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3

Hint: Choose $m$ such that $n >m$ implies $|1-\frac {\sin (x/n)} {x/n}| <\epsilon$ for all $x \in [0,\pi]$. Then $\frac {\sin x} {(1+\epsilon)x}<f_n(x)<\frac {\sin x} {(1-\epsilon)x}$ for all $x$ if $n >m$. Can you finish? [Check that $|f_n(x)-\frac {\sin x} x| <\frac {\epsilon} {1-\epsilon}$ for $n >m$ using the fact that $0 \leq \frac ... 3 We want to study uniform convergence of$f_k(x)=e^{\frac{x}{k}}$. First we look at th epunctual convergence of the sequence of functions. We initially fix$x_0\in \mathbb R$$$\lim_{k\to\infty}e^{\frac{x_0}{k}}=0:=f(x),\forall x_0\in \mathbb R .$$ For the uniform convergence we have to look at the sup of$|f_k-f|$with$x\in \mathbb R$and$n\in\mathbb N$... 3 For$x \in [-1+\delta,1-\delta]$where$0 < \delta < 1$, we have $$\left|(-1)^k \frac{x^k}{k+1}\right|\leqslant \frac{(1-\delta)^k}{k+1} \leqslant (1-\delta)^k$$ As the geometric series$\sum_{k \geqslant 0} (1-\delta)^k$converges, it follows by the Weierstrass M-test that we have uniform convergence on$[-1+\delta,1+\delta]of the series $$\frac{\... 3 A most straightforward argument for part b) is to notice that$$f_n \left( \frac{1}{n}\right) = \frac{1}{2}$$does not tend to 0, hence (f_n) cannot converge uniformly. 3 Since f is rapidly decaying x^{4} f(x) is bounded. If |f(x)| \leq \frac M {x^{4}} the |f(\sqrt {a^{2}+x^{2}})| is bounded by \frac M {x^{2}} which is integrable in (1,\infty). 2 I'm not a huge fan of the o(1) notation being used here, and this may be hiding where the mistake lies. It is true that for any fixed n you have f_{n}(x + o(1)) = f_{n}(x) + o(1) (using your notation). However, this may not be true for all n sufficiently large. In fact, saying that it is true for all n sufficiently large is virtually the ... 2 Hint: f_ng_n -fg = (f_ng_n -fg_n) + (fg_n -fg). 2 Part 1 If you can show that the algebraic dimension of the space C^\infty(K) is equal to \frak{c}, then since \frak{c}^{\aleph_0}=\frak{c}, this post shows that there does exist a norm that makes it a Banach space: Can every vector space (over \mathbb{R} or \mathbb{C}) can be a Banach space (or Hilbert space)? Part 2 Of course, the above is not ... 2 If (f_n) converges uniformly to f, then you have for every finite-length path \gamma, that$$\left|\int_{\gamma} f_n(z) dz - \int_{\gamma} f(z) dz \right| =\left|\int_{\gamma} f_n(z) -f(z) dz \right| \leq \int_{\gamma} \left| f_n(z) -f(z) \right| dz \leq ||f_n-f||_{\infty} L(\gamma)$$where L(\gamma) denotes the length of the path \gamma. Because |... 2 Since for x\ge 0:$$\sin x \ge x - \frac{x^3}{6}$$we have$$\begin{align} \|f_n(x) - f(x)\| &= \left\|\frac{\sin x}{n\sin(x/n)} - \frac{\sin x}{x}\right\| \\ &= \color{blue}{\left\|\sin x\right\|}\left\| \frac{1}{n\color{red}{\sin(x/n)}} - \frac{1}{x}\right\| \\ &\le \color{blue}1\cdot \left\|\frac{1}{n\color{red}{\left(\frac{x}{n} - \frac{x^3}... 2 Your conclusions and general approach for\sum u_n(x)$are correct but the arguments have a few errors. For$x < 0$: You claim that $$\displaystyle \frac{\exp(-nxt^4)}{n^2}=\frac{1}{e^{nxt^4}n^2}\sim_{n \to \infty}\frac{1}{e^{nxt^4}},$$ but the symbolism$f(n)\sim g(n)$means$\lim_{n \to \infty}f(n)/g(n) = 1$, which is not true in this case. The correct ... 2 It holds$\sin(\frac x \pi) = \pm 1$if and only if$x= \pi(\frac \pi 2+ k\pi)$, for$k\in \mathbb Z$. So for$x_0= \pi(\frac \pi 2+ k\pi)$you got $$\lim_{n\to \infty} \left(\sin \left(\dfrac {x_0} \pi \right)\right)^{2n}= \lim_{n\to \infty} (\pm 1)^{2n}=1.$$ For the other values of$x$it holds$-1<\sin \left(\dfrac {x} \pi \right)<1$, hence$u:=\sin ...

2

First consider uniform convergence on any interval $(0,\delta)$ where $\delta \leqslant1$. We have $$\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}k(x-1)^k\right|= \sum_{k=n+1}^{\infty}\frac{(1-x)^k}k \geqslant\sum_{k=n+1}^{2n}\frac{(1-x)^k}k ,$$ and $$\sup_{x \in (0,\delta)}\left|\sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}}k(x-1)^k\right|\geqslant \sup_{x \in (0,\... 2 It does converge pointwise to 0, for the reason you said. Basically, f_n(x) = 0 for sufficiently large n (where "sufficiently large" depends on x). To prove it formally, suppose x \in [0, 1]. If x = 0, then by definition, f_n(0) = 0 for all n, so f_n(0) \to 0 as n \to \infty. Otherwise x > 0. We can then use the fact that \... 2 Asserting that the pointwise limit of a sequence (f_n)_{n\in\Bbb N} of functions from \Bbb R into \Bbb R is some f(x) unless x=\frac1n makes no sense, since \frac1n is not a fixed number. Besides, for every real number x you do have \lim_{n\to\infty}\frac{nx}{n^2x^2+1}=0. So, your sequence converges pointwise to the null function. However, ... 2 Hint: If x=\frac1{n^2}, then$$\frac n{1+nx}=\frac n{1+1/n}=\frac{n^2}{n+1}.$$1 hint You made a mistake in your \lim_{n\to+\infty}f_n(x)=1. In fact,$$\lim_{n\to +\infty}f_n(x)=0$$and$$M_n=\sup_{0<x<1}|f_n(x)-0|=1$$thus, the convergence is not uniform at (0,1). Or$$M_n\ge f_n(\frac 1n)=\frac 12$$By the same, let$$G_n(x)=|g_n(x)-0|$$then$$G_n'(x)=\frac{1}{(nx+1)^2}$$and$$\sup_{0<x<1}G_n(x)\le g_n(1)=\frac{1}{n+1}...

1

Your solutions are correct. But here is another solution for part b: It's clear that $f_n \to 0$ pointwise on $(0,+\infty)$, so for uniform convergence, we would need the uniform norm to converge to $0$ as well. But: $$\lVert f_n \rVert = \sup_{x \in (0, +\infty)} |f_n(x)|=\sup_{x \in [0,+\infty)} |f_n(x)|=f_n(0)=1$$

1

Having proved the pointwise convergence you are in fact done, because of the following Claim: Let $(h_n)$ be a sequence of non-decreasing functions, each mapping $(0,1)$ into $\Bbb R$ and converging pointwise to a continuous strictly increasing function $h$ mapping $(0,1)$ onto $\Bbb R$. Then the convergence is uniform on compact subsets of $(0,1)$. Fix $0&... 1 Recall the basic inequalities$y- \frac{y^3}{6} \leqslant \sin y \leqslant y$for$y>0$. If you are not familar with the LHS inequality, it is easily derived from the RHS inequality by integrating twice. We have $$x- x \sin \frac{x}{n} = \begin{cases}x \left(1 - \frac{\sin \frac{x}{n}}{\frac{x}{n}}\right), & 0 < x \leqslant 1 \\ 0 , & x = 0\... 1 Hint: For x\ne 0,$$n\sin(x/n) - x = x\left(\frac{\sin(x/n)}{x/n}-1\right).$$1 For U=\mathbb{R^+}$$\underset{n\to\infty}\lim\underset{x\in U}\sup |f_n(x)-f(x)|=\underset{n\to\infty}\lim\underset{x\in U}\sup \int\limits_{x+n}^{+\infty}\dfrac{du}{2e^u+\sin^2u}\leqslant\\ \leqslant\lim\limits_{n \to \infty} \int\limits_{n}^{+\infty}\dfrac{du}{2e^u+\sin^2u}=0$$But for U=\mathbb{R} we have \sup\limits_{x \in \mathbb{R}} \int\limits_{... 1 The definition of uniform convergence may help: you need to verify that$$sup_{x\in[0,1]}\|e^{\frac{x}{k}}-1\| \to 0 $$and it may worth to notice where do you get that supermum (which is maximum in this case). from there you can complete the proof 1 hint For any k>0, the function$$g_k: x\mapsto e^{\frac xk}-1$$is positive and strictly increasing at [0,1].$$M_k=\max_{x\in[0,1]}|g_k(x)|=g_k(1)=e^{\frac 1k}-1\lim_{k\to +\infty}M_k=1-1=0$$So, the convergence is uniform at [0,1]. 1 I'm assuming you meant to write$$\sup_{x\in X}\left\{d_Y\left(f_n\left(x\right){,}\ f\left(x\right)\right)\right\}<\varepsilon\ \Leftrightarrow\ \forall x \in X, d_Y\left(f\left(x\right){,}\ f_n \left(x\right)\right)<\varepsilon,$$since that would facilitate the proof. The reverse implication does not hold with strict inequality, but this presents no ... 1 Set f_k(x)=(1-x)x^k. Note that f_k is positive and since f_k is continuous on the compact interval, it admits its maximum value. We can solve \frac{d}{dx}f_k(x)=0 to find the maximum. We have that \frac{df_k}{dx}(x)=kx^{k-1}-(k+1)x^k, so kx^{k-1}-(k+1)x^k=0 if and only if x=0 or x=\frac{k}{k+1}. It is easily verified that the maximum value of ... 1 Call your sequence f_k(x). Consider first finding the maximum as a function of k. The derivative gives:$$-x^k+(1-x)kx^{k-1}=0$$and when x\neq 0, (1-x)k = x, giving x=k/(1+k). Using a second derivative test, show that this is indeed a maximum. So f_k(x)\leq f_k(k/(1+k)).$$f_k((k/(1+k))=(1/(k+1))(k/(1+k))^k\rightarrow 0$$which should give you ... 1 It converges uniformly on any set [a; +\infty), but doesn't converges uniformly on \mathbb R. It's enough to check case a < 0. If n > -a + 1, |g_n(x)| = \int\limits_{n + a}^n \frac{dt}{\exp(t^3)} < \int\limits_{n + a}^n e^{-t}\,dt < e^{-n - a}. Taking M_n = e^{-n - a} note that \sum\limits_{n=0}^\infty M_n converges, thus \sum g_n(... 1 |\cos y-1| =2 \sin^{2} (\frac y 2) \leq \frac {y^{2}} 2. 1 Using$$\int_c^d |f_i(x,t)| \, dt \leqslant\int_c^d g(t) \, dt,$$and the fact that the integrand on the LHS is nonnegative, you can only conclude pointwise convergence, that is for each x \in A,$$\lim_{d \to \infty} \int_c^d |f_i(x,t)| \, dt = \limsup_{d \to \infty}\int_c^d |f_i(x,t)| \, dt \leqslant \lim_{d \to \infty}\int_c^d g(t) \, dt$\$ To prove ...

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