Let $f:[0,1]\mapsto\mathbb{R}$ be a continuous function. Evaluate  $\lim_{n\to\infty}n\int_0^1x^nf(x)dx$
19232
Suppose $f:[0,1]\to \mathbb{R}$ is continuous. Show that   $$\lim_{n\to\infty} n\int_0^1 f(x)x^n\,dx = f(1).$$  My answer so far: First I want to assume that $f\in C^1$. Then   $$n\int_0^1f(x)x^n\,dx = \left[\frac{n}{n+1}x^{n+1}f(x)\right]_0^1 - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\\ \frac{n}{n+1}f(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx,$$  which goes to $f(1)$ because the last integral goes to zero.  But approximating $f$ by $\phi\in C^1$ won't necessarily work, because $\phi(1)$ may not equal $f(1)$... how can we finish the argument?
18432
Given that $f:[0,1]\to \Bbb{R}$ is a continuous function, I need to show that $$\lim_{n \to \infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=e^{\int_0^1f(x)dx}$$  Writing $$y=\lim_{n \to \infty}\prod_{k=1}^{n}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)$$ and taking the logarithm of both sides and interchanging limit and logarithm(since $f$ is continuous), I get $$\log y = \lim_{n\to \infty}\sum_{k=1}^{n}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)$$ Now I know that the limit of the Riemann sum $\lim_{n\to \infty}\sum_{k=1}^{n}\frac{b-a}{n}f\left(a+\frac{(b-a)}{n}k\right)=\int_a^bf(x)dx$. However if I want it to read $\int_0^1f(x)dx$ in the given problem, I would need the term inside the summation to read $\frac{1}{n}f\left(\frac{k}{n}\right)$. The way to do this it would seem is to take the series expansion of $\log(1+x)$ and ignoring the $x^2$ and higher order terms. Is it justifiable to do so?
5056
Suppose $f: \mathbb{R} \to \mathbb{R}$ is a continuous function which is 1-1, then does $f$ map Borel sets onto Borel sets?
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!
58485
I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help with the following problem.       Let $f:\mathbb{R}\to[0,\infty)$ be measurable and $f\in L^1$. Show that $F(x)=\int_{-\infty}^x f$ is continuous.   I know is isn't much but the only thing a could think of is that given $x, y \in \mathbb{R}$ with $x &lt; y$ we note that $F(x) \leq F(y)$, i.e. $F$ is increasing. So maybe we can apply one of the convergence theorems of Lebesgue integration theory.    I was also wondering if this problem can be solved using only Riemann integration theory.
107421
Give an example of a continuous function $f : [0,  ∞) \mapsto [0, ∞)$ such that  $\int_{0}^{\infty}f(x)dx$ exists but  $f$ is unbounded.   I have been thinking about this. And I have come to the conclusion that I will need to construct a function, $f$, such that $f$ is a sequence of triangles of increasing height, but decreasing base. I obviously need $f$ such that both the height of the triangles, and sum of the bases tends to infinity. But I also need that the $\sum (\text{height} \times \text{base}) \leq \infty $
23871
Suppose that $f$ is a continuous, non-negative function on the interval $[0,1]$. Let $M$ be the maximum of $f$ on the interval. Prove that $$\lim_{n \to \infty}\bigg[\int_0^1f(t)^n \text{dt}\bigg]^{1/n}=M$$  We wrote out some simple examples to show it worked for functions such as $x^2$. We are having trouble finding how to create a general proof. Thanks for any help!
88196
Let $f$ be a continuous function on the unit interval $[0,1]$. Show that $$\lim_{n\to\infty}\int_{0}^{1}\cdots\int_0^1\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)$$  This problem is from Selected Problems in Real Analysis. But the author doesn't include a solution. Maybe there is a method for this sort of problem?  Maybe we can use this: $$\int_0^1\cdots\int_{0}^{1}\int_{0}^{1}\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}?$$  I also found a similar problem in .
37755
Assume that $f: [0,1] \rightarrow \mathbb{R}$ has a continuous $f''$ and $f$ is positive on the interval $(0,1)$ and $0$ at the endpoints.    I want to prove that $$\int_0^1 \Big| \frac{f''(x)}{f(x)} \Big| \,dx &gt;4. $$  I could not make any progress. Since the unit interval is compact, we know that the continuity of $f, f'$ and $f''$ are uniform and they attain their maximum and minimum values. Moreover, I tried to apply the Mean Value Theorem but I failed.
23268
Let $f\colon(0,\infty)\to \mathbb R$ be a continuous function such that $f(x)\le f(nx) , \forall n \in \mathbb N , \forall x &gt;0$ , then is it true that $\lim_{x\to \infty} f(x)$ exists (may be infinite) ?
61825
Suppose $f$ is continuous on $[0,1]$ and   $$ \int\limits_0^1 f(x) x^n d x = 0 $$   for all $n=0,1,2,...$. Then, $f(x) = 0$  I was trying to use by parts  $$ 0 = f(x) \frac{x^{n+1}}{n+1} \bigg|_0^1 - \int\limits_0^1 \frac{x^{n+1}}{n+1} f(x) dx $$  Thus,  $$ \frac{f(1)}{n+1} = \int\limits_0^1 \frac{ x^{n+1} f(x) dx }{n+1} \implies f(1) = \int\limits_0^1 x^{n+1} f(x) dx $$  but here I am stuck. Maybe my idea is not the best one? How would you approach this problem?
23306
Let $f : \mathbb{R} \to \mathbb{R}$ be a function which is differentiable everywhere on $\mathbb{R}$. If we know that  $$\lim_{x\to-\infty}f(x) = L = \lim_{x\to\infty}f(x)$$ How would one go about showing that there must exist $c \in \mathbb{R}$, such that $f'(c) = 0$ for all cases at once?
2876
Let $f:[0,+\infty)\to\Bbb R$ be a function bounded on each finite interval. I want to show that if $\lim\limits_{x\to+\infty}[f(x+1)-f(x)]= L,$ then also $\lim\limits_{x\to+\infty}\dfrac{f(x)}x = L$
18467
Let $f(x)$ be a continuous real-valued function on $[a,b]$ and $M=\max\{|f(x)| \; :\; x \in [a,b]\}$. Is it true that: $$ M= \lim_{n\to\infty}\left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n} ? $$  Thanks!
81925
Question: Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=0$ and $$\int_0^1f(x)dx=0.$$ Prove that $\exists a, b\in(0,1)$ such that $$\int_0^{a} xf(x)dx=0\text{ and }\int_0^bxf(x)dx=\frac{b^2f(b)}{2}.$$  My approach: Let $g:[0,1]\to\mathbb{R}$ be such that $$g(x)=x\int_0^xf(t)dt-\int_0^xtf(t)dt, \forall x\in[0,1].$$ By the first fundamental theorem of calculus we can conclude that $g$ is differentiable on $[0,1]$ and $$g'(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$ Also, observe that $g(0)=0$ and $g(1)=-\int_0^1tf(t)dt$. Thus, by applying MVT to the function $g$ on the interval $[0,1]$, we can conclude that, $\exists c\in(0,1)$ such that $$g'(c)=\int_0^cf(t)dt=-\int_0^1tf(t)dt.$$ Observe that clearly three cases are possible, i.e, either$$\int_0^cf(t)dt&lt;0\text{ or }\int_0^cf(t)dt=0\text{ or }\int_0^cf(t)dt&gt;0.$$ Now let $h:[0,1]\to\mathbb{R}$ be such that $$h(x)=\int_0^xtf(t)dt, \forall x\in[0,1].$$ Please note that the part highlighted below is wrong, but still I have included it to just demonstrate my thinking process, as it might be of some help to others trying out this problem.  Observe that if $\int_0^cf(t)dt&lt;0$, then, $h(1)&gt;0$. This also implies that $\exists$ an open interval $(d,e)\in[0,c]$, such that $f(t)&lt;0, \forall t\in(d,e)$. Now select any point $c_1\in(d,e)$. Applying MVT to the function $h$ on the interval $[0,c_1]$, we can conclude that $\exists c_2\in(0,c_1)$ such that $$h'(c_2).c_1=f(c_2).c_2.c_1=h(c_1)-h(0)=h(c_1)&lt;0.$$ Now $h(c_1)&lt;0$ and $h(1)&gt;0$. Thus, by IVT we can conclude that $\exists a\in(c_1,1)\subseteq(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ A similar reasoning for the case when $\int_0^cf(t)dt&gt;0,$ shows that $\exists a\in(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ Now finally if $\int_0^cf(t)dt=0$, then we will have $h(1)=0$. Now if $f$ is identically equal to $0$ on $[0,c]$, then clearly $tf(t)=0, \forall t\in[0,c]\implies h(x)=0, \forall x\in[0,c].$ Thus choosing any point $x\in(0,c]$ and setting it as $a$, we will have $h(a)=0$ and we will be done in that case. Now if $f$ acquires both positive and negative values on $[0,c]$, then we can conclude that $\exists c_1,c_2\in(0,c)$, such that $f(c_1)&gt;0$ and $f(c_2)&lt;0$. Also, let us assume WLOG that $c_2&gt;c_1$.  I have not been able to make any significant approach other than this. Can someone help me out with this problem? Please note that a solution using integration by parts might not be possible, since $f$ is not a differentiable function.
109939
Let $f$ be a continuous function. I wish to show $$\lim_{n \rightarrow \infty} \int_0^1 n x^n f(x) \; dx = f(1)$$ I can try to split up the integral over intervals $[0, 1-\delta]$, $[1-\delta, 1]$. The integral over $[0, 1-\delta]$ vanishes as $n \rightarrow \infty$, but I am having difficulty estimating the integral over $[1-\delta, 1]$. Thanks!
2654
If $f,g : [a,b] \to (0,\infty)$ are continuous functions, prove that there exists $c \in (a,b)$ such that: $$\int_{a}^{c} f(x) \ \mathrm dx = \int_{c}^{b} g(x) \ \mathrm dx$$