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1

For each $f$ $$ \int f(x)\times \frac {\epsilon }{x^2 + \epsilon ^ 2} dx = \int f(\epsilon y) \times \frac { 1}{y^2 + 1} dy \to \int f(0) \times \frac { dy}{y^2 + 1} = \pi f(0) $$ so $$ \frac {\epsilon }{x^2 + \epsilon ^ 2} \to \pi\delta $$


1

You hit the nail on the head there with your thoughts, except that the interpretation should be that the limit does exist, whenever $x\neq 0$, but does not exist, when $x=0$. The limit will depend on which value of $x$ you are given.


1

You're confusing the notion of a limit of functions with that of a numerical limit. As long as you're willing to work with the extended real numbers, the function $$ f(x) = \begin{cases} \infty & \text{if}~x = 0\\ 0 & \text{otherwise} \end{cases} $$ is a perfectly valid limit function; in fact, a pretty well-known one, the Dirac delta. If you don't ...


1

Simply consider the new functions $$\check f(t):=f(-t),\quad \check g_1(t):=g_1(-t),\quad \check g_2(t):=g_2(-t)\qquad(-b\leq t\leq -a)\ .$$ Then $\check g_1\leq\check f\leq\check g_2$, and $\int_{-b}^{-a}\bigl(\check g_2(t)-\check g_1(t)\bigr)\>dt$ is essentially the same sum of rectangles as before.


0

What the MVT tells us is that for some $0\lt c(x)\lt x$, we have $$ \int_0^xe^{t^2}\,\mathrm{d}t=xe^{c(x)^2}\tag{1} $$ Since all we know is that $0\lt c(x)\lt x$, we only know that $\liminf\limits_{x\to\infty}e^{c(x)^2}\ge1$. Therefore, $$ \begin{align} \liminf_{x\to\infty}\int_0^xe^{t^2}\,\mathrm{d}t ...


0

How do you infer $f(x,y) = k + xy$? I suggest: Pick one choice and then see if it satisfies all the properties. If it does, you are done. If it does not, go back to 1. I think?


0

Hint: Write $$ \| f(x_2)-f(x_1)-Df(x_0)(x_2-x_1)\| \leq \| f(x_2)-f(x_1)-Df(x_1)(x_2-x_1)\| +\| (Df(x_1)-Df(x_0))(x_2-x_1)\|. $$ Then both terms can be made small using the continuity of the derivative.


0

Some thoughts that are too long for a comment: If you consider a function continuous on $[a,b]$ we must have some information about what the function does outside of $[a,b]$ as well in order to define the derivative at the endpoints. But, if we have that information and the derivative is not continuous only at $b$, then it must be only jump discontinuous at ...


2

Here is an outline of a proof (which seems correct to me): First show it is locally constant, and then use the fact that $U$ is connected. Choose some point $y\in U$ and take some point $x$ in some ball centered at $y$ which is contained in $U\,.$ Also take some $x^*\in \mathbb{F}^*.$ Then $x^*[f((1-t)x+t(y-x))]$ can be considered as a function from $[0,1]$ ...


0

Let's take a look at a simpler example, $f(x)=I_{x<0}$. Clearly $f$ is not differentiable at $0$ since it isn't even continuous at $0$. Now what about $f(x)=x I_{x<0}$? Now it's continuous at $0$, but the left derivative is $+1$ and the right derivative is $0$, so it's not differentiable. In general we can take $f(x)=g(x)I_{x<0}$ and if $g$ is ...


1

Firstly, just to clarify, by $\mu\leq \nu$ I expect you mean $\mu(A)\leq \nu(A)$ for $A$. We therefore have that $\mu$ is absolutely continuous wrt $\nu$ (written $\mu\ll\nu$, and why I'm asking about the meaning of your inequality) and so there is a function in $L^1(\nu)$, $\frac{\mathrm d\mu}{\mathrm d\nu}$ such that $$\int_E \frac{\mathrm d\mu}{\mathrm ...


0

Your proof is correct. Depending on the rigor required, the following intuition may also pass: The supremum of a set is always either contained in the set or is a limit point of the set. Since a compact set by definition contains all its limit points, it must contain its supremum. Similar for infimum.


0

You are right, a function cannot have extremes, global or local, and be one-to-one. The only case where $\frac {d}{dx}f(x_0)=0$ does not mess up the function's being one-to-one is when $f(x_0)$ is an inflection point, where concave up meets concave down, i.e. the second derivative crosses the x-axis. Your approach of checking with horizontal lines is good in ...


1

As far as continuity: Let $y=mx$. If $\lim_{x\rightarrow 0}g(x,mx)=0$ for all $m$. (Since $x^2$ and $y$ are in additive relationship, independently of the path, if both tend to $0$ the limit is zero.) So the function is continuous at $(0,0)$. I don't understand why you divided by $x$, though... Considering the definition of the directional derivative we ...


0

A Taylor series converges if and only if its remainder term, $R_n(x)$ tends to zero. This is just a consequence from taylors theorem $\mathbf{Theorem:}$ If $f(x)=T_n(x)+R_n(x)$ , where $T_n$ is the nth degree Taylor polynomial of f at a and $\lim_{n \to \infty} R_n(x)=0$ for $|x-a| < R$ , then f is equal to the sum of its taylor series on the interval ...


0

$g(x,0)= 0 \implies g_x(0,0) = 0, g(0,y) = y \implies g_y(0,0) = 1.$ So if $Dg(0,0)$ exists, it must be the linear transformation $(x,y)\to y.$ The question is then: Is it true that $$g(x,y) = g(0,0) + y + o((x^2+y^2)^{1/2})$$ as $(x,y) \to (0,0)$? Yes! But I'll leave that one for you. (It will require some thought.) Given the above, the directional ...


0

Multiplying both sides of the ODE by the integrating factor $r^{n-1}$ and integrating with respect to $r$ yields $b^{n-1} u'(b) - a^{n-1} u'(a) = -\int_a^b r^{n-1} u^p dr$. The boundary conditions imply $0 = \int_0^R r^{n-1} u^p dr$ so $u$ cannot be strictly positive on $(0,R)$. In particular, if $p$ is even then $u=0$ is the only radial solution. This can ...


1

Let's start with a simpler one because the one you give is more complicated than it needs to be to illustrate the idea. So instead let's just look at $f(x) = x \sin{(x)}$. There are many ways to state the condition for a function to have limit as its argument goes to infinity. One way is something like If, for any x, there exists M such that |f(y) - L| ...


2

Even if we do not determine the primitive explicitly, we can show that it cannot have a limit at $\infty$. If the limit existed, then for each $\epsilon>0$ there woiuld exist an $x_0$ such that for all $x,y>x_0$ we'd have $|F(x)-F(y)|<\epsilon$. If we let $x=2n\pi+\frac\pi3$ (with $n$ to be specified in a minute) and $y=x+\frac\pi3$, we note that ...


0

We have to expect that the anwser will depend on $a.$ You have a little problem with your answer when $a=1.$ Treat that case separately (it converges). For $a>0,a\ne 1,$ what you did is fine. You arrive at a one-variable integral where the integrand appears to have two singularities, at $0$ and $1.$ But the singularity at $0$ is removable. At $1,$ the ...


2

You know the expansion for $\exp(t)$, hence the expansion for $\exp(-t^2/2)$. Plug in, integrate, and by happy: $$\exp(t) = \sum_{n \geq 0} \frac{t^n}{n!} \implies \exp\left(-\frac{t^2}{2}\right) = \sum_{n \geq 0}(-1)^n \frac{t^{2n}}{2^nn!},$$ so we get: $$F(x) = \sum_{n \geq 0}(-1)^n\frac{x^{2n+1}}{(2n+1)2^nn!}.$$ Edit: for the edited question with ...


2

\begin{align} F'(x) &= \exp - \frac {x^2}2 = \sum_{n = 0}^\infty \frac{(-x^2/2)^n}{n!} \\ &= \sum_{n = 0}^\infty \frac{(-1)^n}{2^nn!} x^{2n} \\ \implies F(x) &= \sum_{n = 0}^\infty \frac{(-1)^n}{2^nn!} \frac{x^{2n+1}}{2n+1} \end{align}


0

Sketch: Certainly $1.$ holds, because Lebesgue measure is translation invariant. For 2., try to prove it first for compactly supported continuous functions. This follows nicely from uniform continuity. To finish, use the fact that such functions are dense in $L^1.$


1

Define a new metric $d$ on $\Bbb R$ by $d(x,y)=\min\{|x-y|,1\}$; you can easily check that $d$ generates the usual topology on $\Bbb R$. Every subset of $\Bbb R$ is bounded with respect to $d$, so we need only find a subset that is not totally bounded. $\Bbb N$ will do: if $F\subseteq\Bbb N$, then $$\Bbb N\cap\bigcup_{x\in F}B_d\left(x,\frac12\right)=F\;,$$ ...


0

your answer is in bookgeneral topologyby munkres.page 275 ,year 2000.or another answer is:R with discrete meter is bounded and isn't totally bounded


1

For the function to have an extremum at a certain point $x_0$, it is not sufficient that $f'(x_0)=0$. What is sufficient is that $f'(x)$ alternates signs in a neighborhood of $x_0$, that is $f'(x)<0$ when $x_0-\epsilon<x<x_0$ and $f'(x)>0$ when $x_0<x<x_0+\epsilon$ or vice versa depending on the nature of the extremum. As you can see, ...


2

Take $U=\{e_n|\:n\in\mathbb N\}\subset \ell^\infty (\mathbb R)$. It is obviously bounded since $\forall x\in U \:\|x\|=1$, but $\forall x,y\in \ell^\infty$ we have $d(x,y)=1$, so obviously for $\epsilon=1$ there is no finite number of open balls with radius $\epsilon$ that cover $U$ - cause each ball would contain at most one member of $U$.


1

The function $$ f(x) \stackrel{\text{df}}{=} \begin{cases} x^{2} & \text{if $ x \geq 0 $}; \\ - x^{2} & \text{if $ x < 0 $} \end{cases} $$ provides a counterexample with respect to the point $ x = 0 $.


0

Hint: write for $\epsilon = \pm 1$ $$ f(x+\epsilon h) = f(x) + \epsilon h f'(x) + h^2/2 f''(x) + o(h^2) $$


0

The hypothesis is that $f$ is twice differentiable at $x$, so if you're interested in whether the converse is true, the first thing you do should be look at examples where that hypothesis does not hold. Do you know any examples of functions that are continuous but not twice differentiable at a point?


3

Try to prove that $x, f(x), f(f(x)), f(f(f(x))),\ldots$ is a Cauchy sequence. Let $a$ be its limit. Then see if you can show that for every $\varepsilon>0$, $d(a,f(a))<\varepsilon$. A function $f$ for which there exists $K$ between $0$ and $1$ such that for all points $x,y$ one has $d(f(x),f(y))\le Kd(x,y)$ is called a contraction.


1

How about $(x,y,z)=(\cos t,\sin t,-\sin t\cos t)$ Check the derivative is never the zero vector.


0

For a matrix $A\in\Bbb R^{m\times n}$ and $1\leq p_1,p_2\leq \infty$ define $$\|A\|_{p_1,p_2}:=\max_{x_2\neq 0}\frac{\|Ax_2\|_{p_1}}{\|x_2\|_{p_2}}.$$ Partial answer to $i)$: Using a generalization of the Perron-Frobenius theorem you can compute some of these norms. This is all explained in this paper (mainly Theorem 1 and 2). Actually they treat the more ...


0

You can create a set of any Lebesgue measure by taking uncountable unions. For example, suppose that we wanted a set of measure $1$. Consider the following set: $$ \bigcup_{0\leq x \leq 1} \{x\} $$ Each set $\{x\}$ has measure $0$, but the uncountable union of these sets such that $0\leq x\leq 1$ is the closed interval $[0,1]$, which has measure $1$. So ...


1

$$ F(x) = x \int_0^{x} (1+\cos{t}) \, dt = x[ t + \sin{t} ]_0^x = x^2+x\sin{x}. $$ Now differentiate, $$ F'(x) = 2x+x\cos{x}+\sin{x}. $$ You then have to find the zeros of this function. It is odd, so there is a zero at $x=0$. In fact, that's all, because for $x>0$, $x(2+\cos{x})>x$, and $\sin{x}<x$, so they can't cancel to zero. To do it without ...


1

If $f\colon[0,1]\to\mathbb{R}$ is Riemann integrable, then $$ \lim_{n\to\infty}\frac1n\sum_{k=1}^nf\Bigl(\frac kn\Bigr)=\int_0^1f(x)\,dx. $$ Depending on the definition of Riemann integral you are using, this is either an immediate consequence of the definition or a theorem. If $f$ is bounded, say $|f(x)|\le M$, then $$ \Bigl|\frac1n\sum_{k=1}^nf\Bigl(\frac ...


1

Here is a generalization of your approach for the one-dimensional case to all cases, making heavy use of the axiom of choice. For convenience, set $\eta(x):= f(x) - f(a) - D_a f(x-a)$. Define $R \colon X \to L(X, Y)$ by choosing at each $x \in X$ an $R(x) \in L(X, \operatorname{span}\{\eta(x)\}) \subset L(X,Y)$ such that $R(x)(x-a) = \eta(x)$ and ...


0

A quick-and-dirty mnemonic is that "by cancellation of differentials", you "should" have $$\frac{dy}{dx} = \frac{\frac{\partial G}{\partial x}}{\frac{\partial G}{\partial y}}$$ You have this, but with an additional minus sign, and with division replaced by a matrix inverse. This mnemonic was part of how I was taught this rule in thermodynamics, where it is ...


0

I always had trouble remembering the implicit function theorem as well. It helped me to formulate it this way:¹ If for banach spaces $E$, $F$ a function $f \colon E \oplus F → F$ is $C^k$ such that $f(0) = 0$ and the tangential of $f∘i_F \colon F → F $ is invertible at $0$ for the inclusion $i_F \colon F → E \oplus F,~x ↦ 0 \oplus x$, … then there ...


1

If $f$ is increasing then $$a_n=\sum_{k=1}^n\left( \underbrace{f(k)-\int_{k-1}^k f(x)dx}_{\ge0}\right)$$ so $(a_n)$ is increasing and $$f(k)-\int_{k-1}^k f(x)dx=\int_{k-1}^k (f(k)-f(x))dx\le f(k)-f(k-1)$$ so by telescoping we get $$a_n\le f(n)-f(0)$$ so if $f$ has a finite limit at $+\infty$ then $a_n$ is bounded above and then it's convergent.


0

Write $\frac{x}{x+1} = 1 - \frac{1}{x+1}$ and realize the problem reduces to $f=-1/(x+1)$. Now $$0 \leq -\frac{1}{1+k} + \int_{k-1}^k \frac{1}{1+x} \leq \frac{1}{k}-\frac{1}{k+1},$$ which can be shown by using the montonicity of $1/(1+x)$. But the above series is $O(1/k^2)$ and converges, so by the comparison test, the series we are interested in ...


1

$\newcommand{\Reals}{\mathbf{R}}$In case it's a helpful mnemonic, the implicit function theorem (in this setting) is essentially a non-linear version of Gaussian elimination from linear algebra. Specifically, let $m$ and $k$ be positive integers, and put $n = m + k$. Suppose $A$ is an $m \times n$ matrix, and let $F:\Reals^{n} \to \Reals^{m}$ be ...


2

In the première partie, chapitre VI, $\S$1, the 1er Théorème is: Lorsque les differens [sic] termes de la série (1) sont des fonctions d'une même variable $x$, continues par rapport à cette variable dans le voisinage d'une valeur particulière pour laquelle la série est convergente, la somme $s$ de la série est aussi, dans le voisinage de cette valeur ...


1

If you take a look at this question, you could adjust it to yours: you already have that $F\circ F = F + c$, right? So, the answer is: $F(x) = x$ and $f(x) = 1$.


1

I have written an answer to explain what is going on, but a little mistake made me loose the LaTeX original file. Fortunately, I have it on PDF ; I include it as a picture : Take care that your $F$ is transformed in $f$ from the $7$-th line! I add a picture which summarizes what I wrote : Hope it will be helpful!


0

$$\|x-y\|<\delta \implies \sqrt{\sum_{i=1}^{n}(x_i-y_i)^2}<\delta\implies|xi-yi|<\delta\implies|y_i-x_i|<\delta \tag{1}$$ Now, $$(\delta\le1-x_i) \land (\delta\le1+x_i)$$ $$\implies (x_i\le1-\delta) \land (-x_i\le1-\delta)$$ $$\implies |xi|\le\delta \tag{2}$$ Using $(1)$ and $(2)$ and triangle inequality: ...


0

Not really an answer to your question, but a suggestion to follow a different and in my view more elegant route (too much for a comment). Denote $A_i=\{x\in\mathbb R^p\mid x_i\in (-1,1)\}$ for $i=1,\dots,p$. If I understand well then it is your goal to prove that $A=\bigcap_{i=1}^pA_i$ is open in $\mathbb R^p$. For this it is enough to prove that the ...


2

We denote $A_{i}=(-1,1)$ for $1\leq i\leq p$ $$ A=\Pi_{i=1}^{p}A_{i} $$ then if $(x_{i})_{i=1}^{p}\in A$ then $x_{i}\in A_{i}$ but $A_{i}$ is open so there is an open ball $B(x_{i},\epsilon_{i})\subseteq A_{i}$ so by letting $\epsilon:=\min\{\epsilon_{i}\mid1\leq i\leq p\}$ we get $$ B(x_{i},\epsilon)\subseteq B(x_{i},\epsilon_{i})\subseteq A_{i} $$ so ...


0

Let $\epsilon>0$ and $x_o\in I$. For $n>N$ you get $$ \forall x\in I \ \ |f(x) - f_n(x)| < \frac \epsilon 3 $$ Using the continuity of $f_{N+1}$, for $y\in I\cap [x_0-\eta, x_0 + \eta]$ you have $$ |f_{N+1}(y) - f_{N+1}(x_0)| < \frac \epsilon 3 $$ Then: $$ |f(y) - f(x_0)| = |f(y) - f_{N+1}(y) + f_{N+1}(y) - f_{N+1}(x_0) + f_{N+1}(x_0) - ...


1

Assume that $f'$ is bounded by some $M$, you get $$\left|\int _{\frac{k-1}n}^{\frac kn} f(t) dt - \frac 1n f\left( \frac kn \right)\right| =\left| \int _{\frac{k-1}n}^{\frac kn} \left[f(t) - f\left( \frac kn \right)\right]dt\right| \le \int _{\frac{k-1}n}^{\frac kn} \left| f(t) - f\left( \frac kn \right) \right|dt \\ \le M\int _{\frac{k-1}n}^{\frac kn} ...



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