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As you correctly pointed out, every automorphism of $\mathbb C$ is of the form $az+b$. Then, let $f$ be a non-constant automorphism: $|f(z)-f(w)|=|az+b-(aw+b)|=|a||z-w|$. Now, let $|z-w| < \frac {\epsilon}{2|a||z-w|}$. Can we use a single value of $2|a||z-w|$ for all $z,w \in \mathbb C$ ? Hint: what happens with the value of |z-w| if we consider $2z,2w ... 0 Hint: is$z \mapsto 2z$uniformly continuous? 1 What you are trying to show is false. The first problem is how to interpret convexity of a function defined on the natural numbers. Normally you want the domain of a convex function to be convex. Here, I will assume that you mean that$V$can be extended to a convex, non-decreasing function on all of$[0,\infty)$. Any other interpretation I can think of ... 12 Note that since$a_n\geq0$, you easily have $$\frac{n^2a_n+1}{n^2+a_n^2}= \frac{n^2a_n+1}{n^2(1+a_n^2/n^2)}\leq\frac{n^2a_n+1}{n^2}=a_n+\frac{1}{n^2}.$$ Now just use a comparison test. 9 The function$x\mapsto f(\sqrt{|x^2-a|})-f(x)$is continuous, maps$0\mapsto f(\sqrt a)-f(0)$and$\sqrt a\mapsto f(0)-f(\sqrt a)$. Now use the IVT. 0 it can be factorized into $$(z+2) \left(z^2-4 z+13\right) \left(z^2-2 z+2\right)$$ 2 Tip:$(z-1-i)(z-1+i)=z^2-2z+2$Then, long division gives:$$\frac{z^5−4z^4+11z^3+12z^2−42z+52}{z^2-2z+2} = z^3-2 z^2+5 z+26$$ So, now, can you find a factor$(z+a)$that divides that to produce a quadratic? vis: find$a,b,c$so$(z+a)(z^2+bz+c) \\ = z^3 +(a+b)z^2+ (ab+c) z+ ac \\ = z^3-2 z^2+5 z+26$1 Counterexample: Let$a=0$,$f(x)=x$,$g(x)=x^2$and for$x\ne 0$let$h_1(x)=\frac1{x^2}$,$h_2(x)=\frac1x$. 2 Suppose that$V$is a vector space over$\mathbb R$,$\|\cdot\|$is a norm on it, and $$U\equiv\{v\in V\,|\,\|v\|<1\}$$ is the open unit ball with respect to the norm. Fix$x\in V$and$x\neq 0$. Claim 1: There exists some$t>0$such that$tx\in U$. Proof: Since$x\neq 0$, one has$\|x\|>0$. Put$t\equiv 1/(2\|x\|)>0$. Then, ... 0 Norms are inspired from the Euclidean distance function and refer to a generalized class of metrics$d$which for a normed linear space$V$, satisfy the properties:$d(a,b) = d(a-b,0) = d(0,b-a)~\forall~a,b \in Vd(\lambda u,0) = |\lambda| d(u,0)~\forall u \in Vd(a,b) \le d(a,c)+d(c,b)d(a,b) \ge 0$with equality$\iff a=b.$1 First consider$[-r,r]$for large$r$. Note$f$is uniform continuous on$[-r,r]$. So for any$\epsilon>0$, there is$\delta$, for any$|x-y|<\delta$, there is$|f(x)-f(y)|<\epsilon$. Since$u_n\to1$, there is$N$, such that for all$n>N$,$|u_n-1|<\delta$. And for all$k$,$|u_n\dfrac{k}{\sqrt{n}}-\dfrac{k}{\sqrt{n}}|<2r\delta$, and $$... 0 PROBLEM Answer Explanation Demonstration Given the real line \mathbb{R}. Consider the metrics:$$d(x,y):=|y-x|d'(x,y):=\arctan|y-x|$$Then one obtains:$$\mathcal{N}=\mathcal{N}'\quad\mathcal{U}\neq\mathcal{U}$$Concluding problem. PROOF Identification Given normed spaces \Omega and \Omega'. Regard the category: ... 0 You can show that$$\int_{C_{r_2}(z_0)} \frac{dw}{w-z} = 2\pi i \quad \text{and} \quad \int_{C_{r_1}(z_0)} \frac{dw}{w-z} = 0$$using the residue theorem. Note that f(w) = 1/(w-z) has a single pole of order one at the point z inside the annulus R_1 < |z - z_0| < R_2. By our hypothesis, because r_1 < |z - z_0| we know that the contour ... 5 I don't really see what surjectivity has to do with this: there is no map f from \mathbb{Q} to \mathbb{N} satisfying m<n\implies f(m)<f(n), at all. Suppose there were. Then f(0)> f(-1)> f(-2)>. . .. But this yields an infinite descending chain of natural numbers, which can't happen. Note that restricting to positive rationals ... 0 The proof has to proceed by induction on n because then you are always working with neighborhoods of P0. F1(x1,x2,y1,y2)=0 F2(x1,x2,y1,y2)=0 Have a solution in a neighborhood of P0. (F1->F2) F1(x1,x2,y1,y2)=0 F2(x1,x2,y1,y2,y3)=0 P01 is entirely different than P02. (F1->’F2) See appendix of “Rutherford Aris; Vectors, Tensors, and the Basic Equations of ... 1 Suppose f(0)=m and f(1)=n, so k=n-m+1 \ge 2 is an integer. What values could you assign for the following?$$f\bigl(\frac{1}{k}\bigr) \quad, \quad f\bigl(\frac{2}{k}\bigr) \quad, \quad \ldots\quad , \quad f(\frac{k-1}{k})=f(\frac{n-m}{k}) $$Those are supposed to be n-m different integers between m and n. But the number of integers between m ... 1 The answer to the first question is no. Say K is compact, has empty interior, no isolated points, and positive measure (for example a fat Cantor set). Let f=\chi_K. Then f is in Baire class 1. But f is discontinuous at every point of K. So you can't include all the discontinuities of f in B if \epsilon<m(K). The second question doesn't ... 1 Ok, I think I found an example of this. Take a solution of$$(1) \qquad \partial_t u + L u=0$$Assuming we can justify the following steps, we integrate in t to get$$\int_0^\infty (\partial_t u + Lu) d\tau = 0\lim_{t\rightarrow \infty} u(x,t) - u(x,0) + L \int_0^\infty u d \tau = 0-u(x,0)+L\int_0^\infty u d \tau = 0,$$where in the last step I ... 1 The current title makes no sense. The previous title was much better. What you really meant was this: "If f:\Bbb Q\to\Bbb N is such that x<y implies f(x)<f(y) show f is not surjective." Also note that the question should be in the body of the post, not just the title. Anyway, there is no such function to begin with, surjective or not. Suppose ... 8 Assume the your function f would be surjective, so we hit the whole \mathbb{N}, that means especially for some x,y\in\mathbb{Q} with x<y we have$$ f(x)=0 \text{ and } f(y)=1 $$now we choose z:=\frac{x+y}2 for which holds x<z<y,f(z)\in\mathbb{N} , so for this it must hold since we have an increasing function f$$ ... 1 You can start with "the norms induce the same topology". Then use the fact that a linear transformation is continuous if and only if it is bounded. And this is one of your inequalities. For the other direction, use the inverse of that linear transformation. 1 Maybe it will be useful to consider an example of two norms$F$and$G$of a vector space$X$not being equivalent to each other. What it means is that at least one of the quantities$\sup\limits_{x \in X}\frac{F(x)}{G(x)}$or$\sup\limits_{x \in X}\frac{G(x)}{F(x)}$is unbounded, i.e. there is a sequence$(x_n)_{n \geq 0}$of vectors in the space such that ... 4$f(x)=(x^2-1)^n$is an even function with two zeroes of multiplicity$n$at$x=\pm 1$. It follows that$f'(x)$has two zeroes of multiplicity$n-1$at$x=\pm 1$and a zero at$x=0$. By Rolle's theorem,$f''(x)$has a zero in$(-1,0)$and a zero in$(0,1)$. Moreover, there are two zeroes of multiplicity$n-2$at$x=\pm 1$. Continuing that way, we may see that ... 0 Hint: You can use induction to show that the$k$-th derivative has$k$real roots in$(-1,1)$. 12 Note that $$\lim_{y\to 0}\frac{y}{\cos\left(\frac{\pi}{2}(1+y)\right)} = \lim_{y\to 0}\frac{y}{\cos\left(\frac{\pi}{2}+\frac{\pi y}{2}\right)} = \lim_{y\to 0}\frac{y}{-\sin\left(\frac{\pi y}{2}\right)} \\ = -\frac{2}{\pi}\lim_{y\to 0}\frac{\left(\frac{\pi y}{2}\right)}{\sin\left(\frac{\pi y}{2}\right)} = -\frac{2}{\pi}(1) = -\frac{2}{\pi}$$ 15 HINT: $$\cos\left(\dfrac\pi2+A\right)=-\sin A$$ and $$\lim_{h\to0}\dfrac{\sin h}h=\text{ ?}$$ 1 Suppose you have a sequence which converges in$G $. The lower bound implies it converges in$F $to the same limit. Suppose you have a sequence which ddoes not converge in$G $. The upper bound implies it does not converge in$F $. That's all you need, since metric spaces are sequential spaces. 2 If$t_n \to l\in \mathbb{R}^+$and$x_n \to x\in C$then$t_nx_n\to lx$. Since $$\left\|t_nx_n - lx \right\|= \left\|t_nx_n - lx_n+lx_n-lx \right\|\le \left\|t_nx_n - lx_n\right\|+\left\|lx_n-lx \right\| \le |t_n-l|\left\|x_n\right\|+|l|\left\|x_n-x \right\|.$$ 0 Presumably$I_1$is an interval such that the lower limit is less than some element of$A$and the upper limit is greater than every element of$A$. Now you bisect$I_1$into two intervals with the midpoint$c$. You ask if$c$is an upper bound for$A$. If so,$I_2$is the lower subinterval and$b$will be in it. If not,$I_2$is the upper subinterval ... 0 Let's apply the bisection method. Since$A$is bounded above, there exists an upper bound of$A$, say$b_1$; also, since$A$is non void, choose$a_1=a-1$where$a \in A$. So$a_1$is not an upper bound of$A$. If$m$is the midpoint of$[a_1,b_1]$, let$a_2=a_1$and$b_2=m$if$m$is an upper bound of$A$, let$a_2=m$and$b_2=b_1$otherwise. Continuing ... 0 Let$b_k=3^ka_k, then Cauchy-Schwarz says \begin{align} \sum_{k=1}^na_k2^k &=\sum_{k=1}^nb_k\left(\frac23\right)^k\\ &\le\left(\sum_{k=1}^nb_k^2\right)^{1/2}\left(\sum_{k=1}^n\left(\frac49\right)^k\right)^{1/2}\\ &=\left(\sum_{k=1}^na_k^29^k\right)^{1/2}\frac2{\sqrt5}\sqrt{1-\left(\frac49\right)^n}\\ ... 0 Since you are talking about functions on \mathbb{R} and infinity is not actually an element of the reals, “vanish at infinity” is always only defined in a limit sense. Now you might say this requires the statement to be “continuous functions with compact support include those that vanish at \infty” – however, since all ... 0 Let G\in \Sigma with \mu(G) = \infty. Choose X_i \in \Sigma with \cup_{i=1}^\infty X_i = X and \mu(X_i) < \infty for all i by the definition of \sigma-finite. Without loss of generality assume the E_i are increasing (replace E_i with \cup_{j\leq i} E_i) Then G = \cup_{i=1}^\infty (G \cap X_i) $$so$$ \infty = \mu(G) = \lim_{i} ... 12 Note thatC_c \subset C_0$, but$C_c \neq C_0$. For example,$f(x) = \dfrac{1}{x^2+1}$belongs to$C_0$but not$C_c$. What you seem to be assuming is that$\lim_{|x|\to\infty}f(x) = 0$implies that there is some$N > 0$with$f(x) = 0$for all$|x| > N$. This is not true, as the above example demonstrates. That is, a function can limit to zero at ... 0 The first one contains functions which are nonzero over the whole real line. The second one does not. 1$f(\lambda)=I_2-2au+2a^2u^2+(-(4/3)a^3-(2/3)a)u^3+((4/3)a^2+(2/3)a^4)u^4+(-(2/5)a-(4/3)a^3-(4/15)a^5)u^5+((46/45)a^2+(8/9)a^4+(4/45)a^6)u^6+(-(2/7)a-(56/45)a^3-(4/9)a^5-(8/315)a^7)u^7+((88/105)a^2+(44/45)a^4+(8/45)a^6+(2/315)a^8)u^8+(-(2/9)a-(3272/2835)a^3-(76/135)a^5-(8/135)a^7-(4/2835)a^9)u^9+O(u^{10})$where$u=\dfrac{1}{\lambda}$and ... 0 Here is a solution avoiding the residu theorem or any change of variables over the complex plane. Let$g(\xi)$be the left hand side :$$g(\xi)=\int_{\mathbb R} e^{-\pi x^2}e^{-2i\pi x\xi}dx.$$ One can easily check that$f:\mathbb R\times \mathbb C\rightarrow \mathbb C$given by$f:(x,\xi)\mapsto e^{-\pi x^2}e^{-2i\pi x\xi}$satisfies :$fis measruable, ... 2 By noticing that $$a x^{2} + b x = \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2} - \frac{b^{2}}{4 a}$$ then \begin{align} I &= \int_{-\infty}^{\infty} e^{-a x^{2} - b x} \, dx \\ &= e^{\frac{b^{2}}{4 a}} \, \int_{-\infty}^{\infty} e^{- \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2}} \, dx \end{align} Making the changet = \sqrt{a} x + ...

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Hint: $$-\pi x^2 - 2\pi i x \xi = -\pi(x^2+2ix\xi) = -\pi(x+i\xi)^2+(?)$$ Try to figure out what $(?)$ should be, and you should get something resembling a known integral. You might have to do some substitutions and/or contour deformations.

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It is certainly true, if $T'$ is continuous at $x^*$. Assume $T$ admits a Lipschitz constant $L<1$ w.r.t the norm $\|\cdot\|$ on a neighborhood $U$ of $x^*$. Let $h$ such that $\|h\|=1$ and $\| T'(x^*)h \| = \| T'(x^*) \|$. Then, for $t > 0$ and sufficiently small, by mean value theorem, we have some $\tau\in(0,t)$ such that $$1 > L \ge ... 1 How about trying Cauchy-Schwarz:$$ \sum_{n=1}^\infty\frac{\sqrt{a_n}}n \le\left(\sum_{n=1}^\infty a_n\right)^{1/2} \left(\sum_{n=1}^\infty\frac1{n^2}\right)^{1/2} $$You can use Cauchy-Schwarz to make the estimate in your question.$$ \begin{align} \sum_{k=n}^m\frac{\sqrt{a_k}}k &\le\left(\sum_{k=n}^m a_k\right)^{1/2} ...

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Uniform closure, of course not. Closure in the topology of pointwise convergence yes, as has been pointed out, but this is trivial (immediate from the fact that a polynomial can take any values you want on any finite set) and I don't see how it's good for anything. There's a closure for which the answer is yes, and which also actually allows one to prove ...

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$$f_n(t)=\frac {\lfloor t\rfloor}n.$$ Each $f_n$ is continuous on all intervals $(m,m+1)$ and discontinuous at $m$ for every integer $m$. The sequence converges uniformly to zero.

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Let $f_1$ be the traditional function continuous at every irrational and discontinuous at every rational ($f(x)=1/q$ if $x=p/q$ in lowest terms, $f(x)=0$ if $x$ is irrational). Let $f_n=f_1/n$. Then $f_n\to0$ uniformly.

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All polynomials are dense in borel functions in pointwise convergence topology because polynomials are dense in continous functions in uniform convergence topology, and those are dense in borel in pointwise convergence topology. However if you restrict to polynomials vanishing at zero you will only get borel functions vanishing at zero.

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$P_K$ acting on any element $Ay$ of $K$ yields $A(A^\top A)^{-1}A^\top Ay=Ay$, so $P_K$ has the correct action on $K$. Any element $z$ of the orthogonal complement of $K$ is orthogonal to the columns of $A$, so $A^\top z=0$. Thus $K$ also has the correct action on the orthogonal complement of $K$, and thus by linearity on the entire space.

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I think there is no simple method to minimize over $\mathbb{N}$. This is a very complex topic in math. Using the computation it will be much more simple to find minimum (and sometimes even more faster). Anyway, I am familiar with some problems of this type: For example find minimum of $f(n) = (x-0.25)^2$ we use $f'(n) =0$ and find that on $\mathbb{R}$ ...

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This is too long for a comment and it could be off-topic. There is no doubt that Ron Gordon's solution is the most elegant for this problem. I have been thinking about sine and cosine integral functions (see my early comment) just because $$\frac x{1+x^2}=\frac 12 \Big(\frac 1{x-i}+\frac 1{x+i}\Big)$$ and $$\int \frac{\sin(x)}{x+a}\,dx=\cos (a) ... 3 Consider the contour integral$$ \oint_C dz \frac{z e^{i z}}{1+z^2}$$where C is a semicircle in the upper half plane of radius R. Then the contour integral is equal to$$\int_{-R}^R dx \frac{x e^{i x}}{1+x^2} + i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{R e^{i \theta} \, e^{i R e^{i \theta}}}{1+R^2 e^{i 2 \theta}}$$As R \to \infty, the ... 0 An alternative approach is the following: the function$$ g_n(x) = \frac{2^{2n+1}}{(2n+1)\binom{2n}{n}}(1-x^2)^{n} $$is a polynomial of degree 2n, with integral over [-1,1] equal to one. g_n(x) is concentrated around the origin, so that g_n(x) converges in distribution to a Dirac delta function. If x_0 is a Lebesgue point for f(x), we have:$$ ...

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