(x) + o(1) \to f'(x) \end{align*} as $h \to 0$.


0

I read this from the book called "The Malliavin Calculus and related Topics", p.27. In fact, it is an expectation question. $(E[|F|^{p}]+\sum_{i}E[||D^{i}F||_{H^{i}}^{p}])^{1/p} \leq (E[|F|^{q}]+\sum_{i}E[||D^{i}F||_{H^{i}}^{q}])^{1/q}$ After simplification, it suffices to prove $[\int \left(|f(x)|^{p} + \sum_{i=1}^n |f^{(i)}(x)|^{p} ...


0

Well my solution looks basically the same as that of A.G; perhaps it's a little more elementary. I'll include it since I just spent some time on it! Suppose $$\sum_{k=1}^{\infty}|a_k| < \infty, \sum_{k=1}^{\infty}|b_k|^2 < \infty.$$ Let $f(z) = \sum_{k=1}^{\infty}a_kz^k, g(z) = \sum_{k=1}^{\infty}b_kz^k.$ Then in the open unit disc, $$f(z)g(z) = ...


1

We need to prove that $$ \sum_{k=1}^\infty\left|\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}\right|^2\frac{1}{\lambda^2}<+\infty. $$ Denote $x_k=\sum_{j=1}^ky_j\left(\frac{1}{\lambda}\right)^{k-j}$, which gives $$ x_{k+1}=\frac{1}{\lambda}x_k+y_{k+1}, \qquad x_0=0,\ k\ge 0.\tag1 $$ Do $z$-transform of the equation, i.e. multiply by $z^{k+1}$ and ...


5

The one dimensional case The result is true if we assume that some integrals are finite. Below I will assume that $\int_0^\infty x \,\mu(dx)=\int_0^\infty x \,\nu(dx)<\infty.$ Emanuele Paolini's argument shows that some such assumption is necessary. Fix $z\in\mathbb{R}$ and $m>0$. The functions $f(x)=(m(x-z)+1)_+$ and $g(x)=m(x-z)_+$ are both ...


1

For a bounded example, try $f_n(x) = (-1)^n$.


1

Write out the first few terms. We have: $$a_1=y$$ $$a_2=q(y+x)=qy+qx$$ $$a_3=q(qy+qx+x)=q^2y+q^2x+qx$$ $$a_4=q(q^2y+q^2x+qx+x)=q^3y+q^3x+q^2x+qx$$ And so on. As n grows we see that the term in y goes to $0$ (it is $q^{n-1}y$). The terms in x are a geometric series which we can easily sum to deduce that the sequence converges to $$\frac {qx}{1-q}$$


2

If $a_{n+1}=q(a_n+x) $, then $\frac{a_{n+1}}{q}=a_n+x $ or, dividing by $q^n$, $\frac{a_{n+1}}{q^{n+1}}=\frac{a_n}{q^n}+\frac{x}{q^n} $. Let $b_n =\frac{a_n}{q^n} $. Then $b_{n+1} =b_n+\frac{x}{q^n} $ or $b_{n+1}-b_n =\frac{x}{q^n} $. Summing from $n=1$ to $m-1$, $b_m-b_1 =\sum_{n=1}^{m-1} \frac{x}{q^n} =x \frac{\frac1{q^m}-\frac1{q}}{\frac1{q}-1} ...


1

Let $E_R =[-R,R]\times [1/R,\infty).$ Claim: There exists a constant $C_R>0$ such that $$f(x,y,t)\le C_R, \text {for}\,(x,y) \in E_R, t\in \mathbb {R}.$$ Suppose the claim is true. Fix $(a,b)$ in the upper half plane. Then there is $R> 0$ such that $D((a,b),b/2)\subset E_R.$ Let $(x_n,y_n) \to (a,b).$ Then for large $n, (x_n,y_n) \in E_R.$ For such ...


1

Denote $$ X_k=\left\{x\in\mathbb{R}\colon\ \mu(f^{-1}(x)\cap[-k,k])\ge\frac1k\right\}. $$ Then $$ X=\{x\in\mathbb{R}\colon\ \mu(f^{-1}(x))>0\}=\bigcup_{k=1}^{\infty} X_k. $$ Fix $k$ and let $x_1, \ldots,x_m$ be different points from $X_k$. Since $f^{-1}(x_j)$ are disjoint we have $$ 2k=\mu([-k,k])\ge\sum_{j=1}^m\mu(f^{-1}(x_j)\cap[-k,k])\ge\frac{m}{k}, ...


2

Let $z=e^{i\theta}$, so that $dz=ie^{i\theta}d\theta$ and the integral $I$ corresponds to the contour integral $$ \oint_C \frac{1}{z^{n}e^{z\rho\log n}}\frac{dz}{iz}=\frac{1}{i}\oint_C\frac{dz}{z^{n+1}e^{z\rho\log n}} $$ where $C$ is the unit circle. The function $f(z)=\frac{1}{z^{n+1}e^{z\rho\log n}}$ is meromorphic with a pole of order $n+1$ at the origin, ...


1

Your set looks countable to me. If I parsed it correctly, the hint is that any family of pairwire disjoint sets of positive measure is countable.


1

Actually I forgot to add the $\cos nx$ in the RHS. Hence: $$\cosh ax = \frac{\sinh \pi a}{\pi a}+ \frac{2 a\sinh \pi a}{\pi}\sum_{n=1}^{\infty}\frac{ (-1)^n \cos nx}{n^2 +a^2} \overset{x=\pi}{\implies}$$ $$\overset{x=\pi}{\implies} \pi \coth \pi a= \frac{1}{a}+ \sum_{n=1}^{\infty}\frac{2a}{n^2+a^2}$$ and the identity is proven. @Aretino, thanks for the ...


6

Take on $\mathbb R$ the measure $d\mu = \frac{dx}{1+x^2}$. Then for every non constant convex function $f(x)$ if the integral is well defined one has $$ \int f(x) d\mu(x) = +\infty $$ since $f(x)>mx$ for either $x\to +\infty$ or $x\to-\infty$. For constant functions the integral only depends on the total mass of the measure $\mu$. Hence you cannot ...


2

Basically what you're looking for is $u'(x)$ $$a*u(x-2h)=a*u(x)-a*u'(x)*(2h)+ a*u''(x)*(2h)^2+...$$ $$b*u(x-h)=b*u(x)-b*u'(x)*(h)+ b*u''(x)*(h)^2+...$$ $$c*u(x+2h)=c*u(x)+c*u'(x)*(2h)+ c*u''(x)*(2h)^2+...$$ Hence: $$a+b+c=0$$ $$-2a-b+2c=1$$ (or any non zero value other than $1$ but be careful later) $$4a+b+4c=0$$ Solve for $a,b$and $c$ and you're done. ...


0

$$ \mathfrak{F}(\int_0^x f(y)g(x-y)dy)=\mathfrak{F}(f(t))\times\mathfrak{F}(g(t))=0$$ In your case $g(x)=e^{-x^2/2}$. $$\mathfrak{F}(g(t))=e^{-x^2/2}>0$$ $$\mathfrak{F}(f(t))=0$$ $$f(t)=0$$


0

Since all constants are positive real numbers, and v0 and v1 are also strictly greater than 0, we can consider two cases. Either, 1) (1-x)>0 or 2) (1-x)=0. In the first case it is easy to see that the function must be greater than 0, since we are multiplying and adding positive numbers. For case number 2 the two terms with (1-x) must still be greater than ...


1

$$f(x)=\sin^2 (x)$$ $$f(0)=0$$ $$f'(x)=\cos(x)2\sin(x)=\sin(2x)$$ $$f'(0)=0$$



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