# Tag Info

3

Let $f(x)$ be zero except for occasional bumps up to $1$ and $g(x)$ the same. If the bumps don't line up, the product will be identically $0$, but neither $f$ nor $g$ will have a limit.

0

$$G(a,b) = \sqrt{a\cdot b}\quad\forall a\cdot b \in [0,\infty),\\ \text{ and } A(a,b) = \frac{a + b}{2} \quad\forall a,b \in \mathbb R$$ Now, we are given that $a = b$, and we need to prove that $G(a,b) = A(a,b)$ $$a = b \\ \implies a - b = 0 \\ \implies (a-b)^2 = 0 \\ \implies a^2 - 2ab + b^2 = 0\\ \implies a^2 + b^2 = 2ab\\ \implies a^2 + b^2 + 2ab = ... 2$$a^2-2ab+b^2=(a-b)^2=0\implies a-b=0\implies a=b$$0 Replacing that axiom would make certain structures into models that were originally not models. For example take standard \mathbb{R} but define every number in the interval [-10,-5) to be greater than every number in the interval [-5,-4). We see that < is still asymmetric, dense, and D-complete. Addition is unchanged. The statement that "x \le y ... 1 I don't see how you would solve the first equation in its own; the system is coupled. Picard iteration for first-order systems works the same as for first-order equations; you just integrate a vector-valued function. In your case, writing z = (x,y), we have$$z(t) = (0,1)+\int_0^t f(z(s))\,ds \qquad \text{where } \ f(x,y) = (y,-\sin x) $$So, starting ... 0$$f(x) = \frac{1}{1-x} - \frac{1}{1+x}$$has nonremovable singularities at x = \pm 1. As x becomes closer and closer to these points, |f(x)| grows without bound. This is true no matter how you rearrange the formula for f. However, for values of x near 1, \frac{1}{1+x} is very small in magnitude compared with \frac{1}{1-x} so you could simply ... 0 Convolution with a mollifier, suggested by Jack D'Aurizio, works well on Euclidean spaces. Another approach, which applies to general metric spaces, is to use the upper or lower L-Lipschitz envelope of given function f:X\mapsto \mathbb R. Specifically,$$g_L (x) = \sup_{y\in X}( f(y)-L\,d(x,y)) \tag{upper envelope}h_L (x) = \inf_{y\in X}( ...

0

Take for example $V = \ell^2(\Bbb{N})$ and $a_n = 1/n \cdot \delta_n$ with $(\delta_n)_m =1$ for $n=m$ and $0$ otherwise. Verification of the claimed properties is an exercise.

1

It is not possible because the set $[0,0.5) \cup (0.5, 1]$ is the union of two disjoint closed sets. So this space is not connected.

4

No, because a continuous function sends connected sets into connected sets (connection is a topological property).

1

(Expanded from my comments) Theorem: $\;D(f)\;$ is $\;G_{\delta \sigma}$ First, Henning Makholm's accepted answer to Continuous functions are differentiable on a measurable set? shows that $D(f)$ is $F_{\sigma \delta \sigma}\;$ (and not $G_{\delta \sigma \delta}).$ Note that Henning Makholm wrote $$\forall\varepsilon:\exists \delta:\exists Y:\forall ... 0 Here I use the Beta function to evaluate this integral and the calculation is a little bit tedious. As Omran Kouba did, \begin{eqnarray} I&=&\int_{0}^{\Large\frac{\pi}{2}} \ln^2 {(\sin x)}\cdot\ln {(\cos x)}\tan xdx. \end{eqnarray} Let$$ J(a,b)=\int_0^{\frac{\pi}{2}}\sin^ax\cos^bx\tan xdx. $$Then clearly$$ \frac{\partial^3 J(0,0)}{\partial ...

1

$$\lim_{n\to\infty}\sum^n_{k=1}\frac{1}{k}-\ln(n) = \gamma$$ It is widely known as the Euler–Mascheroni constant which is defined as the limiting difference between the harmonic series and the natural logarithm. $$\gamma \approx 0.5772\space1566\space4901\space5328\space6060\space6512\space090(1)$$ The radius of curvature is not valid for this series as it ...

0

Since $F(t)>0$ for $t>0$ we have $$\frac{F'(t)}{F(t)}\ge \frac{\theta}{t} \quad \forall t>0.$$ Hence: $$\ln\frac{F(s)}{F(1)}=\int_1^s\frac{F'(t)}{F(t)}\,dt \ge \int_1^s\frac{\theta}{t}\,dt=\theta\ln s=\ln s^\theta.$$ It follows that $$F(s)\ge F(1)s^\theta \quad \forall s>0.$$

0

Hint: take $f(x) = \begin{cases} |x|, &|x|<1 \\ 1, &|x| \geq 1 \end{cases}$ and $f_n$ as its smooth approximation

2

Put $\displaystyle g(x)=\frac{f(x)-f(a)}{x-a}$, with $g(a)=f^{\prime}(a)=0$. We compute : $\displaystyle g^{\prime}(x)=\frac{(x-a)f^{\prime}(x)-(f(x)-f(a))}{(x-a)^2}$. Now suppose that the assertion is false. Then $g^{\prime}(x)$ is not zero for $x\in (a,b)$. By Darboux's theorem, $g^{\prime}$ has a constant sign on $(a,b)$. Suppose for example that ...

2

Take $f_n(x) = \sqrt{x^2+{1 \over n}}$. We have $f_n \to | \cdot |$ uniformly, and all derivatives are bounded (by 1), but the limit is not $C^1$. Jonas has pointed out a flaw in my example. I think the following is a fix. Let $f_n(x) = \arctan \sqrt{x^2+{1 \over n}}$, then $f_n \to \arctan \circ (| \cdot |)$ uniformly, and all quantities are bounded.

2

This is the Euler-Mascheroni constant, $\gamma \approx 0.5772 \ldots$

0

Your condition does hold in the case where $x/n=1/2$, due to the symmetry in $f(p)$. That is, $$f(1/2 + h)=f(1/2 - h).$$ In the general case, I don't think there is a nice condition, since the growth/decay on one side of $x/n$ will be different than the decay/growth on the other side.

1

Let $f:{\mathbb R} \rightarrow {\mathbb R}$ be an arbitrary function. Then the set of points where $f$ does not possess a finite derivative is a $G_{\delta \sigma}$. See relevant question: Continuous functions are differentiable on a measurable set?

0

There is an interesting related result (which unfortunately does not answer your question): The pointwise limit of continuous functions can only fail to be continuous on a meagre set. Thus if a function on the real line is differentiable everywhere, its derivative is continuous outside a meagre set.

1

Let $g(x)=\lvert x\rvert$ in $[-1,1]$, and extend $g$ to be periodic in $\mathbb R$, with period $2$. Set $$g_{k,\ell}(x)=\frac{1}{k} g(\ell x).$$ It is not hard to see that $$g_{k,\ell}\in E_n \quad\text{iff}\quad n\ge \frac{\ell}{k}.$$ Fix now $n\in\mathbb N$. We shall show that $E_n$, which is a closed subset of $C[0,1]$, has empty interior. Let ...

0

We may assume that $f$ has no multiple roots as in that case one of the critical points would be zero and not have the same sign as the other. Assume there are at least two real roots $x_1,x_2$ of $f(x)=x^2+ax^2+bx+c$. Then by polynomial division you find $f(x)=(x-x_1)(x-x_2)g(x)$, where $g$ is linear, i.e. $f$ has three (distinct!) real roots, wlog ...

0

Let $f(x) = x^3 + p x + q$. Then $f'(x) = 3x^2 + p$. Let $a = -\sqrt{\frac{-p}{3}}$ and $b = \sqrt{\frac{-p}{3}}$, so $f$ is increasing when $x \le a$ or $x \ge b$. In the interval $a \le x \le b$ the function is decreasing. Assume first that $f(a),f(b) \le 0$. Then because $f$ is (strictly) increasing to the left of $a$, we must have $f(x) < f(a) \le 0$ ...

2

I assume you mean $C[0,1]$ with sup-norm. Hint: Try to show that for each $n$ and arbitrary $\varepsilon>0$ there exists $g\notin E_n$ such that $\|g\|_\infty<\varepsilon$. Try to show that $f\in E_n$ and $g\notin E_{2n}$ implies $f+g\notin E_n$. Using these two facts you should be able that if $f\in E_n$ then in any ball $B(f,r)$ there is a ...

0

Let's start with along the lines of the standard proof. Let us divide $[0,1]$ into $2k$ intervals of length $1/k$; i.e. $[0,1/k]$, $[1/k,2/k]$, $[2/k,3/k]$, etc. Now by Dirichlet principle there are two numbers $a\ne b$ such that $\{a\alpha\}$, $\{b\alpha\}$ which are in the same interval. If $b>a$, then $(b-a)$ is a positive integer and either ...

0

Let $Lf = -if'$ be defined on the linear space $\mathcal{D}(L)$ of absolutely continuous functions $f \in L^{2}[a,b]$ for which $f(a)=f(b)$ and $f' \in L^{2}[a,b]$. $L$ is symmetric on its domain, i.e., $(Lf,g)=(f,Lg)$ for all $f,g\in\mathcal{D}(L)$. It is not hard to show that $(L-\lambda I)$ is surjective for $\lambda \ne n\frac{b-a}{2\pi}$ for $n=0,\pm ... 2 Adding a few things to the good answers you already received, using the change of variable suggested by TeeJay, you obtain $$\int{\frac{e^{-nx}}{\sqrt{x}}}dx=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{n} \sqrt{x}\right)}{\sqrt{n}}$$ from which $$\int_{0}^{\infty}{\frac{e^{-nx}}{\sqrt{x}}}dx=\frac{\sqrt{\pi }}{\sqrt{n}}$$ provided$n \gt 0$. 0 Suppose not: then for some$\delta$and some$n_k\uparrow\infty$, $$\lVert f_{n_k}-f\rVert_p\geqslant\delta,\quad k\geqslant 1.$$ Using the definition of convergence in measure, we can construct$m_k\uparrow\infty$such that$\lambda\{x, |f_{n_{m_k}}(x)-f(x)|>2^{-k}\}\leqslant 2^{-k}.$The sequence$(f_{n_{m_k}})_{k\geqslant 1}$converges almost ... 1 Yes. This is maybe easier to see if we phrase convergence in terms of sequences. $$\lim_{x\to\infty} g(x) = b$$ Is equivalent to the statement " For every sequence$\{a_n\}$going to infinity,$\{g(a_n)\}$converges to$b$", and similarly, $$\lim_{x\to b}f(x) = f(b)$$ is true only if for every sequence$\{b_n\}$converging to$b$,$f(b_n)$converges to ... 1 For a gentle introduction to functions of bounded variations, I recommend A First Course in Sobolev Spaces By Leoni, where Chapter 2 concerns one-variable BV space and Chapter 13 deals with several variables. There are many exercises throughout the text. Special functions of bounded variation (SBV) is a recent topic, not even 30 years old yet. The book ... 4 The integrand is less than$1/\sqrt{x}$, so that$0<\int_\varepsilon^1 \frac{e^{-nx}dx}{\sqrt{x}}<\int_0^1 \frac{dx}{\sqrt{x}}=2$, and$\int_0^1 \frac{e^{-nx}dx}{\sqrt{x}}$exists. As to$\int_1^\infty \frac{e^{-nx}dx}{\sqrt{x}}$, the integrand rapidly decreases to$0$as$x\rightarrow\infty$, in particular,$0<\int_1^N ...

2

Regarding $\int_{0}^{\infty}e^{-nx}/\sqrt{x} dx$, the tail is not a problem because for $x > 1$ we have $$0 < \frac{e^{-nx}}{\sqrt{x}} < e^{-nx}$$ and the right hand side is integrable. The only other cause for concern is the singularity at $x=0$. But this is also not a problem, because for all $x > 0$, we have $e^{-nx} \leq 1$, and so $$0 < ... 1 Without giving too much away, there is a way to simply evaluate the integral. It is related to the Gamma function \Gamma(\frac{1}{2}) but there is a subtlety involing the variable x in place of nx. By performing a variable substitution, you should get a Gaussian integral which can be directly evaluated. For starters, let u=\sqrt x and go from there. ... 2 Injective: suppose that$$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_2^2+1}}\ .\tag{$*$}$$Squaring both sides and multiplying out denominators,$$x_1^2(x_2^2+1)=x_2^2(x_1^2+1)\quad\Rightarrow\quad x_1^2=x_2^2\ .$$Now substituting back into the denominator on the RHS of (*),$$\frac{x_1}{\sqrt{x_1^2+1}}=\frac{x_2}{\sqrt{x_1^2+1}}\quad\Rightarrow\quad ...

1


2

This proof cover all cases: $a,b$ finite or not. In the case that $a,b$ are not finite, $u(a)$ is understood as $\lim_{x\to-\infty}u(x)$. Analogous for $b$. Also, if $a,b$ are not finite, then we weill consider locally things, i.e. $BV_{loc}((a,b))$. I am also assuming that $H_0^1((a,b))$ is the closure of $C_0^1((a,b))$ with respect to the $H^1((a,b))$ ...

1

In fact, $\alpha\beta$ is optimal. To prove it, let $f,g:[0,1]\to\mathbb{R}$ with $f(x)=x^\alpha$ and $g(x)=x^\beta$. Note that $f\circ g\in C^{\alpha\beta}$, however, for all $\gamma>\alpha\beta$, $f\circ g$ does not belong to $C^\gamma$.

1

There's a simpler explanation of why $R'(0) = 0$: You assumed your solution is radially symmetric, and performed a separation of variables. If $R'(0) \neq 0$ you have a cusp at the origin, where the function $u$ is in fact not differentiable. So you require $R'(0) = 0$. The power series explanation is basically the same: you need $$R'' + R'/r = c R$$ ...

4

This question is answered in the affirmative in Abraham, Robbin, Transversal mappings and flows, Ch.1, $\S$2, A criterion for smoothness. They prove this converse to Taylor's theorem for functions between Banach spaces and attribute the one-dimensional case to Marcinkiewicz, Zygmund, On the differentiability of functions and summability of trigonometrical ...

1

You're assuming the series $\sum\limits_{n=1}^\infty |a_n|$ converges. This, by definition means that the sequence of partial sums, $(S_m)$, given by $S_m=\sum\limits_{n=1}^m |a_n|$, converges. If you look closely, you should be able to see that your condition is just saying $(S_m)$ is a Cauchy sequence (which it is, of course).

1

Just inspect the definition of $\lim \dfrac{\sqrt{x}}{\log x} = \infty$. It says given any $M \gt 0$ there is a $m \gt 0$ such that $x \ge m \implies \dfrac{\sqrt{x}}{\log x} \gt M$. Notice the definition for an infinite limit of a sequence is similar. Let $M \gt 0$ be arbitrary. Then there is a natural number $n' \gt m$. Then for every $n \gt n' \gt m$ we ...

0

I saw the other answer. Though I'm a beginner myself in this subject, I'd like to try my hand at it. Let us define the terms first. $\lim\sup s_n :=\lim_{N \rightarrow \infty} \sup\{s_n:n>N\}$ Let $A_N := \{s_n:n>N\}$, $v_N := \sup A_N$ Let's talk about the property of the sequence of $v_N$. By the definition of the sets $A_N$, it can be seen that ...

0

Define the set $NE = \{n x \ | \ x \in E\}$. Now clearly $x \le \frac {K}{n}$ for every $x \in E$ and hence $nx \le K$ whence it follows that $K$ is an upper bound for $NE$. Then $l = \sup NE$ exists. Now consider the set $A = \{ n \in \Bbb N \ | n \ge l \ \}$. This set is non-empty since it contains $K$. By the Well-Ordering principle this set ...

4

Given that: $$\frac{x_{n+1}}{y_{n+1}}=\frac{x_n}{y_n}+\frac{1}{x_n y_n}$$ we have: $$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n y_n}+\frac{1}{x_{n-1} y_{n-1}}+\ldots+\frac{1}{x_1 y_1}+\frac{x_1}{y_1}$$ or just: $$\frac{x_{n+1}}{y_{n+1}}=\frac{1}{x_n\cdot\ldots\cdot x_1}+\frac{1}{x_{n-1}\cdot\ldots\cdot x_1}+\ldots+\frac{1}{x_1}+2.$$ Hence we need to prove that: ...

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