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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