Let $g$ be a function on $\mathbb R$ to $\mathbb R$ which is not identically zero and which satisfies the equation $g(x+y)=g(x)g(y)$ for $x$,$y$ in $\mathbb R$.   $g(0)=1$. If $a=g(1)$,then $a&gt;0$ and $g(r)&gt;a^r$ for all $r$ in $\mathbb Q$.   Show that the function is strictly increasing if $g(1)$ is greater than $1$, constant if $g(1)$ is equal to $1$ or strictly decreasing if $g(1)$ is between zero and one, when $g$ is continuous.
18361
This is part of a homework assignment for a real analysis course taught out of "Baby Rudin."  Just looking for a push in the right direction, not a full-blown solution.  We are to suppose that $f(x)f(y)=f(x+y)$ for all real x and y, and that f is continuous and not zero.  The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that $f(x)=e^{cx}$ where c is a real constant.  I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers.  Is this an approach worth taking?
30995
If $f:[0,1]\to\mathbb R$ is continuous and $\int_0^x f(t)dt = ∫_x^1f(t)dt$ then prove that $f(x)=0$.  I suppose that $f(x)$ isn't zero, set $g(x)= \int_y^xf(t)dt$ with $y\in[0,1]$ and use Bolzano but I don't know if it's right.
89065
suppose that $f$ is a mapping from $\mathbb R^n \to \mathbb R$.  a positively homogeneous of degree $n$ and suppose $f_1, f_2, .... f_N$ are continuous for $a \neq 0$. then   $$\sum_{i=1}^n a_if_i(a)= n*f(a)$$   now, they asked me to prove this. (Which is a total no no to me)  I write my proof as;  $f$ is a mapping from $\mathbb R^n \to \mathbb R$  A homogeneous function of degree $n$ if for all $a \in \mathbb R^n$ and all $L&gt;0$, $f(La)=L^n\,f(a)$. let $a \in \mathbb R^n$ and define function $g$ be a mapping from $[0,\infty) \to \mathbb R$. so I write $g(L) = f(La)-L^nf(a)$. then $L≥0$ let $L=0$ then, $g(L)=0$ and therefore $g'(x) =0$?  then i don't know what's goes next. can anyone help me about this?
46530
Suppose $f$ and $g$ are continuous on $[a,b]$ and that $f(a) &lt; g(a)$ but $f(b) &gt; g(b)$.   Prove that $f(x) = g(x)$ for some $x \in [a,b]$.     I did try and actually did get as far as coming up with $f-g$. Would I set $h(x)= f(x)- g(x)$ and say $h(a) &lt;x&lt; h(b)$?
122184
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a monotone increasing function. We know from a general fact that the set of discontinuities of $f$ is countable. Denote the set of discontinuities by $D$.  Define $g:\mathbb{R}\rightarrow\mathbb{R}$ as follows: $$g(x)=f(x^-)-\sum_{y\in D \text{ and }y&lt;x}(f(y^+)-f(y^-))$$ It should be relatively clear that $g$ is monotone increasing. But is $g$ necessarily continuous?
18326
Let $a, b$ be positive real numbers. Let $0 &lt; x &lt; 1$. Show that  $(a+b)^x &lt; a^x + b^x$.  The function $f(x) = (a+b)^x -a^x - b^x$ has $f(0) = -1$, $f(1) = 0$ and it seems to be an increasing function. But I am unable to show that it is increasing function.
103459
Suppose $f:\mathbb{R}\to\mathbb{R}$ has period 1, and for some $q\in(0,1)$:  $$|f(x)-f(y)|\leq q|x-y|\quad \forall x,y$$  Let $g(x)=x+f(x)$, for any $a\in\mathbb{R}$,define the following sequence:  $$a_1=a,\quad a_2=g(a_1),\quad a_3=g(a_2),\quad \dots,\quad a_{n+1}=g(a_n)$$  Show that  $$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$  exists and does not depend on the choice of $a$.  I don't even know what is the possible approach, any hints? Thanks!
106543
I don't know how to prove this example. Let $y(t):\mathbb{R}^+\to \mathbb{R}$ be a nonincreasing function. Assume that there is a constant $A&gt;0$ such that \begin{equation} \int_{s}^{+\infty}y(t)dt \leq Ay(s), \quad 0\leq s&lt;+\infty \end{equation} Then $y(t)\leq y(0)e^{1-\frac{t}{A}}$, for all $t&gt;0$.
17239
Find all functoins $f: \mathbb R \rightarrow \mathbb R$ such that $\forall x,y \in \mathbb R$ the equality:: $$f(xy+f(x))=xf(y)+f(x)$$  My work so far:  1) $f(0)=0$;  Let $f(a)=f(b)\not=0$  $x=a, y=b \Rightarrow f(ab+f(a))=af(b)+f(a)$;  $x=b, y=a \Rightarrow f(ba+f(b))=bf(a)+f(b)$;  Then $af(b)+f(a)=bf(a)+f(b) \Rightarrow a=b$  $x=0: f(f(0))=f(0) \Rightarrow f(0)=0$  2) $f(x)=x -$ solution. I have tried to prove that $g(x)=f(x)-x = 0 (\forall x \in \mathbb R)$ $$g(xy+x+g(x))=xg(y)$$
89121
Let $g:\mathbb{R}\to\mathbb{R}$ be a measurable function such that    $$g(x+y) =g(x)+g(y).$$   How to prove that $g(x) = cx$ for some $c\in \mathbb{R}?$     The main thing to do here relies upon the fact that such function should be continuous and therefore by natural argument the answer will follow.  Using this    Therefore I found out that there is nothing missing in this question.
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!
89560
I am having difficulties showing that this is true. How would you prove this statement?  Given that the function $h: \mathbb{R}\rightarrow\mathbb{R}$ is increasing. Show that $h$ is continuous if and only if  $h(\mathbb{R})$ is connected.
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.
73521
Let $f$ be defined on $\mathbb{R}$ and suppose that |$f(x)$ - $f(y)$| $\leq$ $(x-y)^2$ $x,y \in\mathbb{R}$. Here I have to show that $f$ is a constant function.  I think I have to show that $f'(x)$ = 0 for all $x$. But I don't know from where to start this. I tried taking it as (|$f(x)$ - $f(y)$|/|$x$-$y$|) $\leq$ |$x$ - $y$|. Am I right in doing so? Any hint or suggestion will be helpful. Thanks
105264
So, I was given the following problem as part of a homework assignment.     Suppose $f'(x) &gt; 0$ in $(a,b)$. Prove that $f$ is strictly increasing in $(a,b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that   $$g'(f(x)) = \frac{1}{f'(x)}$$   I have proven that $f$ is strictly increasing in $(a,b)$, and I could prove that $g'(f(x)) = 1/f'(x)$ if I could prove that $g$ is differentiable. The problem is that I am having trouble with a proof of that. Any advice?   Also, as a reference, this is exercise 5.2 from Baby Rudin.
115513
Suppose $f,g$: $\Bbb R$ → $\Bbb R$ are continuous functions such that $f(r) = g(r)$ for all $r$ in $\Bbb Q$. That is to say, $f$ and $g$ are equal on the rational numbers.   Prove that  $f(x) = g(x)$  for all $x$ in $\Bbb R$
62291
If $f(x)$ is a quadratic expression such that $f(x)&gt;0\;\forall x\in\mathbb{R},$ and if $g(x)=f(x)+f'(x)+f''(x),$  Then prove that $g(x)&gt;0\; \forall \; x\in \mathbb{R}$.  $\bf{My\; Trial \; Solution::}$ If $f(x)&gt;0\;\forall x\in \mathbb{R}$. Then function $f(x)$ has Minimum value .  Let Minimum occur at $x=x_{0}$. Then $f(x)_{Min} = f(x_{0})&gt;0$  Now Given $g(x) = f(x)+f'(x)+f''(x)$. Then $g'(x) = f'(x)+f''(x)+f'''(x).$  Now at $x=x_{0}\;\;,$ Value of $g(x_{0}) = f'(x_{0})+f''(x_{0})+f'''(x_{0}) = 0+f''(x_{0})+f'''(x_{0})$  Now I did not understand How can i solve after that,   Help me   Thanks