Let $f:[0,1]\rightarrow\mathbb{R}$ be a continuous function and let $\{a_n\}_n$ and $\{b_n\}_n$ be sequences of real numbers such that: $$\lim_{n\to\infty}\int_0^1|f(x)-a_nx-b_n|\text{dx}=0$$ Prove that: a$)$ The sequences $\{a_n\}_n$ and $\{b_n\}_n$ are convergent b$)$ The function $f$ is linear  In this particular problem  which was given in an entrance examination of $10+2$ level,my approach was to do the epsilon delta definition of limit with integration ,but of no avail. I cannot start and arrive at the result... Kindly help anyone.
122905
Let $f:[0,1] \to \mathbb R$ be a continuous function and the sequences $(a_n)_n,(b_n)_n$ s.t. $$\lim_{n\to \infty}  \int_0^1 |f(x)-a_nx-b_n| dx=0.$$ Prove that $(a_n)_n,(b_n)_n$ are convergent.   I know that $$\left|\int_0^1f(x)dx\right| \le \int_0^1|f(x)|dx.$$ Can somebody help me, please?
55215
Supose that $f\colon [0,1] \rightarrow \mathbb R$ is a continuous function. Show that: $$\lim \limits_{n \rightarrow+\infty} \int_{0}^{1}{f(x^n)dx}=f(0)$$  I'm studying the Riemann Integral theory and I have not idea to how do this. I need some hint to start because I do not see how to apply the Riemann sum or other theory concept to be able to help me.
37572
I'm preparing for a test for real analysis and I came across this problem in Royden's book:  Let $\{f_n\}$ be a sequence of real valued functions on $[a,b]$ that converges pointwisely on $[a,b]$ to the real valued function $f$. Show that $TV(f) \leq \liminf  ~TV(f_n)?$  This looks quite similar in form to Fatou's Lemma to me, but can't find any way to establish TV with integration, can anybody please help?  (TV is short for total variation)
82145
Prove that $\lim_{n \to \infty} \sqrt[n]a = 1$ using only the basic properties of sequences  I don't have an idea on how to tackle this problem - I thought about the Squeeze Theorem but I can't think of a sequence of n which approaches 1.  Could you give me some hints on how to solve this? If it is possible, please, choose the least sophisticated method to solve this as I am entirely new to Calculus.
93641
Let $f_n:[a.b]\rightarrow \mathbb{R}$ be sequence of $L-$Lipschitz functions, that is: $$\forall x,y\in[a,b]: |f_n(x)-f_n(y)|\leq L|x-y|$$   Suppose $f_n \rightarrow f$ pointwise, prove $f_n \rightrightarrows f$   I have all the parts of the puzzle for the proof, and I'm trying to put them all together, I'm using  in my answer.  I would appreciate is you could correct my proof, and if you have an alternative proof, I would be more then happy to see it.  My proof:   Let $\epsilon&gt;0.$  $f_n$ are uniformly continuous on $[a,b]:$  $\tag{1} \exists \delta&gt;0\ \forall x,y\in[a,b]: |f_n(x)-f_n(y)|&lt;\frac{\epsilon}{3}$  $f$ is also $L-$Lipschitz:  $\tag{2} \forall x,y\in[a,b]:|f(x)-f(y)|&lt;L|x-y|=\frac{\epsilon}{3}$  Let us set a partition of $[a,b]$ such as Stephen Montgomery-Smith suggests:     Pick points $x_1,\dots,x_m \in [a,b]$ which are distance   $\frac{\epsilon}{3}$ from each other.        For each $1 \le i \le m$, find a number $N_i$ so that for all $n \ge  N_i$ we have $|f_n(x_i)-f(x_i)| \le \epsilon/3$.   Let $N = \max_{1 \le i \le m} N_i$      Now given any $x \in [a,b]$, pick $1 \le i \le m$ such that $|x-x_i| &lt;\frac{\epsilon}{3L}:  |f_n(x)-f(x)|&lt;\frac{\epsilon}{3} \tag{3}$   $$\begin{align}|f_n(y)-f(y)| &amp;=|f_n(y)-f_n(x)+f_n(x)-f(x)+f(x)-f(y)| \\ &amp;\leq |f_n(y)-f_n(x)| + |f_n(x)-f(x)|+|f(x)-f(y)| \\ &amp;&lt; \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \\\end{align}$$
30401
I want to prove the following theorem     Let $\{a_{n}\}$ be a convergent sequence. Prove that $\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}$   Do I need to show that it is monotone increasing and then use the relation  $$0\leq \lim_{n\to \infty}a_{n+1}- \lim_{n\to \infty}a_{n}\leq 0 \;?$$  Please, I need help on this! Various proofs are welcome! Thank you very much for your time!
20736
I want to show that a sequence $(a_n:n \in \mathbb N)$ converges to a limit $L \in \mathbb R$ if, and only if, an N-th tail of the sequences converges to $L$ for some $N \in \mathbb N$. For each $N \in \mathbb N$, an N-th tail sequence can be defined as $(t_n:n \in \mathbb N)$ where $t_n := a_{N+n}$.  Using the definition of a limit for the sequence $(a_n:n \in \mathbb N)$, we have the following:  For every $0 \lt \epsilon$, there is a natural number $N_{\epsilon} \in \mathbb N$ such that for all $N_{\epsilon} \lt n$, $|a_n - L| \lt \epsilon$. But I'm not sure what to do next in order to show that the $N$-th tail converges. Which natural number should I choose for the $N$-tail? Any help would be much appreciated.
49119
The statement I am trying to prove:  Let $\{ f_n \}$ be a sequence of equicontinuous, real valued, uniformly bounded   continuous functions on $\mathbb{R}$. Show that $\{ f_n \}$ has a convergent subsequence   which converges uniformly on any bounded subset of $\mathbb{R}$ and pointwise on all   of $\mathbb{R}$ to a continuous function. (Royden pg 210, problem 9).  I believe the first portion follows directly by Arzela Ascoli, once it is noted that if we are given some bounded subset $E$, we can look at $\overline{E}$ a compact set on which the family is equicontinuous and uniformly bounded.  For pointwise convergence on all of $\mathbb{R}$, the question seems to require some sort of diagonalization argument: Take some dense sequence of $\{x_i\}$ (the reals are separable) and write columns of $f_n$ applied to each individual point. There will be a convergent subsequence by Bolzano-Weirstrass and since the outputs are real and bounded by the uniform boundedness of the family. I am a bit fuzzy on this part: Repeating this literally pointwise and moving far enough diagonally, we can correct for different convergence "speeds" pluck out a single convergent subsequence for any given epsilon and have $\sup|f_n(x)-f(x)|&lt;\epsilon$ for any given epsilon. Finally, continuity should follow from an $\epsilon/3$ argument and equicontinuity of the family.  However, in formalizing the above, I am getting caught up in the notation of that sort of argument and am hoping for a more elegant solution. Is there a better way? Am I just generally confused and that's why I'm having trouble writing things down formally?   Thanks!
75107
I have this question :  Proof if $f$ continuous in $x_0$ then there is a neighbourhood of $x_0$ so $f$ bounded.  I want to know if my proof is valid :  If continuous in $x_0$ then :  $$\lim_{x \to x_0} f(x)= f(x_0)$$  Therefore :  All $\epsilon&gt;0$ exists $\delta&gt;0$ so any $x$ that implies $|x-x_0|&lt;\delta$ implies $|f(x)-f(x_0)|&lt;\epsilon$.  Therefore for $\epsilon=1$ there is $\delta&gt;0$ so any $x$ that implies $|x-x_0|&lt;\delta$ implies $|f(x)-f(x_0)|&lt;1$.  $|x-x_0|&lt;\delta \rightarrow x_0-\delta&lt;x&lt;x_0+\delta$.  Lets choose $x_1,x_2$ that implies $x_0-\delta&lt;x_1&lt;x_2&lt;x_0+\delta$  Therefore in $[x_1,x_2]$ from Weierstrass from theorem bounded there.
76391
If $\{x_n\}$ is a sequence which satisfies $\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c$  where $c$ is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of $\{x_n/n\}$.  Attempt: $\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c$ where $c &gt;0$  => $x_n$ is unbounded and divergent.  However, I am stuck on how to relate this to convergence/divergence of $x_n/n$ . Thanks for the help.
43844
Hope this is a meaningful question, but I'm curious if is possible to show that   $$\lim_{n\to\infty} a^{\frac{1}{n}}=1, \text{where  }a&gt;0$$  using $\delta-\epsilon$ directly or other methods. One method that I am aware of is to use the following:     If $\{s_n\}$ is a nonzero sequence, then $\liminf\bigl|\frac{s_{n+1}}{s_n}\bigr|\le \liminf |s_n|^{\frac{1}{n}}\le \limsup |s_n|^{\frac{1}{n}}\le\limsup\bigl|\frac{s_{n+1}}{s_n}\bigr|$
81593
Let $f$ be a real, continuous function defined for all $0\leq x\leq 1$ such that $f(0)=1$, $f(1/2)=2$, and $f(1)=3$. Show that  $$\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx$$  exists and compute the limit.  Attempt:   Since $f$ is real-valued continuous on $0\leq x\leq 1$ and the boundaries $f(0)=1$ and $f(1)=3$, $f$ is bounded on the interval  and the integral $\displaystyle\int_0^1 f(x^n)dx$ exists for any positive $n$. Thus, we can interchange the limit and the integral and compose the limit, \begin{align*} \lim_{n\rightarrow\infty} \int_0^1 f(x^n)dx&amp;=\int_0^1\lim_{n\rightarrow\infty}f(x^n)dx\\ &amp;=\int_0^1 f\left(\lim_{n\rightarrow\infty} x^n \right)dx\\ &amp;=\int_0^1 f(0)dx=1. \end{align*}  Questions: So, my first question concerns whether interchanging the limit and the integral is correct. It seems it would be justified by the dominated convergence theorem, where $f(x^n)$ is dominated by a function $g(x)=\displaystyle\sup_{0\leq x\leq 1} f(x^n)$ and $f_n(x^n)$, on $0\leq x\leq 1$, converges pointwise to a function $f$ that takes value 1 for $x\neq 1$ and 3 for $x=1$.  My next question is whether the interchange between the limit and composition of the function is allowed. Since the function is continuous and the limit exists at that point, it seems the interchange would be justified.
104844
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?
100496
Not sure if my proof as is is correct:  Since $a_n \leq b_n$ for all $n$, we have $a_n - b_n \leq 0$. As such, by the limit location theorem it holds that $\lim_{n\to\infty} (a_n - b_n) \leq 0$. By linearity, we have $\lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n \leq 0$, which implies $lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$.  My one concern is using the LL Theorem on the sequence $a_n - b_n$: do I need to prove this sequence is convergent first?
103954
If $f$ Continuous bounded on $\mathbb{R}$，and $$\lim_{h\to0}\sup_{x\in\mathbb{R}}\left|f(x+h)-2f(x)+f(x-h)\right|=0.$$ prove that $f$ uniformly continuous on $\mathbb{R}$. I try to find a $\delta&gt;0$ to make it arbitrary $x_1,x_2\in \mathbb{R}$and$\left|x_1-x_2\right|&lt;\delta$have$$\left|f(x_1)-f(x_2)\right|&lt;\varepsilon.$$but I don't know how to use this complex limit expression.I hope you can give me some tips, thank you!
73794
Suppose that $f:[0,\infty)\rightarrow\mathbb{R}$ is continuous. Is it true that   $$\int_{0}^\infty|f(t)|dt&lt;\infty\Rightarrow \lim_{t\rightarrow\infty}f(t)=0?$$  If so can you provide a proof, otherwise a counter example. Thank you.
87362
Suppose $F$ is a nondecreasing and right continuous function, and the sequence $\{x_n\}_{n\geq1}$ converges to $x$. Then $\liminf\limits_{n\to\infty}F(x_n)\leq F(x)$.    How can I prove this?