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0

Notice simply that $$\left\|e^{At}x_0 + \int\limits_0^t e^{A(t-s)}b(s)ds\right\|\le \|e^{At}\|\left(\|x_0\| + \int\limits_0^\infty \|e^{-As}\|\cdot\|b(s)\|ds\right).$$ Since $\|e^{At}\|$ tends to zero exponentially when $t\to\infty$, the expression inside parenthesis is bounded (because $b$ is bounded) and so the right-hand side tends to zero exponentially ...

1

The idea: if $$f(x)\approx f(\hat x) + f'(\hat x)(x-\hat x)\qquad\text{and}\qquad g(x)\approx g(\hat x) + g'(\hat x)(x-\hat x)$$ then $$f(x)g(x)\approx f(\hat x)g(\hat x) + g(\hat x)f'(\hat x)(x-\hat x) + f(\hat x)g'(\hat x)(x-\hat x) + f'(\hat x)(x-\hat x)g'(\hat x)(x-\hat x)$$ with the last term much smaller than the other terms.

0

Looking at each $x_i$ individually, we know that the derivative of $x\mapsto |x_i-x|$ is $x \mapsto \left\{ \begin{array}{ll} -1 & (x< x_i) \\ 1 & (x> x_i) \\ \end{array} \right.$ Therefore, we know that the solution isn't at $]-\infty,x_i[$ or at $]x_i,+\infty[$, as $|x_i-x|$ will grow indefinitely in those regions. Let's ...

0

Your weak formulation is $\int\limits_{\Omega}{\nabla u\cdot\nabla vdx}+\int\limits_{\Omega}{Vuvdx}=\int\limits_{\Omega}{fvdx},\,\forall v\in H_0^1(\Omega)$. So if you denote $$a(u,v)=\int\limits_{\Omega}{\nabla u(x)\cdot\nabla v(x)dx}+\int\limits_{\Omega}{V(x)u(x)v(x)dx}$$ you have $$a(u,u)\ge C\|u\|_{H_0^1(\Omega)}+\int\limits_{\Omega}{Vu^2dx}\ge ... 1 Squeeze theorem (like the OP wrote) we know that that lim_{y\to 0} {siny\over y} = 1 Using the epsilon delta definition (am going to use it in for lim_{y\to 0} {siny\over y} = l, l \in \mathbb R) gives the following: Given a \epsilon > 0 \exists \delta such that |{siny\over y} - l| < \epsilon for y \in [y_0-\delta, y_0+\delta], l \in ... 1 Let u = x \ln x. We have$$\lim_{x \to 0^+} u = 0,$$and$$\lim_{u \to 0}\frac{\sin u}{u} = 1.$$You can prove both limits with L'hôpital's rule. For the first one, apply the rule to \ln x/(1/x). Therefore$$\lim_{x \to 0^+}\frac{\sin(x \ln x)}{x \ln x} = 1.$$0 Expanding on André Nicolas's comment, \begin{array}\\ (x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2 &=((x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2) \dfrac{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{(x_nn^4 + y_nn^2)- x_nn^4}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{ y_nn^2}{(x_nn^4 + y_nn^2)^{1/2} + ... 0 For 2.: Around each point t\in[0,1] such that f(t)\in V:=\mathbb{R}^2\setminus(F_1\cup F_2), you can take a maximal open interval (c_1,c_2) such that f((c_1,_2))\subseteq V (since V is open). By maximality, we have that f(c_1),f(c_2)\in F_1\cup F_2. Now take a to be the \sup of all the endpoints c such that f(c)\in F_1. This must be less ... 2 Since \cos(1/\|x\|) approaches all values of [-1,1] when x\to0, it is necessary and sufficient for differentiability at 0 that (\sin\|x\|)^p/\|x\|\to0 when x\to0. But you know that$$\frac{(\sin\|x\|)^p}{\|x\|^p}\to1$$when x\to0. Since$$ \frac{(\sin\|x\|)^p}{\|x\|}=\frac{(\sin\|x\|)^p}{\|x\|^p}\|x\|^{p-1}, $$you find readily that it ... 0 It is not possible to have all the minimums from the second part. As you asked I am just giving hint. To have all the minimums from the second part the following inequalities hold$$ t-\mid x-y\mid\leq \mid x-y\mid\\ t-\mid x-z\mid \leq \mid x-z\mid \\ t-\mid z-y\mid\leq \mid z-y\mid $$Now these three imply all differences and x,y,z\in[0,t) would give ... 1 Note that if you replace x^2 by x' and y^3 by y', then the expression reduces to \frac{x'y'}{x'^2 + y'^2}. If you now take x' = y', then you get 1/2. This shows that the limit does in fact not exist, as you found correctly that some approaches lead to 0. If you do not like the replacement: when you approach via (t^3, t^2) you get ... 2 There is no limit. Along the semicubical parabola x=t^3, y=t^2 the limit is equal to 1/2. S.G. 0 Hint: Let \mathcal P _{\mathbb Q} be the set of polynomials with rational coefficients. Let f \in C^1. Then as you probably know, we can approximate f' in C^0 by polynomials in \mathcal P _{\mathbb Q}. We should then be able to approximate f in C^1 polynomials of the form$$r+ \int_0^x p(t)\,dt,\,\,r\in \mathbb Q, \, p \in \mathcal P _{\mathbb ...

0

For ones who read German, I strongly recommend Harro Heuser's 'Lehrbuch der Analysis Teil I'. There is also 'Teil II'. I tried couple of other German text books, but gave up continuing due to many errors or lack of completeness, etc. Then a person recommended me this book. This book is self-contained and proofs are quite error-free as well as well-written ...

1

Starting like A. Jiménez, $$D_uf(0,0) = \frac{u^2_1u_2}{u_1^2+u_2^2}.$$ But can be easily proved that if $f$ is differentiable in $(0,0)$: $$D_uf(0,0) = Df(0,0)u,$$ and this is impossible because in this case $D_uf(0,0)$ is a nonlinear funcion of $u$.

1

If it were Frechet, then there would be a (bounded) linear operator $A$ so that $$\|v\|=A(v)+o(\|v\|).$$ Now insert $-v$ and compare.

5

To check whether a function $f: \Bbb{R^2}\to \Bbb{R}$ is differentiable at $(0,0)$, we need to find a linear map $Df(x_0)$, called differential. Now, let's take $u=(u_1,u_2)\in \Bbb{R^2-\{(0,0)\}}$. The directional derivatives along $u$ are: $$D_uf(0,0)=\lim_{t\to 0}\frac{f(tu_1,tu_2)-f(0,0)}{t}=\lim_{t\to 0}\frac{f(tu_1,tu_2)}{t}=\lim_{t\to ... 1 Write functions in C^1 as$$ f(x)=f(0)+\int_0^xf'(t)dt $$with f'\in C^0 to find that this provides a parametrization of C^1 by \Bbb R×C^0. Show that this is an homeomorphism. Only then can you use the separability of C^0. 0 The absolute value of the expression is$$\left |\frac{\arctan (xy)}{\sqrt {x^2+y^2}}\right| = \left |\frac{\arctan (xy)}{xy}\right|\cdot |y|\cdot \left |\frac{x}{\sqrt {x^2+y^2}}\right |\le \left |\frac{\arctan (xy)}{xy}\right|\cdot |y|\cdot 1.$$Because \lim_{u\to 0}(\arctan u)/u =1 and |y|\to 0, the desired limit is 0. 2 From where you left off: Assume$$n \geq m \Rightarrow |x_n-x_m| \leq M\left(|q|^{m+1}+|q|^{m+2} + ...+ |q|^n\right)= M|q|^{m+1}\left(1+ |q|+ |q|^2+...+|q|^{n-m-1}\right)= \dfrac{M|q|^{m+1}\left(1-|q|^{n-m}\right)}{1-|q|} < \dfrac{M|q|^{m+1}}{1-|q|}$$. Thus let \epsilon > 0 be given, and choose N \in \mathbb{N} such that if m > N we have: ... 2 Use polar coordinates and L'Hospital's rule:$$\lim\limits_{(x,y) \to (0,0)} \frac{\arctan(xy)}{\sqrt{x^2+y^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{\sqrt{(r \cos(\theta))^2+(r \sin(\theta))^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{r} = \lim\limits_{r \to 0} \frac{2r \cos(\theta) ...

3

$-\frac{x^2+y^2}{2}\leq xy\leq \frac{x^2+y^2}{2}$ and $\arctan$ is a monotonic increasing function, so $$\frac{\arctan\frac{-r^2}{2}}{r}\leq \frac{\arctan{(xy)}}{\sqrt{x^2+y^2}}\leq \frac{\arctan\frac{r^2}{2}}{r}$$ where $r=\sqrt{x^2+y^2}$. Now, using L'Hospital's rule you get that both limits in the sandwich are $0$. Or, you can use the inequality Oliver ...

7

Observe that you have $$\left|\arctan u\right|\leq|u|, \quad |u|\leq1,$$ then, switching to polar coordinates with $r=\sqrt{x^2+y^2}$, as $r\to 0$, you get $$\left|\frac{\arctan(xy)}{\sqrt{x^2+y^2}}\right|=\frac{\left|\arctan(r^2 \sin \theta \cos \theta)\right|}{r}\leq \frac{\left|r^2 \sin \theta \cos \theta\right|}{r}\leq r.$$ The sought limit ...

6

A set $U\subset \mathbb R$ is open if and only if for every $x\in U$, there exists some $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon)$ is a subset of $U$. For $U=\mathbb Z$, this is clearly not the case: Take $x=0$ Take any $\epsilon > 0$. Then, $\min\{x+\frac\epsilon2, x+\frac12\}$ is an element of $(x-\epsilon, x+\epsilon)$, but it is not an ...

2

$\mathbb{Z}$ is not open in $\mathbb{R}$. One way to see this is that given any $n\in \mathbb{Z}$ we have for every $\epsilon>0$ that $(n-\epsilon, n+\epsilon)$ is not contained in $\mathbb{Z}$.

1

We have that If a sequence of functions $(f_n)_{n\geq 1}$ converges pointwise to a function $f$ and uniformly to a function $g$, then $f=g$. and If a sequence of continuous functions $(f_n)_{n\geq 1}$ converges uniformly to a function $f$, then $f$ is continuous. Prove convergence pointwise to the function $f\colon[-1,1]\mapsto ... 1 The two are unrelated mathematical structures. However, every topological space can be given a particular useful$\sigma$-algebra, called the Borel$\sigma$-algebra. This is the smallest$\sigma$-algebra containing all the open sets of the space. One useful property is that continuous real-valued functions are measurable with respect to this ... 1 Notice that the series is defined as $$\sum_{n=0}^{\infty}{a_n} = \lim_{n \to \infty} \sum_{k=0}^{n}{a_k}.$$ Your example does not fit into the definition, because you would have $$a_k = \binom{n}{k}x^k$$ and your$a_k$would depend on$n$, but$n$is different for different partial sums. So this is not description of infinite series and you cannot use ... 0 I am proving this result for positive rationals; proof for negative rationals will be similar. For convenience, I am using$(a)$to denote the fractional part of$a$. Let$\frac{p}{q}$be a rational. Let$r=\frac{p}{q}$. Now we write the function as $$f(r)=\sum_{n=1}^{\infty}\frac{(nr)}{n^2}=\sum_{n:\ q\mid n}\frac{(nr)}{n^2}+\sum_{n:\ q\nmid ... 0 (I don't think you have to use Weierstrass M-test, since one cannot conclude that a series is not uniformly convergent from that test) Hint: Calculate the series, both for x\neq 0 and x=0. Do you have any knowledge of what kind of function a series of continuous functions that converge uniformly must converge to? Update If you have a series of ... 0 Yet another formulation for topological spaces: If f:X \to Y continuous and f(x_\iota) is a net in f(X), then x_\iota has a converging subnet, say x_\tau \to x. Then f(x_\tau) \to f(x), hence f(x_\iota) has a converging subnet, so f(X) is compact. 0 I was recently in your situation and encountered this book http://www.amazon.com/Calculus-Algebra-Differential-Unified-Approach/dp/0971576653/ref=sr_1_1?ie=UTF8&qid=1455361217&sr=8-1&keywords=vector+calculus+linear+algebra+and+differential+forms+a+unified+approach It is written at a beginner level and is very easy to read for some with your ... 0 Starting from e^x\ge 1+x and thus also e^{-x}\ge 1-x one gets$$ -\ln(1-x)\ge x \ge \ln(1+x) $$Now set x=\frac1n to get$$ \ln(n)-\ln(n-1)\ge\frac1n\ge\ln(n+1)-\ln(n) $$The left and right terms are telescoping in sums, so$$ \ln(n)-\ln(m)\ge\sum_{k=m+1}^n\frac1k\ge\ln(n+1)-\ln(m+1) $$Since \ln(n)\to\infty, this shows that the sequence of partial ... 0 For your first question, C_c(X) is a subspace of L^1(\mu), and \phi is dominated by the L^1(\mu) norm on C_c(X). By the Hahn-Banach theorems, \phi extends to some continuous linear functional \Lambda on L^1(\mu). For any f\in L^1(\mu), we can find a sequence \{f_n\} in C_c(X) such that \|f_n-f\|_1 \rightarrow 0 as n\rightarrow ... 1 However, since there exists an element cx \in \{xn\}, and cx>c, it is a contradiction. This is false, and I cannot figure out how you contrived this claim. \def\nn{\mathbb{N}} Instead, since c = \sup(\{x_n:n\in\nn\}) \ge x > 1, let y \in \{x_n:n\in\nn\} such that y>\frac{c}{x} since \frac{c}{x} < c. Then xy > c, which is ... 2 Note first that x>0. Moreover p\neq 0 because this is only possible when x=1, which would not comply with the 3 roots assumption. Next, we have$$ p = \frac{|\log (x)|}{x} $$When 0<x<1, the right hand side derivative w.r.t x is$$ \frac{\log (x)-1}{x^2} $$Now, \log(x) is negative, thus this function is increasing to the left of ... 0 According to the hint in "Problems in Mathematical Analysis", Biler and Witkowski, problem 4.112. Define g_n as the function which is 0 on the intervals (-\infty, 0] and [2/n,\infty), 1 at 1/n, and linear otherwise. Then g_n converges pointwise to 0 but is not uniform in any interval containing 0. Now define$$f_n(x) = ... 1 It helps to prove that a function is continuous in that point iff the oscilation is zero. Also you can aproach as much as you want to the discontinuity, and measure the skip. 1 Here is another approach. Since$x >1$, we have$x=1+t$for some$t>0$. Then$x^n = (1+t)^n = \sum_k \binom{n}{k} t^k \ge 1 + n t$. Since$1+nt \to \infty$, we have the desired result. 5 It's the closure of$C^\infty_c(\Omega)$with respect to$ ||.||_{1,2}$and is thought of as the space of Sobolev functions with zero boundary value. As closure of a linear subspace in a Hilbert space it is a Hilbert space, too. See, e.g., this Wikipedia entry 0 YES. An infinite sum of irrational numbers can be rational. PROOF: Let the set A be all the positive irrational numbers and the set B be the negative irrational numbers. Take each positive irrational number and add it to the matching negative irrational number to get 0. The sum of all these 0 numbers is 0 which is a rational number. 2 They are equivalent. Suppose your first statement holds, then$\forall y_1,y_2 \in S$, $$d(y_1,y_2) \leq d(x,y_1)+d(x,y_2) < 2r.$$ Conversely, suppose your second statement holds, then$\forall x \in S$,$S\subset B_x(2r)$. Your second definition is more commonly seen in literature. There is, however, a different definition of boundedness in general ... 6 We use the definition of Cauchy sequence to show the sequence$(a_n)$is not Cauchy. Let$\epsilon=1/2$. We will show that there does not exist an$m$such that for any$n\gt m$, we have$|a_n-a_m|\lt \epsilon$. For let$m$be given, and let$2^k$be the smallest power of$2$that is$\gt m$. Let$n=2^{k+1}-1$. Then $$a_n-a_m\ge ... 0 Hint: Showing it doesn't converge (specifically that it goes to infinity) would help. 2 The problem is, the proof is proceeding by contradiction. That is why it says "Then {m_1+1\over N}>b" -- because it's assumed there's no number of the form m/N between a and b. It then proves that this leads to a contradiction, just as you proved that in fact (m_1+1)/N in your situation was actually <b and not >b as assumed. The ... 1 I am assuming you want V to actually be the image of U. In this case, there is no such map satisfying your second condition. If m = n, this follows from invariance of domain, since the image of U will necessarily be open. If m < n, there is no continuous injective map from \mathbb{R}^n to \mathbb{R}^m (let alone to a closed subset). You ... 0 If possible , let, f is not uniformly continuous.Then there exists a Cauchy sequence {x_n} in \mathbb R such that {f(x_n)} is not a Cauchy sequence. (If not, i.e for every Cauchy sequence {x_n}, {f(x_n)} is a cauchy sequence, then f is uniformly continuous, which contradicts our hypothesis) So there exists \epsilon > 0 corresponding to a ... 1 If you want to analyze differentiability away from (0,0), you need not worry about the f(0,0) part, just about the formula. In general, writing x=(x_1,x_2) (I'm using notation different from your (x_1,y_1) to make indexing easier, so I can compute all the partial derivatives at once), we have that:$$\frac{\partial}{\partial x_i}(\|x\|) = ... 1 You shouldn't start with what you're trying to prove. Note that the result you're trying to prove is intuitively obvious if you divide both sides by$(z-y)(z-x)$: The average value over$[y,z]$is larger than the average value over$[x,z]$because the function is bigger over$[x,y]$than$[y,z]$. Basically this is like saying the average of 2, 4, and 5 is ... 1 Here is why there always exists a sequence in$K$converging to$\sup K$, provided its existence (which is guaranteed by the fact that bounded sets always have a least upper bound in$\Bbb R$). Let$A\subseteq \Bbb R$and let$s=\sup A<+\infty$. If there were a$n>0$such that$\left(s-\frac1n,+\infty\right)\cap A=\emptyset$, then$s-\frac1{2n}\$ ...

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