I would appreciate if somebody could help me with the following problem:  Q: Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function and $f(x+y)=f(x)f(y)$. Then $f(x)\geq 0$
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Let $g: \mathbf R \to \mathbf R$ be a function which is not identically zero and which satisfies the equation $$ g(x+y)=g(x)g(y) \quad\text{for all } x,y \in \mathbf{R}. $$ Show that $g(x)\gt0$ for all $x \in \mathbf{R}$.
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If $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x)=x$ has no real solution, then show that $f(f(x))=x$ has no real solution either.   Is the proof trivial as it seems or does it need an analytical approach?
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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.
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I tried to solve this problem but I could not continue  Let $f: [a, b​​]\to \mathbb{R}$ be a continuous function.  Prove that if $x_1, x_2,\ldots,x_n  \in (a,b​)$, then exists $x_0 \in (a, b​​)$ such that $$f (x_0) =\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}$$  the function is continuous, so $\forall\ x_1, x_2,\ldots,x_n  \in(a, b​​)$ $$\begin{align*} \lim_{x \to \ x_1}f(x)&amp;=f (x_1)=L_1\\ \lim_{x \to \ x_2}f(x)&amp;=f (x_2)=L_2\\ \lim_{x \to \ x_3}f(x)&amp;=f (x_3)=L_3\\ &amp;\vdots\\ \lim_{x \to \ x_n}f(x)&amp;=f (x_n)=L_n\\ \end{align*}$$  We have that:  $$\lim_{x \to \ x_0}f(x)=f (x_0)=\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}= \frac{1}{n}\sum_{j=1}^n f(x_j)$$ Now $$\min\left \{ L_1,L_2,L_3,\cdots,L_n \right \}\le\frac{1}{n}\sum_{j=1}^n f(x_j)\le\max\left \{ L_1,L_2,L_3,\cdots,L_n \right \}$$  and at this point I'm lost ...
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Is there a counterexample?  Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function  such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta&gt;0$.  Any hints would be appreciated.
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The statement is: $f$ is a real fucntion on $\mathbb R$. Then if $f'(x)=f(x)$ and $f(0)=1$, then $f(x)\neq 0.$
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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?
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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.
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I have really no idea about this:  Problem: Show that there exists a function $f:[0,1]\rightarrow\mathbb{R}$ such that:  $f$ is discontinuous in all $x\in \mathbb Q$. $f$ is increasing in $[0,1]$. $f$ is integrable.   EDIT: Sorry, it is not discontinuous in all $x\in \mathbb R \setminus \mathbb Q$, just in $\mathbb Q$.
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This is question B4 in 1999 Putnam Exam. I would appreciate if somebody could help me with this.      Let $f(x),f'(x),f''(x),f'''(x)&gt;0$ , $f'''(x)$ is a continuous function and $f'''(x)&lt;f(x)$ on $\mathbb{R}$. Then show that   $$f'(x)&lt;2f(x),~ \forall x\in \mathbb{R}.$$
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Suppose $f:[0,\infty)$ $\rightarrow$ R and has continuous $f'(x)$ and $f''(x) $ .  $f(x) $$\rightarrow $0 as $x$ $\rightarrow$ $\infty$. Given $f'(x)\rightarrow b$ as $x \rightarrow \infty$  show that $b=0$  Get stuck in the first place, but I tried to use Mean Value Theorem and then connected it to the limits.  Any hint?
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Do there exist functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y),$ but which aren't linear? I bet you they exist, but I can't think of any examples.  Furthermore, what hypotheses do we need to put on $f$ before no such functions exist? I feel continuity should be enough.
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Prove that every continuous function $f$ can be written $f = g−h$, where   $g$ and $h$ are nonnegative and continuous.   If I define $g=\max\{f,0\}$ and $h=\max\{0,-f\}$, do I need to prove that $g$ and $h$ are still continuous and if so how do I do that?
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Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function. We denote, for every $x\in\mathbb{R}$, with $m(x)$ and $M(x)$ the minimum and maximum of $f$ on the interval $[x-1,x]$. Show that, if $m(x)+M(x)=0,\forall x\in\mathbb{R}$, then the functions $m$ and $M$ are constant.  I have tried to prove that $f$ is periodic, but I didn't managed to do it; I'm not sure it's even true...
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Any help with this problem is appreciated. It comes up in the context of Lusin's theorem, where it was assumed to be true.     Suppose $E$ is a closed subset of $\mathbb{R}$ and $f:E\rightarrow \mathbb{R}$ is continuous. There exists a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x)=g(x)$ for $x \in E$ and $g$ satisfies $$\sup_{x \in \mathbb{R}} |g(x)| \leq \sup_{x \in E} |f(x)|\;.$$
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Given a function on a closed interval $f\colon I\subset \mathbb{R}\to \mathbb{R}$ with $$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 $\lambda \geq 1$.
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Problem: Given $f:[0,1] \rightarrow \mathbb{R}$ ($f$ continuous ) and $f(x^2) = f(x)$ $\forall x \in [0,1]$. Show that function $f$ is a const.