Let be $f:[0,\infty[ \rightarrow \mathbb{R}$ a decreasing and continuos, $f(x)&gt;0 , \forall x$. If $\displaystyle \int_0^{\infty}f(x)dx$ converges then $\displaystyle \lim_{x \rightarrow \infty} xf(x) = 0.$  How to prove that?  Ok, I know that if $G(x) = \displaystyle \int_0^{x}f(t)dt $ then $ \displaystyle \lim_{x \rightarrow \infty} G(x) = L$, How can I use that for arrive in $xf(x)$??
47797
Let $f:\mathbb{R}_+ \to \mathbb{R}_+$ be a monotone decreasing function defined on the positive real numbers with $$\int_0^\infty f(x)dx &lt;\infty.$$ Show that $$\lim_{x\to\infty} xf(x)=0.$$   This is my proof: Suppose not. Then there is $\varepsilon$  such that for any $M&gt;0$ there exists $x\geq M$ such that $xf(x)\geq \varepsilon$. So we can construct a sequence $(x_n)$ such that $x_n \to \infty $ and $x_n f(x_n ) \geq \varepsilon$. So $$\frac{\varepsilon}{x_n}\leq f(x_n) \implies \sum_{n\in\mathbb{N}}\frac{\varepsilon}{x_n} \leq  \sum_{n\in\mathbb{N}} f(x_n) \leq \int_0^1 f(x)dx.$$ So we get a contradiction.  I feal like I have the correct idea but some details are wrong.  Any help would be appreciated.
59864
Let $l \in\mathbb{R} $ and $ f:(0,\infty) \to \Bbb R $ be a function such that $ \lim_{x\to \infty} xf(x)=l $ .  Prove that  $ \lim_{x\to \infty} f(x)=0 $.  Any help would be appreciated - I found this difficult to prove
76888
Let $f: [0,\infty) \to [0,\infty)$ be a continuous function such that $f(f(x)) = x^2, \forall x \in [0,\infty)$. Prove that $\displaystyle{\int_{0}^{1}{(f(x))^2dx} \geq \frac{3}{13}}$.  All I know about this function is that $f$ is bijective, it is strictly increasing*, $f(0) = 0, f(1) = 1, f(x^2) = (f(x))^2, \forall x \in [0, \infty)$ and $f(x) \leq x, \forall x \in [0, 1]$**. With all these, I am not able to show that  $\displaystyle{\int_{0}^{1}{(f(x))^2dx} \geq \frac{3}{13}}$.  *Suppose that $f$ is strictly decreasing. Then, $\forall x \in (0,1), x^2 &lt; x \implies f(x^2) &gt; f(x) \iff (f(x))^2 &gt; f(x) \iff f(x) &gt; 1$, which is false because, if we substitute $x$ with $0$ and with $1$ in $f(x^2) = (f(x))^2$ we get that $f(0) \in \{0,1\}$ and $f(1) \in \{0,1\}$. So $f$ is strictly increasing.  **Suppose that there exists $x_0 \in [0, 1]$ such that $f(x_0) &gt; x_0$. Then, $x_0^2 = f(f(x_0)) &gt; f(x_0) &gt; x_0$, which is false. Then $f(x) \leq x, \forall x \in [0,1]$.  Edit:  I have come up with an idea to use Riemann sums, but I reach a point where I cannot continue.  Let $\epsilon &lt; 1$. Then $f(\epsilon) = x_1$ and $f(x_1) = \epsilon^2$. And now $(f(\epsilon))^2 = f(\epsilon^2) = x_2$ and so on. Now we will use the Riemann sum:  We will take the partition $\Delta = (1 &gt; \epsilon &gt; \epsilon ^2 &gt; ... \epsilon ^{2^n} &gt;0 ) $and the intermediate points will be the left margin of each interval. Then we have:  $\displaystyle{\int_{0}^{1}{(f(x))^2 dx}}  = \displaystyle{ \lim_{\epsilon \to 1}{\lim_{n \to \infty}{\sum_{k = 0}^{n}{(\epsilon^{2^k} - \epsilon^{2^{k+1}})\epsilon^{2^{k+1}}}}}}$  I do not know how to compute this.
43867
I ran into the following problem when I was doing my homework, and I have no thoughts on where I should start with:  (1) If $f\in L^{2}$, show that $\displaystyle \lim_{p \rightarrow 1^{+}}\int_{[0,1]}|f|^{p}=\int_{[0,1]}|f|$  (2) If $0&lt;p$, show that $\displaystyle \lim_{q\rightarrow p^{-}}||f||_{q}=||f||_{p}$  My first thought was Generalized LDCT, but it didn't seem to work. I also made some other attempts but none of them were successful... Can anybody give me some hints on how I should look at this question?  Also, I know if $p\rightarrow\infty$ then $||f||_{p}\rightarrow||f||_{\infty}$ on $[0,1]$, but does similar continuity in p holds for other $L^{p}[0,1]$ norms in general?  Thank you!  Edit:   Sorry if I did not make it clear enough in the question. All $L^{p}$ refers to $L^p[0,1]$.  The first question is found  (thanks to t.b.), but the second question remains, mainly because $f$ is not guaranteed to be in any $L^{p}$.
62646
$$\mathop {\lim }\limits_{n \to \infty } {{{a_n}} \over {{b_n}}} = 1$$ Prove the statement implies $\sum {{a_n},\sum {{b_n}} } $ converge or diverge together. My guess the statement is true.    if $\sum{{a_n}}$ diverges, then $\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0$    So,  $$\eqalign{   &amp; \mathop {\lim }\limits_{n \to \infty } {a_n} = L \ne 0  \cr    &amp; {{\mathop {\lim }\limits_{n \to \infty } {a_n}} \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow {L \over {\mathop {\lim }\limits_{n \to \infty } {b_n}}} = 1 \Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {b_n} \ne 0 \cr} $$  therefore,  $\sum {b_n}$ also diverges.    What I was not managed to do is proving that the two series converges together. Or maybe the statement is not always true?
22753
Prove that if one of $\displaystyle\lim_{x \rightarrow 0 }{f(x)}$ and $\displaystyle\lim_{x \rightarrow 0}{f(x^3)}$ exists, then the other one also exists. Can anyone guide me on this ? I have no idea on how to start .
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.
105530
I'm doing this exercise:     Prove that $F(a)=\displaystyle\int \limits _{0}^{\infty}\frac{e^{-2x}-e^{-ax}}{x}\text{d}x$ converges for every $a&gt;0$. Calculate $F'(a)$ and deduce $F(a)$.   I've separated the integral in  $$F(a)=\displaystyle\int \limits _{0}^{\infty}\frac{e^{-2x}-e^{-ax}}{x}\text{d}x=\displaystyle\int \limits _{0}^{\infty}\frac{e^{-2x}}{x}\text{d}x-\displaystyle\int \limits _{0}^{\infty}\frac{e^{-ax}}{x}\text{d}x$$ and I've tried to identify them with the gamma function, but $\Gamma(0)$ does not exist.  Edit: To decide whether it's convergent or not, can I use Dirichlet's test?  Any hint would be really appreciated. Thanks.
78547
Can anyone please give an example of why the following definition of $\displaystyle{\lim_{x \to a} f(x) =L}$ is NOT correct?:  $\forall$ $\delta &gt;0$ $\exists$ $\epsilon&gt;0$ such that if $0&lt;|x-a|&lt;\delta$ then $|f(x)-L|&lt;\epsilon$  I've been trying to solve this for a while, and I think it would give me a greater understanding of why the limit definition is what it is, because this alternative definition seems quite logical and similar to the real one, yet it supposedly shouldn't work.
76619
Let $f$ be continuous at $0$. Suppose $\displaystyle \lim _{x\rightarrow 0} \frac{f(2x)-f(x)}{x} =L$. Prove that $f'(0)=L$.   My Work:  $\displaystyle \Bigg|\frac{f(x)-f(0)}{x}-L\Bigg|=\Bigg|\frac{f(x)-f(2x)+f(2x)-f(0)}{x}-L\Bigg|$  $\displaystyle \leq \Bigg|\frac{f(x)-f(2x)}{x}-L\Bigg|+\Bigg|\frac{f(2x)-f(0)}{x}\Bigg|$. Since $f$ is continuous at $0, |f(2x)-f(0)|$ goes to $0$ as $x$ goes to $0$. Now I want to get rid of $|x|$ of the second term. How can I do it? Can somebody please help me?
40686
Let $f$ be $f\in C([0,\infty ])$, such that $\lim_{x \to \infty} f(x) = L $.  Calculate  $$ \int _{0}^\infty \frac{f(x)-f(2x)}{x}dx $$  Help?
80936
$f(x)$ is a continuos fuction in $[0, 1]$ and differentiable twice on $0$.  $U_n=(-1)^{n}f(\frac{1}{n})$. I need to prove that:   $1.$if $\sum_{n=1}^\infty U_n$ converge then $f(0)=0$.  $2.$if $f(0)=f'(0)=0$ then $\sum_{n=1}^\infty U_n$ converge absolutely.  Now, from the detail that $f(x)$ is differentiable twice on $0$ I think I need to write $f(1/n)$ with Maclaurin series with 2 degree but it doesn't work very nice. can you guide me please? thanks!
76626
Prove that:  {$f_n$} converges unifromly to $f$ on $E$ as $n \rightarrow \infty$ if and only if $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$  I don't understand the approach for the Forward proof.  My approach is:  By definition, for $\epsilon &gt; 0$, there exist $N \in \mathbb{N}$ such that $|f_n(x) - f(x)| &lt; \epsilon$; hence, the smallest upper bound should also be less than the $\epsilon$ (i.e., ${sup}_{x \in E}|f_n(x)-f(x)| &lt; \epsilon$). Thus, $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$.  However, the solution uses $\frac{\epsilon}{2}$ as the following:     $|f_n(x)-f| \leq {sup}_{x \in E}|f_n(x)-f(x)| &lt; \epsilon $      ${sup}_{x \in E}|f_n(x)-f(x)| \leq \frac{\epsilon}{2}$      ${sup}_{x \in E}|f_n(x)-f(x)| \leq \frac{\epsilon}{2} &lt; \epsilon$   Hence, $\lim_{n\rightarrow \infty} ||f_n-f||_u = 0$.  Why?
79153
I want to prove or disprove this problem: If there exist $\lim\limits_{x\rightarrow \infty} (f'(x)+f(x))=L&lt;\infty$ then $\lim\limits_{x\rightarrow\infty} f(x) =L$.  When I assume problem below:  If there exist $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x)) =L&lt;\infty$, There exists $\lim\limits_{x\rightarrow\infty} f(x)$?  I can use mean-value theorem to show that.  So my question is:     If $\lim\limits_{x\rightarrow\infty} (f'(x)+f(x))=L&lt;\infty$, does $\lim\limits_{x\rightarrow\infty} f(x)$ exist?
7416
I'd like a hint to solve the following problem:     Given a $f$ continuos and positive on $[a,b]$ prove that $$\lim_{n\to \infty}  \left[\int_a^b (f(x))^n dx\right]^{{1}/{n}}=\sup_{x\in [a,b]} f(x).$$   Thanks a lot!
1268
Let $f:[0,\infty)\rightarrow \mathbb{R}$ be a continuous and strictly decreasing function. If $\displaystyle\lim_{x\rightarrow\infty}f(x)=0$, prove that the following integral is divergent :   $$\int_0^\infty \frac{f(x)-f(x+1)}{f(x)}dx$$   First, it's not hard to see that $f(0)&gt;0$ and also $f(x)&gt;0$ for each $x$.  I think we have to show that behavior of $\frac{f(x)-f(x+1)}{f(x)}$ is "similar" to $\frac1{x+1}$  or $\frac{f(x+1)}{f(x)}$ is "similar" to $\frac{x}{x+1}$ for big values of $x$.  But meaning of "similar" is not clear yet !  The most naive and rough approach is to show $\frac{f(x+1)}{f(x)}&lt;1-k$ for some $0&lt;k&lt;1$ and therefore $f(x)-f(x+1)&gt;kf(x)$.  One step forward, it's good to compare $\frac{f(x+1)}{f(x)}$ and a multiple of $\frac{x}{x+1}$ or equivalently to compare $C (x+1)f(x+1)$ and $xf(x)$. But still I can't extend this idea .  I tried more ways but I couldn't find a proper way to solve the problem.  Any hint is appreciated !
23404
If $$\lim_{x\to\infty}(f(x)+f'(x))=l$$ then prove that $$\lim_{x\to\infty}f(x)=l \text{ and } \lim_{x\to\infty}f'(x)=0 $$    I assume four cases $$\begin{array}{c|c|c|c|}   &amp; f(x) &amp; f'(x) \\ \hline \text{1} &amp; \infty &amp; -\infty \\ \hline \text{2} &amp; -\infty &amp; \infty \\ \hline \text{3} &amp; l &amp; 0 \\ \hline \text{4} &amp; 0 &amp; l \\ \hline \end{array}$$ and elimination(1,2,4) of not possible cases can give the answer(3).    My work is not a correct/perfect or flaw proof.What would be a correct one or is this correct.