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No, not all neighbourhoods are connected. For example, in $\mathbb R$ with the usual metric, $(-1,1) \cup (2,3)$ is a disconnected neighbourhood of $0$. A neighbourhood of $x$ is not $N_r(x)$, it's any set that contains $N_r(x)$. But it's also not true that $N_r(x)$ is connected. For example, in $\mathbb Z$ with the usual metric, $N_r(x)$ is disconnected ...

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$$\left|\frac{n}{n^3 - 3} \right| < \epsilon \implies n > N$$ For some $N$. Since $n > 1$ (we can set this because you are searching convergence), $$n^3 - 3 < n^3 \implies \frac{1}{n^3 - 3} > \frac{1}{n^3}$$ Giving, $$\left| \frac{n}{n^3} \right| < \left|\frac{n}{n^3 - 3} \right| < \epsilon$$ $$\to \left| \frac{1}{n^2} \right| < ... 1 No. Take A = (1,0), B=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})  , C = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}) and D=(-1,0) Take d_1((x,y),(x',y')) = |x-x'|+|y-y'| and d_2 = \sup( |x-x'|,|y-y'|) For d_1, the diameter is realized for B and C and for d_2 the diameter is realized for A and D 0 No. You could have a metric space consisting of finite set of points (but more than one) in \mathbb R. With the usual metric, the diameter of this set is the largest point minus the smallest point. Under the discrete metric, the diameter is 1, and could correspond to any pair of distinct points. 0 I can't say anything about the applications to Hausdorff metrics, but there's one case where a multi-valued mapping comes up in a big way, and for a good reason: assigning, to a nonzero point of the plane, an "angle", with points on the positive x-axis getting an angle of 0, the positive y-axis getting angle \pi/2, etc. The problem with trying to do ... 3 By the assumption that f(0) = f'(0) = \dotsc = f^{(n-1)}(0) = 0, the Taylor series of f centred at 0 is$$f(z) = \sum_{k = n}^\infty a_k z^k.$$Hence the function$$g(z) = \frac{f(z)}{z^n} = \sum_{k = n}^\infty a_k z^{k-n} = \sum_{k = 0}^\infty a_{n+m}\cdot z^m$$is holomorphic on B(0,1). By the maximum modulus principle, for 0 < r < 1, ... 2 Differentiation is a real operator, that is, it commutes with conjugation,$$d\overline{\omega} = \overline{d\omega}$$for every k-form. Since you know that \overline{f(z)}\,dz is closed, i.e. d\bigl(\overline{f(z)}\,dz\bigr) = 0, the result follows. Alternatively, use the Wirtinger derivatives,$$dg = \frac{\partial g}{\partial z}\,dz + ...

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@Ian has provided a very efficient way to tackle a problem such as this. I thought it might be instructive to see how tight a bound we can achieve. To that end, let's write for $n\ge2$ \begin{align} \left|\frac{n}{n^2-3}\right|<\epsilon \implies n&>\frac{1}{2\epsilon}+\frac{1}{2\epsilon}(1+12\epsilon^2)^{1/2} \tag 1\\\\ ... 3 If n \geq 3 then n^2-3>n^2/2 hence the quotient is less than 2/n . Can you finish from here? The point of this initial step was to get rid of the subtraction by absorbing it into a constant factor. This is desirable because, as you noticed, the subtraction in the denominator makes it harder to get a good upper bound. You would do the opposite to ... 1 I think the correct statement to prove for a circle of radius r, circumference C, and area A is:A = \frac{1}{2} r \, C$$(otherwise you run into question of the type "... but what is \pi?" ) Now you can have your favorite intuitive proof ( using say inscribed and circumscribed polygons). I'll try a "blind" proof using integrals. It's enough ... 1 As pointed out by David C. Ullrich, much depends on the desired level of rigour in the definitions of area and arc length. For expositions that would be acceptable at school level, see Geometría elemental by Pogorelov, or Geometry by Lang and Murrow, with the first being more rigorous than the second. Naturally, the level of rigour increases when you look ... 2 I'm starting with the definition of 2\pi as the total angle around a point, or, equivalently, as twice the measure of a straight angle. As I often do, I'll play around with an idea and see where it leads. Divide a circle or radius r into n congruent triangles with one vertex on the center and two on the circle. For each of these triangles, half the ... 1 Let A(r) be the area of a disk with radius r and C(r) the length of its boundary. Then:$$ \frac{d}{dr} A(r) = C(r)\tag{1} $$since the difference between two concentric disks with radii r,r+\varepsilon is a thin annulus with width \varepsilon and length \approx C(r). On the other hand,$$ A(r) = K r^2 \tag{2} $$for some constant K that equals ... 1 Suppose we already know that C = 2\pi r. Let A be the area of a circle of radius r and let A+\Delta A be the area of a circle of radius r+\Delta r, whose circumference is C+\Delta C= 2\pi(r+\Delta r). Here's a step that I'm not sure how to prove simply and quickly while still being rigorous:$$ C\,\Delta r \le \Delta A \le (C+\Delta C)\,\Delta ...

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This is not at all correct. Try e.g. $f(x) = x^2$. Then $(f(x)-f(y))/(x-y) = x + y$, and the requirement on $g$ is that $\int_{\mathbb R} g(y)\; dy = 0$ and $\int_{\mathbb R} y g(y)\; dy = 0$. Somewhat more generally, if $f$ is a polynomial of degree $d$ you need the moments $\int_{\mathbb R} y^k g(y)\; dy = 0$ for $0 \le k \le d-1$.

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It's true if $\xi$ is continuous at $0$ and $1$. Alternatively, it's true for arbitrary $\xi$ if we restrict to $f$ with $f(0)=0=f(1)$. I'm assuming here that $\xi$ is defined on $\mathbb R$, so that the convolution is defined. If $\xi$ is originally defined only on $[0,1]$ you can extend it. There's a simple proof if you know about Schwarz functions and ...

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That seems false to me. Look at $$f(x) = \begin{cases} x \sin \frac{1}{x^2} & x \ne 0 \\ 0 & x = 0 \end{cases}$$ On the interval $[-1, 1]$ it's bounded and continuous, and can be extended to a continuous function by constants outside that interval. But I don't believe it's locally Lipschitz near zero, because the derivative, as you approach ...

2

You should check the Lawrence Evans' book about PDEs. More precisely, you can check its fifth chapter. There, you will see that, under some really general assumptions on the open set $\Omega \subset \mathbb{R}^n$, you have that $C^{\infty}$ is dense on $W^{k,p}$. About the other question, you generally have that $C^{\infty}_c$ is not dense on the ...

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Here is one way. Let $$I(a)=\int_0^{\infty}\frac{\sin x}{x}e^{-ax}dx$$ Taking the derivative of $I$ yields $$I'(a)=-\int_0^{\infty}e^{-ax}\sin x\,dx=-\frac{1}{a^2+1}$$ Integrating, we find that $$I(a)=-\arctan(a)+C$$ Noting that $\lim_{a\to \infty}I(a)=0$, we see that $C=\pi/2$. Thus, $$I(0)=\pi/2$$ and we're done!

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Let $f$ have an uncountable number of zeros. Assume that each disk $D_n=\{z:|z|\leq n\}$ contains only a finite number of zeros of $f(z)$. Call the set of these zeros $A_n=\{z\in D_n: f(z)=0\}$. Then the set of all zeros of $f$ is $$A=\bigcup_{n\in\mathbb N}A_n$$ is countable, at most. This is a contradiction, so there exists $D_n$ which contains an infinite ...

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We will build an uncountable discrete set of functions belonging to $BVC([0,1])$. Let $\alpha=\alpha _1\alpha _2...$ is infinite binary sequence. We will build the analogue of Cantor function for every such sequence. Let us consider the next construct (or see informal description below). Let $F_\alpha (0)=0, F_\alpha (1)=1$; Step 1. For $x\in ... 0 I don't see why you think you need to show they are the same for all Lebesgue measurable sets. Consider the example$\alpha$being the Cantor function https://en.wikipedia.org/wiki/Cantor_function. Then the Cantor set has$\mu$measure equal to$1$, but Lebesgue measure zero. Then you should be able to find a subset$E$of the Cantor set which is ... 0 Hard to beat the simple solution of @Keith: We'll give a proof for$X$infinite dimensional Banach space (extra condition). First, show that there exists a sequence$x_n$in$X$such that$||x_n|| =1$and$d(x_n, \langle x_1, \ldots x_{n-1}\rangle ) \ge \frac{1}{2}$. One constructs the sequence inductively. Once$x_1$,$\ldots x_{n-1}$are obtained, take ... 0 I heartily recommend the part 23, Haar Measure. Convolution, of the book Bourbaki. Elements of the History of Mathematics, Springer. I have the book but in Spanish. This chapter 23, not numbered as such in the original version, is truly extraordinary and bright. I copy here from Internet the beginning of this part 23 which unfortunately does not reach the ... 0 This is not what is expected, but a different method which allows to compare the respective results. $$\frac{d^2x}{dt^2}+\left(\frac{dx}{dt} \right)^2-2x=0$$ This is an autonomous ODE. So, the change of function is$\frac{dx}{dt}=f(x)$and$\frac{d^2x}{dt^2}=\frac{df}{dx}\frac{dx}{dt}=f'f$$$f'f+f^2=2x$$ With$F(x)=f(x)^2$: $$F'+2F=4x$$ ... 1 A sufficient condition to have finite local extrema is to ask that each local extreme point is a strict one (since they are in a compact set [a,b] then they are finite) 0 I guess that what you mean is that every non-increasing sequence of intervals for which the length converges to zero defines a unique real number$s$. This is a kind of hybrid of the Dedekind and Cantor (Cauchy sequence) constructions for the real numbers in terms of the rational numbers. The way you have to think about this is that real numbers are not yet ... 2 Here are some references that might help: Wikipedia page: https://en.wikipedia.org/wiki/Nash%E2%80%93Moser_theorem "An Inverse Function Theorem in Frechet Spaces" by Ivar Ekeland: https://www.ceremade.dauphine.fr/~ekeland/Articles/InverseFunctionTheorem.pdf "On the Nash-Moser Implicit Function Theorem" by Lars Hormander: ... 1 What about the set of points where$f$has local extrema is the finite sum of points and intervals? 6 Every infinite-dimensional normed space has a non-closed subspace. Let$X$be an infinite-dimensional normed space, let$a$be a nonzero vector. Assume by induction that we have found vectors$x_1, x_2, \dots, x_{n-1}$for which$|x_i - a| < 1/i$and$a \not\in V_{n-1} = \Sigma_{i=1}^{n-1} \mathbf{R}x_i$. We will extend this sequence by finding an$n$th ... 3 Just take$(x,0,0,\ldots,0)$where$x\in\Bbb R\setminus\Bbb Q$. Each point is a ball of radius$0$, clearly disjoint from every other ball, and the complement is certainly path connected. 1 The smoothness and rank of the differential is a red herring.$B$is compact, so a minimum is attained somewhere on$B$.$\lVert f(0)\rVert < \,?\lVert x\rVert \geqslant 1 \implies \lVert f(x)\rVert > \,?$3 Ooh, a tauberian theorem with a simple elementary proof, cool. Assume$x$is always so large below that$xf(x)$is increasing. Fix$\delta>0$. It follows that $$\int_x^{(1+\delta)x}f(y)\,dy\sim A((1+\delta)^\alpha-1)x^\alpha.$$In particular, if$x$is large enough we have $$\int_x^{(1+\delta)x}f(y)\,dy\le(1+\delta)A((1+\delta)^\alpha-1)x^\alpha.$$But ... 2 Well, the integral need not even exist for every continuous$f$. If, say, we restrict to continuous$f$with compact support, then yes we can find such$\xi^n$. For example, say$\phi\ge0$is smooth, has compact support, and$\int\phi=1$. Define$\phi_n(t)=n\phi(nt)$, and now define$\xi^n\$ as the convolution $$\xi^n=\xi*\phi_n.$$

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Note: I really like the other answer. This is just a side note to add a little intuition as to why convolution has a useful function in Mathematics. To see why it gets so much attention, consider the product of two power series \begin{align} \sum_{n=0}^{\infty}f(n)z^{n}\sum_{n=0}^{\infty}g(n)z^{n} & = ...

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What are the accumulation points of: $$\{.11,.101,.1001,.10001,\dots,.1\}$$ What about those of: $$\{.011,.0101,.01001,.010001,\dots,.01\}$$ Or those of: $$\{.0011,.00101,.001001,.0010001,\dots,.001\}$$ Finally, let's union these all together! We end up with (and you should verify that this actually is the union): $$\{x\in(0,1):x\text{ has at most two ... 2 Citing from jeff560.tripod.com: CONVOLUTION. Expressions that would now be described as “convolutions” appear in Laplace’s earliest work on sums of independent random variables, “Mémoire sur l’inclinaison moyenne des orbites des comètes, sur la figure de la terre, et sur les functions,” Mém. Acad. R. Sci. Paris (Savants Étrangers), 7, (1773), ... 3 It means (ii), which is clearly equivalent to (iii). It's also clear that (ii) implies (i): Given \delta>0, if P has mesh less than \delta and P' is a refinement of P then P' has mesh less than \delta. But in this generality (i) does not imply (ii). Silly example: Say [a,b]=[-1,1]. Define F(P) by saying that F(P)=1 if 0\in P, ... 0 Suppose not i.e., F is differentiable at some rational point x_0 \in [0,1] and therefore f is continuous at this x_0. Let x_0 \in [0,x] with 0\le x\le 1. Since 0\le x_0 \le x then x-x_0\ge 0 so by Archimedean property there exists a positive integer n \ge 1 such that n(x-x_0)>1 i.e., x-x_0 > \frac{1}{n}. Also, since f is ... 1 On the very next page, Proposition 5.11(e) shows that it is always a unit vector field. 0 Yes it is. Assume you have to go from point A to point B\neq A. If you just "go straight", you may intersect a disk D_n, but the intersection between your path P and D_n is homotopy equivalent to a small path on \partial D_n (just consider the central projection with respect to the centre of D_k as homeomorphism). Since the disks D_m are ... 2 Hint: What are the accumulation points of the countable set \{\,a+\frac1n\mid n\in\mathbb N\,\}? 3 Let x\in\Bbb R and \delta>0. For every \epsilon\in(0,\delta) there is some integer n such that x-\delta<n\epsilon<x+\delta. Proof: Using the Archimedean property, we see that there exists some u\in\Bbb Z such that u\epsilon\ge x+\delta and some v_0\in\Bbb Z such that v\epsilon\le x-\delta for any integer v\le v_0. Then the ... 0 The elements of your set are of the form 0.a_1a_2\dots a_m where a_1a_2\dots a_m is the decimal representation of a square number, let's call this set S. We see that a_1\neq 0 so S\subseteq [\frac{1}{10},1] so S can't be dense (0,1). It is however dense in (\frac{1}{10},1). To see this take some element a\in (\frac{1}{10},1) and let ... 2 The answer is a partial YES and a partial NO. If you split the Riemann improper integral into two pieces:$$\int_0^\infty \frac{f(x)}{x} dx \stackrel{def}{=} \lim_{\Lambda \to \infty, \lambda \to 0} \int_\lambda^\Lambda \frac{f(x)}{x} dx = \lim_{\Lambda\to\infty} \int_1^\Lambda \frac{f(x)}{x} dx + \lim_{\lambda\to 0} \int_\lambda^1 \frac{f(x)}{x}dx $$The ... 1 Observe that :  10^{m-1} < S_{m,n} < 10^m \Rightarrow 1/10 < S_{m,n}/10^m < 1 which means that it cannot be dense in the interval (0,1/10). So it cannot be dense in (0,1) or any (\epsilon, 1) for \epsilon < 1/10. 2 Other example. Take f(x)=e^{-x}. 2 No. Consider f=1_{[0,1]}, that is, f(x)=1 when x\in[0,1], and 0 otherwise. 6 No, for example, a_n = n^{-3/4}. 0 After a while I figured it out. Consider the function$$\begin{align}\psi: [0,1] &\to \mathbb R\\ t &\mapsto \langle f(a + tv), v \rangle\end{align}$$defined at [0,1], differentiable with$$\begin{align}\psi' (t) \cdot v &= \langle f'(a + tv)\cdot v ,v\rangle + \langle f(a+tv),0 \rangle \\ &= \langle f'(a + tv)\cdot v \rangle > ...

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