I'm trying to study this sequence of functions: $f_n(x) = nxe^{-nx}$ for $x \in [0,1]$ i ended up with the  point convergence -&gt; $f(x) = 0$  uniform convergence The function doesn't have the uniform convergence to $f(x) = 0$ in $[0, 1]$. Becouse during the study of the convergence if found that the $sup_{x \in [0,1]}|f_n(x) - f(x)| = \frac{1}{n}$ and for this $x=\frac{1}{n}$ the $\lim\limits_{n \to +inf}e^{-1} \not= 0$   I think that that's ok but I know that a correct and complete study of the sequences of functions include the analisys of the uniform convergence in other sub-ranges $[\alpha, \beta] ⊂ [0, 1]$. I don't know how to do this in a step-by-step way, but i think that in this case the uniform convergence is present in $[0, a] ⊂ [0, 1]$ with $a&lt;\frac{1}{n}$ $[b, 1] ⊂ [0, 1]$  with $b&gt;\frac{1}{n}$ where $\frac{1}{n}$ is the max value assumed by the function at index $n$ of the sequence. Am i right or thats a wrong idea?
122661
Why isn't $nxe^{-nx}$ uniformly convergent for $x \geq 0$?   The definition of uniform convergence states that $f_n$ is uniformly convergent if $$\lim_{n\to\infty} ||f_n-f|| = 0.$$  I apply the definition with $f = 0$ and get $$\lim_{n\to\infty}||\frac{nx}{e^{nx}} - 0 || = 0,$$  since $e^{nx}$ grows much faster than $nx$ regardless of $x$ (except when $x = 0$, but if that's the case everything is $0$).   Where am I wrong?
43399
I was able to prove, hopefully correctly, that the sum converges uniformly. But, I'm not sure how to show it converges uniformly specifically to $f$.  The way I proved it converges uniformly was by using the Dirichlet convergence test, by taking $a_k(x) = \frac1n$ and $b_k(x) = f_k(x)$. $\sum_{k=1}^n{b_k(x)}$ is uniformly bounded on $[0,1]$ by $n+2$ for instance, and $a_n$ is monotone and uniformly convergent to 0 because it does not depend on the value of $x$.  I got rather stuck after this. I guess I have to use the continuity of the functions and the uniform convergence of both, but I can't figure out how to show that they converge to the same thing. Any help will be greatly appreciated.
94761
I'd like your help with the following claim to prove:  $$\lim_{n \to \infty} \int_{0}^{\sqrt n}\left(1-\frac{x^2}{n}\right)^ndx=\int_{0}^{\infty} e^{-x^2}dx.$$  I think I should use the claim:     Let $a,b$ two real numbers and $\{f_n\}$ a sequence of continuous functions on $\left[a,b\right]$  which converges uniformly to $f$ on $[a,b]$. Then    $$\lim_{n\to\infty}\int_a^bf_n(t)dt=\int_a^bf(t)dt.$$   But for this, I must prove that $(1-\frac{x^2}{n})^n$ uniformly converges to $e^{-x^2}$.  How do I do that? One of the hardest and trickiest things to do is to prove this uniformly converges.. every function has its own way. I believe that the more examples I'll see the easiest it will be. Thanks again!
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Test the sequence of functions for uniform convergence       $$f_n(x)=\frac{nx}{1+n^2x^2}$$      $a) \text{on} \ [0,1]$      $b) \text{on} \ [1,{\infty})$   a)  If we differentiate this with respect to x we get the following :  $$(\frac{nx}{1+n^2x^2})'=\frac{n(1-n^2x^2)}{(1+n^2x^2)^2}$$  This is equal to $0$ for $x=^{+}_{-}\frac{1}{n}$  From here we can either check these two values in the original function and take the larger one as the supremum or we can find the second derivative and check whether it's negative. In either case we find that for $x=\frac{1}{n}$ we get the value $f_n(x)=\frac{1}{2}$ and for $x=-\frac{1}{n}$ we get $f_n(x)=-\frac{1}{2}$. Note that $x=-\frac{1}{n}$ doesn't even belong in our interval. Now we can finish the problem.  $$\lim_{n\to \infty}{\frac{nx}{1+n^2x^2}}=0$$  So it is piecewise converging to $f(x)=0$  $$\lim_{n\to \infty}\sup{\left \lvert{f_n(x)-f(x)}\right \rvert}=\lim_{n\to \infty}{\frac{n\frac{1}{n}}{1+n^2 \frac{1}{n^2}}}=\frac{1}{2}$$  So $f_n(x)$ does not convergence uniformly on $[0,1]$ to $f(x)=0$  b)  Now we need to test the on the interval $[1,\infty)$.From the first derivative we notice that if $x\in [1,\infty)$ then the numerator seems to be negative for all $x&gt;1$. So $f_n(x)$ is monotonically decreasing. That means that the supremum will be at the first point of the interval or $x=1$. Taking the limit :  $$\lim_{n\to \infty}{\frac{n}{1+n^2}}=0$$  We can see that the function convergences uniformly on this interval.  Is there something wrong with either of these solutions?  I'm having doubts mostly about the second interval since $x=\frac{1}{n}$ is a supremum but only for $n=1$. However I believe that we need to have the supremum that's satisfied for all $n \in N$. Pointing even the tiniest mistake in my solutions would be very helpful. Thank you.
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Problem Statement:  Let $\{f_n\}$ converge uniformly on a set $E$. Suppose that each $f_n$ is bounded. Prove that there exists an $M &gt; 0$ s.t. $|f_n(x)| \leq M$ for all $x \in E$ and all $n = 1,2, ...$.  Proof:  If each $f_n$ is bounded by some $M_n &gt; 0$ then can't we just take $M = \text{max}\{M_n\}$?  I don't see what's so difficult here. Am I not using the fact that $\{f_n\}$ converges uniformly?
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I want to prove that $xe^{-x}$ is bounded by some constant for all the $x \in [0, \infty)$. So I take derivative of this guy, and found that on $x \in [0, \infty)$, its derivative is positive from $[0, 1)$, $0$ at $x=1$, and negative from (1, in infinity). So  $xe^{-x}$ achieves its maximum in $[0, \infty)$ at $x=1$, which is $1/e$. However, to prove that it's bounded by $1/e$, I need to prove that as $x$ goes to infinity, $xe^{-x}$ converges to $0$. Because otherwise, if it went to some value below $(-1/e)$, $xe^{-x}$ would not be bounded by $1/e$. How can I prove that $xe^{-x}$ converges to $0$? I mean, as $x$ goes to infinity, $e^{-x}$ converges to zero, but at the same time, x diverges to infinity. How can I formally prove that the product term converges to $0$? Do I need to use the epsilon definition of convergence to prove it?  Could you just show me the formal proof of $xe^{-x}$ converges to $0$? Or is there's any other way to approach this question? My professor suggested that I should use the power series for $e^{-x}$...?   Your help is greatly appreciated! Thank you so much!  I hope you enjoy your Saturday!
121684
Suppose that $x_0 = \alpha \in \mathbb{R}$ and $x_{n+1} = x_n ^2-x_n +1$.  I am asked to study the boundedness of $(x_n)$ and then asked if $(x_n)$ converges. How can I show that $(x_n)$ is bounded?  I have noted that $$x_{n+1}-x_n = (x_n-1)^2\geq 0$$ so $x_n$ is increasing.  Suppose that $(x_n)$ converges, then $l = l^2-l+1$ so $l = 1$. This means that the sequence converges only when $\alpha \leq 1$. And so $(x_n)$ is bounded above by $1$ if $\alpha\leq 1$ (since an increasing sequence converges to its supremum) otherwise the sequence is not bounded above.  Is my reasoning correct?
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Exercise 5.3.1 in Durrett's "Probability Theory and Examples" states  Let $X_n$, $n\ge 0$, be a submartingale with $\sup X_n &lt; \infty$. Let $\xi_n=X_n-X_{n-1}$, and suppose $E(\sup \xi_n^+)&lt;\infty$. Show that $X_n$ converges almost surely.  Here are my thoughts: It looks like we need to use theorem 5.2.8 (Martingale convergence theorem), which states that $X_n$ converges almost surely if $\sup EX_n^+&lt;\infty$. I am trying to understand why we can just do the following (or at least I'm guessing we can't):  Suppose $X_n &gt; 0$ (edit: on some set with positive measure), then $0&lt; \sup X_n &lt; \infty$, and therefore $EX_n^+ \le \sup X_n &lt; \infty$. So we're done.  Otherwise, if $X_n \le 0$ a.e., then $\sup X_n^+ = 0$ a.e. Once again, we're done.  Am I missing something? Since I didn't use all the information, I probably am.  Thanks for the help.
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Let $(x_n)$ be a sequence of real numbers.  Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ .              Thanks for any help.
121002
Let $f:[-1,1]\rightarrow \mathbb{R}$ be continuous with $f(0)=0$ and continuously differentiable in $]-1,1[$. Suppose that  $$|f'(x)|\leq |f(x)|,\;\mbox{ for all }x\in]-1,1[.$$  Prove that $f(x)=0$ for all $x\in[-1,1]$.  My idea is take one point (arbitrary) $x_0&gt;0$ in $]-1,1[$, then using mean value theorem, exist $y_0\in[0,x_0]$ such that   $$|f(y_0)|\geq|f'(y_0)|=\frac{|f(x_0)-f(0)|}{|x_0-0|} \mbox{ then }|f(x_0)|\leq x_0|f(y_0)|,$$  So if $f(y_0)=0$ implies $f(x_0)=0$ and we ended. But is $f(y_0)\neq0$ then I can follow and obtain a decreasing sequence but I not sure that converge to zero.  Thank you by some hints.
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If the sequence of functions $f_{n}: X \longrightarrow \mathbb{R}$  is such that $f_{1} \geq ... \geq f_{k} \geq ...$ and $f_{n} \longrightarrow 0$ uniformly in $X$. Prove that $\displaystyle \sum(-1)^{n}f_{n}$ uniformly converges in $X$   Since $f_{n}$ is uniformly convergent, the convergence of $f$ not dependes of $x$. Thus, is the proof reduced to the Leibniz test?
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For the sequence 1/(n-1): x1 is undefined. I think that makes the sequence unbounded. However I can prove from definition that the sequence is convergent. And there is the theorem that all convergent sequences are bounded. So what am I missing?
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Suppose $f_n:[0,1]\to\mathbb{R}$ such that $f_n=nx(1-x)^n$, then I need to show that $f_n$ converges pointwise but not uniformly. I can see that since $f$ is defined on $[0,1]$, the part $(1-x)^n$ is smaller than $1$ and therefore will converge, but what about $nx$? This will go to infinity...  Therefore, I can't even see that it converges pointwise, but let's suppose it converges to some function, how do I show that it doesn't converge uniformly? I should find an $\epsilon$ such that it's not valid for all $n&gt;n_0(\epsilon)$ that $|f_n-something|&lt;\epsilon$, right?
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Let $y_n$ be a sequence of real numbers. Prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges.  Let us suppose that $y_{n}$ goes to a limit $L$. Then for all $\varepsilon &gt;0$, for sufficiently large $n$, $|y_n -L| &lt; \varepsilon$ .  But $|y_n - L| &lt; 2|x_n-\frac{L}{3}|+|x_n - \frac{L}{3}|$  But then how to proceed?
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I would like some advice on how to solve problems like the following: Let $(x_n)$ be a sequence defined by $x_1= 3$ and $x_{n+1} = \frac{1}{4-x_n}$. Prove that the sequence converges.  My strategy is to use the Monotone Convergence Theorem, but I am having trouble showing that the sequence is decreasing and bounded below. Here's my work so far:  Decreasing: The first 4 values are $3,1,1/3,3/11$, so let's assume $x_n \leq x_{n-1} \leq \ldots \leq x_1$. Want to show $x_{n+1} \leq x_n$. We have $x_{n+1} = \frac{1}{4-x_n}$. I want to have an upper bound for the RHS, but can't find one and don't really know where to go from here.  Bounded below: I wanted to show that all values are positive, but if we assume $x_n &gt; 0$, that doesn't rule out $x_{n+1}$ from being negative.
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Let $a_n(t)=\frac{1}{n} \text{1}_{ [0,n)}(t)$. Does this sequence converge in $L^1/$pointwise?  My ideas: We have for all $t$: $\mathbf{1}(t)\in\{0,1\}$ and $$0 \leq \frac{1}{n}\textbf{1}(t) \leq 1 \Rightarrow 0 \leq \left| \frac{1}{n}\textbf{1}(t)\right|^p \leq 1 \quad \forall p\ge1$$   So it's a null sequence and we have dominated convergence with dominated function $g(t)=1$. $g(t)=1$ is integrable so all $a_n$ is integrable for all $n$.
121934
I have tried a little bit which as follows- Since $f(x_n)\in[0,1]$, $\{f(x_n)\}$ has a convergent subsequence say $y_n=f(x_{r_n})\ \forall n\in\Bbb{N}$ Let, $\lim y_n=l\implies \lim \frac{y_1+y_2+\cdots+y_n}{n}=l\implies \lim \frac{f(x_{r_1})+f(x_{r_2})+\cdots+f(x_{r_n})}{n}=l$ But I am getting no idea to proceed and prove the convergence of $\{x_n\}$. I have also tried to prove the sequence to be cauchy which goes- $x_{m+1}-x_{n+1}={\sum_{i=1}^m f(x_i)\over m}-{\sum_{i=1}^n f(x_i)\over n}\le {\sum_{i=1}^m f(x_i)\over n}-{\sum_{i=1}^n f(x_i)\over n}$ (since $m\ge n)$ $\implies |x_{m+1}-x_{n+1}|\le \frac{|f(x_m)|+|f(x_{m-1})+\cdots+|f(x_{n+1})|}{n}\le\frac{m-n}{n}$ (since $f([0,1])\subseteq [0,1])$ Now, what will I get if I tend $m,n\to\infty$? But I need to do something more in the 2nd case since I have nowhere used the continuity of $f$.  Can anybody give an idea to prove it? Thanks for assistance in advance.