Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function and   $f(x)=f(2x)$ is true $\forall{x\in{\mathbb{R}}}$. If $f(1)=3$, then   the value of $$\int_{-1}^{1}f(f(x))dx$$ equals.   I don't have any idea how to proceed. Any hints would be helpful.  Thank you.
122584
If $f(x)$ is a continuous function such that $f(x) = f(2x)$ and $f(1) = 3\;,$ Then $\displaystyle \int_{-1}^{1}f(f(x))\,dx$   $\bf{My\; Try::}$ Here $-\infty &lt;x&lt;\infty$ and Given $f(x) = f(2x)$  So Using Recursively $$\displaystyle f(x) = f(2^1x)=f(2^2x)=..........=\lim_{n\rightarrow \infty}f\left(2^{n-1}x\right)$$   OR we can write it as $$f(x) = f\left(\frac{x}{2}\right)=f\left(\frac{x}{2^2}\right)=........=\lim_{n\rightarrow \infty}f\left(\frac{x}{2^{n-1}}\right)$$  Now How can i prove that $f(x)$ is a constant function.  Help me  Thanks
31188
Suppose that $f\colon\mathbb{R}\to\mathbb{R}$ without any further restriction. If $f(f(x))=x^2-x+1$, how can one find $f(0)$?   Thanks in advance.
98516
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous and differentiable function in all $\mathbb{R}$. If $f(0)=0$ and $|f'(x)|\leq |f(x)|$ for all  $x\in\mathbb{R}$, then $f\equiv 0$.  I've been trying to prove this using the Mean Value Theorem, but I can't get to the result. Can someone help?
9297
Let $f : [0, 2] → \mathbb{R}$ be a continuous, positive function and  $$\int_{0}^1 f(x)\,dx = \int_{1}^2 f(x)\,dx= 1$$ For each $x \in [0, 1]$, prove that there exists a unique function $g(x) \in [1, 2]$ such that $$\int_{x}^{g(x)} f(t)\,dt = 1$$ and $g \in C^1$. Well, I think in $g(x) = x +1$, but I'm not sure. Any hint is appreciated.  Thank you.
1790
Let $f$ be a real valued and continuous function over $[a,b]$ with $a &lt;b$ such that $f(a)=f(b)$ , My question here is :     Question (claim): How to show that : there  exists  $c \in [a, \frac{a+b}{2} ]$ such that : $f(c)=f(c+\frac{b-a}{2})$.   Note: I have tried intermediate values but I didn't succeed.
53589
If $f$ is continous in $[a,b] \subset \mathbb{R}$ and exist real numbers $\alpha \not= \beta$ such that $$\alpha\int_{a}^{c} f(x) dx + \beta\int_{c}^{b} f(x) dx  =0$$ for all $c \in [a,b]$ show that $f (x)=0$ for all $x \in [a,b].$
50389
Suppose $f$ is continuous, $f(x) \ge 0$ on $[a,b]$ and $\int_a^b f(x) dx=0$. Show that $f(x)=0$ on $[a,b]$.   I know that if $f$ in continuous at $x_0$ and $f(x_0) &gt; 0$, then there exists $\delta &gt; 0$ s.t. $f(x) \ge \frac{1}{2} f(x_0)$ for $|x-x_0| &lt; \delta$ and I think it should help me but I'm stuck at this point.
116800
I am trying to prove that if I have $f:\mathbb{R} \to \mathbb{R}$ satisfying $\forall x,y\in\mathbb{R},f(x+y) = f(x) + f(y)$. Which is assumed continuous at $0$, that $f$ is continuous on $\mathbb{R}$  I am fairly sure it is, for that property seems to be a property of polynomials, and we know polynomials are continuous where defined(all reals)  Any ideas for rigor?  $|x-a| \lt \delta \implies |f(x+a)-f(x)-f(a)| \lt \epsilon$  No idea where to go from here, this is my first time doing $\epsilon-\delta$ stuff.
101432
I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times.  However, I can't see how to proceed.  I went down what I thought were the typical paths and just reached dead ends.  Does someone maybe have a hint?  I think that with the appropriate nudge I should be able to solve this question.  If not, I'll edit and ask if someone knows how to solve the whole thing.  Here's the question:  Let $f: \mathbf R \to \mathbf R$ be such that $f$, $f'$, $f''$, and $f'''$ exist and are continuous on $\mathbf R$ and satisfying $f(-3)=-1$, $f(0)=0=f'(0)$, and $f(3)=8$.  Prove that there exists $\xi \in (-3, 3)$ such that $f'''(\xi) \ge 1$.
98869
How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$?  I have already recognized one problem caused from this: it  follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?  Best regards and thanks, John
11660
I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:  $$\lim_{|x|\rightarrow \infty} f(x)= 0$$  Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?  Thanks in advance.
87970
Given a continuous function $f:\mathbb{Q}\to\mathbb{Q}$ ,does there exist a continuous function  $g:\mathbb{R}\to\mathbb{R}$ such that   $g|_{\Bbb Q} = f$?  What I have no Idea about how to attempt this Question! Any suggestion will be very helpful.
68668
If $f:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $f(0)=f(1)$, then there exists $ x\in [0,1]$ such that  $f(x) = f(x+\frac{1}{n})$, where $n$ is any natural number.  Let $g(x)=f(x+\frac{1}{n})-f(x)$ on $[0,1-\frac{1}{n}]$. Now, $f$ is continuous function on closed interval.So, it must attain its bounds. Let $m$ be minima at $c_1 \in [0,1]$ and $M$ be maxima at $c_2 \in [0,1]$.  Now if $c_1,c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $g(x)$ on $[c_1,c_2]$.  If $c_1$ or $c_2 \in [0,1-\frac{1}{n}]$, apply IVT to $[c_1,c_1-\frac{1}{n}]$ or  $[c_2,c_2-\frac{1}{n}]$ respectively. If  $c_1,c_2$ both do not belong to $[0,1-\frac{1}{n}]$, then apply IVT to $g$ on $[c_1-\frac{1}{n},c_2-\frac{1}{n}]$.  Is this proof correct? Thanks.
67821
Assume  $f : I \rightarrow \mathbb{R}$ is a non-decreasing on an open interval $I$ and that $f$ satisfies the Intermediate value property or  on $I$ (that is, for any $a &lt; b$ in $I$ and any $L$ between $f(a)$ and $f(b)$, there exists $c \in [a, b]$ such that $ f (c) = L)$. Then, prove that $f$ is continuous.  However, I know that a function can be discontinuous and also satisfy the IVT at the same time. Could someone point me in the right direction?
117310
Let $f_n, f \colon [a,b] \to \mathbb{R}.$ Show that, if $f_n \to f$ uniformly, then the Lebesgue integrals are equal, i.e. $\int f = \lim \int f_n$.  This is clearly true for continuous functions, but how do I handle the case of non-continuous functions?
69661
Let $f(x)$ be continuous on $[a,b]$ and suppose $\int_a^b f(x)g(x)dx = 0$ for every continuous function $g$ on $[a,b]$. Prove that $f(x)=0$ on $[a,b]$.   I understand that $f(x)$ must be zero otherwise the integral can't possibly be zero for all possible $g$. But, I am not sure how to approach this systematically. Any suggestions would be appreciated. Thanks!
51298
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$(x+y)(f(x)-f(y))=(x-y)f(x+y)$$  My attempt:   If $x=-y \not = 0$ then $0= 2x f(0)$ so $f(0)=0$.   Suppose for the sake of contradiction that $f(x)=f(x+\epsilon)$ for some $x$ and $\epsilon&gt;0$. Let $y=x+\epsilon$. Then $$0=\epsilon \cdot f(2x+\epsilon)$$ therefore as $2x+ \epsilon$ can take any real value $f$ is either strictly increasing or strictly decreasing or $f(x)=0 \; \; \forall \; \;x \in \mathbb{R}$.   Note that $f(x)=ax$ is a solution $\forall \; \;a \in \mathbb{R}$. Thanks so much for any help!