I'm struggling to solve this problem: Let $f$ be a continuous function defined on $[0,+\infty[$, derivable in every point such that $$ f'(x)=2f(2x)-f(x), \ \ \ \ \ \ \forall x&gt;0, $$ and $$ M_n=\int_{0}^{\infty}x^nf(x)dx&lt;+\infty, \ \ \ \ \ \ \forall n\in\Bbb{N}. $$ Prova that exists a non zero function that satisfies this conditions. For which sequences of real numbers $\{a_n\}$ it's true that $a_n=M_n, \ \ \forall n\in\Bbb{N}$.   All I managed to get is that $$ M_n=n!\prod_{i=1}^{n}\frac{2^i}{2^i-1}M_0=n!\prod_{i=1}^{n}\frac{2^i}{2^i-1}\int_{0}^{\infty}f(x)dx. $$
121887
Show that there is at least a nonzero function $f$, differentiable on $[0,+\infty)$, satisfying $$f'(x)=2f(2x)-f(x) \qquad \forall x&gt;0 $$   $$M_n:=\int_{0}^{\infty}x^nf(x)dx&lt;\infty \qquad \forall n\in  \mathbb{N} $$   My best idea so far is to assume that the solution is a power series, i.e.  $$ f(x)=\sum_{n=0}^{\infty}a_nx^n\qquad \forall x&gt;0$$ Then the equation becomes $$ \sum_{n=0}^{\infty}na_nx^{n-1}=2\sum_{n=0}^{\infty}a_n2^nx^n-\sum_{n=0}^{\infty}a_nx^n$$ equating all the coefficients of the same degree I get $$na_n=(2^{n}-1)a_{n-1}\qquad \forall n\geq 1$$ So setting $a_0=1$, I get $$a_{n}=\frac{1}{n!}\prod_{k=1}^{n}(2^k-1) \qquad \forall n$$ But does the power series actually converge? Using Hadamard's formula, and that $2^{k}-1\geq 2^{k-1}$, $$ |a_n|^{1/n}\geq\frac{1}{(n!)^{1/n}}\left[2^{n(n-1)/2}\right]^{1/n}\sim\frac{e}{n(2\pi n)^{1/2n}}2^{(n-1)/2}\to \infty$$ so the radius of converge of the series is $0$, so it doesn't actually define a solution on $[0,+\infty)$.
2723
Question: Suppose $f: [0,1] \rightarrow \mathbb{R}$ is continuous on $(0,1)$. Prove there is a sequence of step functions $\left\{f_{n}\right\}$ which converges pointwise to $f$ on $[0,1]$.    I saw the same question was asked  but I was still confused regarding the solution to the question.   I had attempted a solution but I now realize this solution is wrong:  Proof: Since $f$ is continuous on $(0,1)$, then for every $x \in (0,1)$ and for every $\epsilon&gt;0$, there exists a $\delta(x)$ such that $|f(x)-f(y)|&lt;\epsilon$ whenever $y \in (0,1)$ and $|y-x|&lt;\delta(x)$. In particular, let $\epsilon_{n} = 1/2^{n}$ and let $\delta_{n}(x)$ be the corresponding $\delta$ which is guaranteed by the continuity of $f$. Fix a value of $n$ and choose a partition $x_{0}=0 &lt; x_{1} &lt; x_{2} &lt; \ldots &lt; x_{m} = 1$ such that $x_{i}-x_{i-1} &lt; \delta_{n}(x)$. Let $f_{n}(x):= f(x_{i-1})$ for all $x \in (x_{i-1}, x_{i})$. Clearly $f_{n}(x)$ is a step function and hence $\left\{f_{n}\right\}$ is a sequence of step functions. Observe that $f_{n}(x)=f_{n}(x_{i-1})=f(x_{i-1})$. Finally, by the continuity of $f$ and by the fact that $x_{i}-x_{i-1} &lt; \delta_{n}(x)$, we may conclude that \begin{align*} |f(x)-f_{n}(x)| \leq |f(x)-f(x_{i-1})| &lt; \epsilon_{n}  \end{align*} and so the sequence converges pointwise to $f$.  End of proof.  I was hoping someone can shed some light on the correct solution to this problem. Any hints would be appreciated. Thanks!
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.
76724
I have proved that if $f \in R[a,b]$ and given $\epsilon &gt; 0$ there exists a continuous function $g$ such that $$\int_a^b |f-g| &lt; \epsilon$$ I was wondering if using this fact there is some way to show that there is also some continuous function $h$ such that $$\int_a^b |f-h|^2 &lt; \epsilon$$ Any help will be appreciated, thanks :)
97927
Prove that the function : $f(x) = x$ when $x$ is rational and $f(x) = 1 - x$ when $x$ is irrational is continuous at $x = \frac{1}{2}$  To prove the continuity it is necessary to prove that for every $\epsilon$ there exists a $\delta$ such that $|f(x) - f(\frac{1}{2})| &lt; \epsilon$ whenever $|x -c| &lt; \delta$ $- (1)$ Now if we choose $a  € (\frac{1}{2} - \delta, \frac{1}{2} + \delta)$ and also $a $ is rational and choose $\epsilon &lt; |a - \frac{1}{2}|$ then condition in $(1)$ is not satisfied. How to move from here?
61445
I was trying to prove an important theorem concerning continuous functions, namely that the product of two continuous functions is continuous. I am nearly at the end of the proof but I do not understand the last step my teacher made... I'm sure some of you is able to help me out :) Thanks in advance!  Theorem  Let $f, g\colon \Bbb R \to\Bbb R $ be continuous. Then $F\colon \Bbb R\to\Bbb R$ defined by $F(x) = f(x)g(x)$ is continuous.  Proof  Let $f, g\colon \Bbb R \to\Bbb R $ be given such that $f$ and $g$ are continuous.  Let $F\colon \Bbb R\to\Bbb R$ be defined by $F(x) = f(x)g(x)$. We want to show: $F$ is continuous, that is, for all $a \in\Bbb R$, for every $\epsilon &gt; 0$, there exists some $\delta &gt; 0$ such that for all $x \in \Bbb R$ with $|x - a| &lt; \delta$, $|F(x)-F(a)| &lt; \epsilon$.  Let $a \in\Bbb R$ be given. Let $\epsilon &gt; 0$ be given.   Let $\delta_f &gt; 0$ be such that for all $x \in\Bbb R$ with $|x-a| &lt; \delta_f$, $|f(x)-f(a)| &lt; \frac\epsilon{2(|g(a)| + \epsilon)}$. Such a $\delta_f$ exists since $f$ is continuous.  Let $\delta_g &gt; 0$ be such that for all $x \in\Bbb R$ with $|x-a| &lt; \delta_g$, $|g(x)-g(a)| &lt; \frac\epsilon{ 2|f(a)| + 1}$. Such a $\delta_g$ exists since $g$ is continuous.  Let $\delta_3 &gt; 0$ be such that for all $x\in\Bbb R$ with $|x-a| &lt; \delta_3$, $|g(x)-g(a)| &lt; \epsilon$, which implies $|g(x)| &lt; |g(a)| + \epsilon$. Such a $\delta_3$ exists since $g$ is continuous.  Choose $\delta = \min\{\delta_f, \delta_g, \delta_3\}$.  Then for all $x \in\Bbb R$ with $|x-a| &lt; \delta$:  $$\begin{align}|F(x) - F(a)| &amp;= |f(x)g(x) - f(a)g(a)| \\ &amp;= |f(x)g(x) - f(a)g(x) + f(a)g(x) - f(a)g(a)| \\ &amp;\le |f(x)g(x) - f(a)g(x)| + |f(a)g(x) - f(a)g(a)|\\ &amp;= |(g(x)[f(x) - f(a)]| + |f(a)[g(x) - g(a)]|\\&amp;= |g(x)| * |f(x)-f(a)| + |f(a)| * |g(x) - g(a)| \\ &amp;&lt; [|g(x)| + \epsilon] \cdot \frac \epsilon {2(|g(a)| + \epsilon)} + |f(a)| \cdot |g(x) - g(a)| \\ &amp;= \frac\epsilon{2 + |f(a)| \cdot |g(x) - g(a)|}\end{align}$$  Problem  Now we need $|f(a)| \cdot |g(x) - g(a)|$ to be equal to $\frac \epsilon 2$ so that we have $\frac\epsilon2+\frac\epsilon 2= \epsilon$ (which completes the proof). The problem is I do not see why/how $|f(a)| \cdot |g(x) - g(a)|$ is equal to $\frac\epsilon2$. Or is there maybe an error somewhere?
61624
Given a function on a closed interval $f\colon I\subset \mathbb{R}\to \mathbb{R}$ with $$f(x+y) \leq f(x) + f(y).$$ Moreover, I know that $f$ is    monotonic increasing continuous on all points except countably many continuous from the right such that the limit from the left at each point exists. $f(nx) \leq nf(x)$ for all $n\in\mathbb{N}$.   Can I say more on this function? In particular, I would like to show that $f(\lambda x) \leq \lambda f(x)$ for all $\lambda \geq 1$ or at least for all rational $\lambda \geq 1$.
114512
One of my friend gave the following problem to me but I am unable to solve it even after trying it for nearly two days.     Problem. Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=f(1)=0$. If $f$ is differentiable on $(0,1)$ then prove that there exists $c\in(0,1)$ such that $f'(c)=f(c)$.   Can anyone help me solving this problem?
63070
$\mathbf{Question}:$ Let $f$ be a continuous function on $[0,1]$. Then prove that the limit $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx$ is equal to $f(1)$.   $\mathbf{Attempt}$: First, we try to show that the sequence of functions $\{nx^nf(x)\}_{x\in [0,1]}$ is uniformly convergent to $0$ on the restricted domain $[0,1-\epsilon]$, $0&lt;\epsilon&lt;1$.   Let $\sup_{x\in[0,1]}f(x)= \mathcal{M}$. Then $|{nx^nf(x)}|\leq n(1-\epsilon)^n|\mathcal{M}|$ for any $x$ and $n(1-\epsilon)^n \to 0$ as $n\to \infty$.  Thereby, $\lim_{n \to \infty}\int_0^{1-\epsilon}nx^nf(x)dx=\int_0^{1-\epsilon}{\lim_{n \to \infty} }nx^nf(x)dx=0$  Now, denote $\sup_{x\in [1-\epsilon,1]} f(x)=M(\epsilon)$ and $\inf_{x\in[1-\epsilon,1]}f(x)=m(\epsilon)$.  $\int_{1-\epsilon}^1nx^n\ m(\epsilon)dx\leq\int_{1-\epsilon}^{1}nx^nf(x)dx\leq \int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx$.  Now,  $\lim_{n \to \infty}\int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx =\displaystyle \lim_{n\to\infty}\bigg[ \frac{nx^{n+1}}{n+1}M(\epsilon)\bigg]_{1-\epsilon}^1=M(\epsilon)$ [Similarly the other one is $m(\epsilon)$]  Clearly, as $\epsilon \to 0 , \ \ M(\epsilon), m(\epsilon) \to f(1)$, so $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx=f(1)$.  Is the procedure correct? Kindly verify.
87288
The question is: Prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x) = 0$ whenever $x$ is rational, then $f(x) = 0$ for all $x\in\mathbb{R}$.  My proof:  Let $ x\in\mathbb{R} $. If $ x\in\mathbb{Q}$, then $ f(x)=0 $.  Otherwise $ x\in\mathbb{R\backslash Q} $. Then there exists a sequence of rational numbers $ (q_n)_{n\in\mathbb{N}}$ (for example $q_n= \frac{\lfloor{nx\rfloor}}{n} ,\forall n\in\mathbb{N}$) such that $q_n\to x$ as $n\to \infty$. By the properties of $f$, we have $f(q_n)=0$ for all $n\in\mathbb{N}$. So $f(q_n)\to 0$ as $n\to \infty$. Since $f$ is continuous, $q_n\to x$ implies that $f(q_n)\to f(x)=0$. So $f(x) = 0$ for all $x\in\mathbb{R}$.   Thank you for any help.
60427
Prove that if the real-valued function $f$ on the interval $[a,b]$ is bounded and is continuous except at a finite number of points then $\int^1_0f(x)dx$ exists.   I know that I can break up the interval $[a,b]$ into subintervals where each subinterval has one discontinuity point. Then if $f$ is integrable on each subinterval then it is integrable on $[a,b]$. So if I let $c$ be my point of discontinuity in the subinterval $[a_1,b_1]\subset [a,b]$, where $c\in[a_1,b_1]$. Then how can I show this subinterval is integrable?
103327
Let $f:(0,\infty)\to \Bbb R$ be continuous and such that for each $x&gt;0$ the sequence $\{f(\frac{x}{n})\}\to 0$.     Does it imply that $\lim_{x\to 0^+} f(x)=0$?   My try:Fix $x&gt;0$.Since $f$ is continuous then $\lim _{n\to \infty}f(\frac{x}{n})=f(\lim_{n\to \infty} \frac{x}{n})=f(0)\implies f(0)=0$  Since $f$ is continuous and $f(0)=0\implies \lim _{x\to 0^+} f(x)=0$  Is the solution correct?In the book it is  given to use the Baire Category Theorem.
39788
If a Continuous function $f(x)$ satisfies $f(x) = f(2x)$, for all real $x$, then does $f(x)$ necessarily have to be constant function? If so, how do you prove it? If not any counter examples?
69015
I'm trying to find an example of a continuous function $f:[1,\infty) \to \Bbb R $ such that  $$f(x) &gt;0 $$  $$\int_{1}^{\infty} f(x) \ dx \ \text{converges and } \int_{1} ^{\infty} f(x)^2 dx \ \text{diverges}.$$
96450
Just wanted to confirm that this is a correct solution:    Proof: Suppose $f(x_0) &gt; 0$ for some $x_0 \in [a,b]$. Then, by continuity of $f$, for $\epsilon &lt; f(x_0)$, there exists $\delta &gt; 0$ so that, if $|x-x_0| &lt; \delta$, we have $|f(x) - f(x_0)| &lt; \epsilon$. That is to say, $f(x)$ is non-zero in a neighborhood $B_\delta (x_0)$. Now,   \begin{eqnarray*} U(f,P) &amp;=&amp; \sum_{P([a,b])} M_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} M_i \Delta_i \\ &amp;=&amp; \sup_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \sup_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \sup_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;&gt;&amp; 0 \end{eqnarray*}  In the same manner,   \begin{eqnarray*} L(f,P) &amp;=&amp; \sum_{P([a,b])} m_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} m_i \Delta_i \\ &amp;=&amp; \inf_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \inf_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \inf_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;\geq&amp; 0 \end{eqnarray*}    We can conclude from here that $\int_{a}^{b} f(x) dx &gt; 0$.  Is this a valid proof?
25113
Suppose $f$ is a continuous function at $x = c$ in $[a,b]$.  Prove that for any sequence ${x_n}$ in $[a,b]$ converting to $c$, the sequence $\{f(x_n)\}$ converges to $f(c)$.  That is, $$ \lim_{n\to\infty}f(x_n)= f\left(\lim_{n\to\infty}x_n\right)$$    This proof seems simple but there are a few things that I need to know first. If $\{x_n\}$ converges to $c$, is it sufficient to substitute $c$ in for $\lim_{n\to\infty}x_n$?  Also needing some guidance on the structure of this proof.  Thanks!
502
I want to show that if $f:(a,b)\to \mathbb R$ is a strictly monotone function and if $x_0\in (a,b)$ such that there exist two sequences $(a_n)$ and $(b_n)$ with $a_n&lt;x_0&lt;b_n$ and $\lim\limits_{n\to \infty}(f(b_n)-f(a_n))=0$, then $f$ is continuous at $x_0$.   By monotonicity we can say that $\lim\limits_{n\to \infty}f(a_n)=f(x_0)=\lim\limits_{n\to \infty}f(b_n)$. Let $(x_n)$ be a sequence in $(a,b)$ such that $x_n\to x$. Here I got stuck. How to show that $f(x_n)\to f(x_0)$? Please help!