Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(x+y)=f(x)+f(y)$. If $f$ is continuous at zero how can I prove that is continuous in $\mathbb{R}$.
22165
The  asks about functions $f \colon \mathbb R \to \mathbb R$ such that $$f(x+y)=f(x)+f(y).$$ It is a very well-known functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for some advanced questions. Solutions of this equation are often called . Also a few other equations related to this equation are often studied. (Equations which can be easily transformed to Cauchy functional equation or can be solved by using similar methods.) Is there some overview of basic facts about Cauchy equation and related functional equations - preferably available online?
61787
Let $f:\mathbb R \to \mathbb R$ be a continuous function such that $f(f(f(x)))=x,\forall x \in \mathbb R$ , then is it true that $f(x)=x , \forall x \in \mathbb R$ ? I can see that $f$ is bijective , hence strictly monotone . But I cannot progress further . Please help. Thanks in advance
64935
How would you prove or disprove that the function given by $$f(x,y) = \begin{cases} \dfrac{x^2y}{x^2 + y^2} &amp; (x,y) \neq (0,0) \\ 0 &amp; (x,y) = (0,0) \end{cases}$$ is continuous at $(0,0$)?
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My exercise says: Let $f:\mathbb{R} \rightarrow \mathbb{R}$ a continuous function e suppose that exists $k$ such that:  $$|f(x)-f(y)|\geq k|x-y|$$  Then $f$ is bijective and  its inverse is continuous.  Well, there's a Theorem , Invariance of domain, that says   "If $U$ is an open subset of $\mathbb {R^n}$ and $f : U \rightarrow\mathbb{R}$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$".  But I'm not knowing how to proceed...need a clue...Thanks for attention!!!
96531
A function $f:D\subset \mathbb R \to \mathbb R$  is lipschitz given that there exists a $L\gt0$ such that $|f(x)-f(y)|\le L|x-y|$  I need to prove this function is then continuous. Is there a best definition of continuous functions to use for this proof given the definition of Lipschitz?  Is the epsilon delta way the only way to do it? can someone help me set it up? I really struggle with implementing that definition.
47071
Let $f:\mathbb{R}\to\mathbb{R}$ be a map sending closed intervals to closed intervals. Prove that $f$ is continuous or find a counter example.  WLOG we just have to prove continuity at $0$ and we can also assume that $f(0)=0$ and $f(1)=1$ (by considering $x\mapsto (f(x)-f(0))/(f(1)-f(0))$ ).  I build up this question while thinking of the question .
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Assume that $f$ be continuous on $\mathbb{R}$, $f'(x)$ exists for all $x\neq 0$, and $\lim_{x\rightarrow 0} f'(x)=1$. We need to show $f'(0)$ exist and is equal to $1$.  $f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$, $\lim_{x\rightarrow 0}f'(x)=1\Rightarrow\lim_{x\rightarrow 0}\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=1$...am I going in the right direction? Please help.
41993
Let $f:\mathbb{R}\to \mathbb{R}$ is a function such that for all real $x$ and $y$, $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$, then prove that $f$ must be one of the two following functions:   $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=0$ for all real $x$   OR   $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x$ for all real $x$   I got to the point where putting the two equations together, you get $f(x+y)f(x)= f(xy) + f(x)^2$ and plugging in $f(x)=x$ checks with it. So am I going in the right direction or am I just doing some guess work? Is there a more elegant way of doing it?   Thanks
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 :)
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I got started recently on proofs about continuity and so on. So to start working with this on $n$-spaces I've selected to prove that every linear function $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous at every $a \in \mathbb{R}^n$. Since I'm just getting started with this kind of proof I just want to know if my proof is okay or if there's any inconsistency. My proof is as follows:  Since $f$ is linear, we know that there's some $k\in \mathbb{R}$ such that $|f(x)|\leq k|x|$ for every $x\in \mathbb{R}^n$, in that case let $a\in \mathbb{R}^n$ and let $\varepsilon &gt;0$. Consider $\delta = \varepsilon /k$ and suppose $|x-a|&lt;\delta$, in that case we have:  $$|f(x)-f(a)|=|f(x-a)|\leq k |x-a|&lt;k \frac{\varepsilon}{k}=\varepsilon$$  And since $|x-a|&lt;\delta$ implies $|f(x)-f(a)|&lt;\varepsilon$ we have that $f$ is continuous at $a \in \mathbb{R}^n$. Since $a$ was arbitrary, $f$ is continous in $\mathbb{R}^n$. Is this proof fine? Or there was something I've missed on the way?
66475
How to prove that a strictly increasing function $f:[a, b]\rightarrow \mathbb{R}$ which has the intermediate value property is continuous on $[a, b]$.
37702
$(i)$ Given a function $f:E \rightarrow \mathbb{R}$ define $|f|:E \rightarrow \mathbb{R}$ by $|f|(x)=|f(x)|$ for $x \in E$ show if $f$ is continuous at $c \in E$ then so is $|f|$.        $(ii)$ Now the converse is not true, find $f$ such that $|f|$ is continuous but $f$ is not.
41637
Can you show me a continuous function $f \colon \mathbb{R}^n\to\mathbb{R}^m$ that satisfies $f(a+b)=f(a)+f(b)$ but is not linear?   We have that  $$f(0)=f(0+0)=2f(0)\implies f(0)=0\\ f(x-x)=f(0)=f(x)+f(-x)=0\implies f(-x)=-f(x)\\ f(nx)=f(x+x+\dots+x)=f(x)+\dots+f(x)=nf(x)\quad \forall n \in \mathbb{N}$$ But $$ f(-nx)=-f(nx)=-nf(x) $$ So: $$ f(ax)=af(x) \quad \forall a \in \mathbb{Z} $$
52466
Let $f:[0,1]\to\mathbb{R}$ be continuous and $f(0)=f(1)$. Show that for any integer $n\geqslant 2$, there is $\xi\in(0,1)$ s.t.$f(\xi)=f\left(\xi+\frac{1}{n}\right).$  I think this requires the intermediate value theorem somewhere.
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Suppose that $f$ is a continuous and real function on $[0,\infty]$. How can we show that if $\lim_{n\rightarrow\infty}(f(na))=0$ for all $a&gt;0$ then $\lim_{x\rightarrow+\infty} f(x)=0$?
40675
The question is from :     Find all continuous functions $f:\mathbb R\to \mathbb R$ such that for any real $x$ and $y$,   $$f(f(x+y))=f(x)+f(y).$$   I'm totally new to functional equations so please correct me if I make a mistake.   I try plugging in simple functions and I find that $f(x)=0$ and $f(x)=x+c$ for some $c \in \mathbb R$ works.  Then I believe the next step is to derive $f(x)=x+c$ as a solution.  Let $y=0$. Then we have  \begin{align} f(f(x))=f(x)+f(0) \end{align}  Then we can make the substitution $u=f(x)$, which produces  $$f(u)=u+c$$  since $f(0)$ is just a constant.   Hence the solutions are $f(x)=0$ and $f(x)=x+c$ for some $c \in \mathbb R$. $\Box$    Is this a valid answer? As some of you may know the AMO is happening in a few days and I would appreciate any help in improving the quality of my answers.
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(a) Let a function $g:I\to\mathbb{R}$ be continuous at $x_0\in I$. Prove that the function $f$ given by $f(x)=(x-x_0)g(x)$ is differentiable at $x_0$.  (b) Let a function $f:I\to\mathbb{R}$ be differentiable at $x_0\in I$ and $f(x_0)=0$. Prove that there exists a function $g:I\to \mathbb{R}$ that is continuous at $x_0 \in I$ such that $f(x)=(x-x_0)g(x)$.    Thought:  (a)  By $Fermat's$, since $g$ is continous, if $g'(x_0)=0$ then $g(x)$ is differentiable?  but how does it connect back with $f(x)$?  (b)  use the definition of every differentiable function is continuous?  Thus $f(x)$ is continuous and also $(x-x_0)$ so $g(x)$ too?  Thank you!!