Given $f:\mathbb{R}\rightarrow\mathbb{R}$, and the function $f$ satisfies $f(x+y)=f(x)+f(y)$ for any $x,y\in S$. Can we say that this function $f$ must be continuous?   I think it is false, but couldn't give an example. Can someone tell me whether it is true or not? Thank you very much.
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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?
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Put precisely, if $f:\mathbb R\to\mathbb R$ is continuous with  $$ \lim_{s\to 0^+}s^{-1}(f(t+s)-f(t-s))=0 \kern20mm (*) $$ for all $t\in\mathbb R$ , is $f$ then constant?  I guess it is but I cannot see the proof. Obviously, if instead of ($*$) we had $\lim_{s\to 0^+}s^{-1}(f(t+s)-f(t))=0$ , i.e. the right-handed derivative zero, then the assertion could be proved indirectly by a simple compactness argument applied to closed intervals $[a,b]$ with $f(a)\not=f(b)$ . However, I cannot see how this argument could be modified to get the above result. Is there possibly a reference to this, or is the result even false?
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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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Given a continuous function $$ f: \mathbb{R} \rightarrow \mathbb{R} \quad \text{with} \quad f(x) = f(x^2) \quad\quad \forall x \in \mathbb{R} $$  How can I show that $f$ must be constant?
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$f(x)$ is continuous on $[a,b]$. Now we define a new function $M(t)$, for every $t\in[a,b]$   $$M(t) = \sup_{a \leq x \leq t} f(x).$$  Prove formally that $M(t)$ is continuous on $[a,b]$. (sup = supremum)   $M(t)$ is monotonically increasing function, but I don't see how to continue from here.   Also I would like to know what are the standard strategy for proving a function is continuous on a given interval (not a point) except showing its a sum/product/composition of another continuous functions.
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I'm wondering if it's possible to have a continuous sequence $f: \mathbb{N} \to \mathbb{R}$? My intuition is telling me no because logically it would be impossible to map the natural numbers onto the real numbers but I'm not sure how to formalize it. What do you think? How would you prove that no function over the natural numbers (no sequence) is continuous?  For example, how would you prove that $f: \mathbb{N} \to \mathbb{R}$ such that $f(n) = n^{2}$ is not continuous over the natural numbers?
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So I had an exam today and one of the questions were: Give an example of a function $f$ which is nowhere continuous but $|f|$ should be continuous at all points. At first I had no idea how to do it then I came up with this, even though I know it's wrong:  $$f(x)= \begin {cases} \ \sqrt{x}&amp;\text{if }x &lt;0\\ \sqrt{-x}\ &amp;\text{if }x &gt; 0\\ \end {cases}$$  Like I said I know it's wrong..but can someone give me an example of this? Thanks.
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I was wondering if there is a name for a function that satisfies the conditions  $f:\mathbb{R} \to \mathbb{R}$ and $f(x+y)=f(x) \cdot f(y)$?  Thanks and regards!
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If $f:\mathbb{R}^+\to\mathbb{R}$ is monotonic and differentiable, does it follow that $f$ is continuously differentiable?  (This question arose from discussion here: )
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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!
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Is a function defined at a single point continuous?  For example $f:\{0\}\to\{0\}$ defined by $f(x)=\sqrt{x}+\sqrt{-x}$ is a sum of two continuous functions and is therefore continuous, however for $f$ to be continuous, it has to have the limit $\lim\limits_{x_0\to x}f(x_0)=f(x)$ but it dosen't seem to have that limit.
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If $ f: \mathbb{R} \to \mathbb{R} $ is continuous, and $\forall x \in \mathbb{R} :\;(f \circ f \circ f)(x) = x  $, show that $ f(x) = x $.  The condition that $f$ is continuous on $\mathbb{R}$ is crucial for the proof. I can find functions such as $\displaystyle\frac{x-3}{x+1}$ that satisfies  $ (f \circ f \circ f)(x) = x $. I have tried to negate the conclusion to see if there's a contradiction, but got stuck.
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Let $f:\mathbb R \to \mathbb R$ be a function such that for any irrational number r, and any real number x we have $f(x)=f(x+r)$. Show that f is a constant function.   It's easy to see any constant function satisfies the original properti... But I don't see how to show this is the only solution.
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Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\mathbb{R}$ and that $f(r)=0$ for all $r \in \mathbb{Q}$. Prove that $f(x)=0$ for all $x \in \mathbb{R}$.   My attempt: Define a sequence $(x_n)$ where $x_n \in \mathbb{Q}$ for all $n \in \mathbb{N}$ and assume that $(x_n) \rightarrow a \not\in \mathbb{Q}$. Since $f$ is continuous, we have $\lim_n{f(x_n)}=f(a)=0$. Since $a$ is arbitrary irrational number, we have $f(a)=0$ for all $a \not\in \mathbb{Q}$. Hence, we proved the statement.  Is my proof valid? or is there any flaw ?
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The question is the following: Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0$, $f'(x)$ exists. If $\lim_{x \rightarrow 0} f'(x) = L$ exists, does it follow that $f'(0)$ exists?  My intuition is that this does not have to hold in general. However, I keep finding counter examples such that $f'(0)$ exists where $\lim_{x \rightarrow 0} f'(x) = L$ does not exist, which is obviously not the original question.
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Possible Duplicate:          Let $f: \mathbb{R} \to \mathbb{R}$ be function satisfying the two conditions: $f(\mathbb{R}\setminus \mathbb{Q}) \subseteq \mathbb{Q}$ and $f(\mathbb{Q}) \subseteq \mathbb{R}\setminus \mathbb{Q}$. Then,     Show that $f$ cannot be continuous.   I'm trying this problem for some time but can't make any useful progress. I will appreciate any help. Even some good hints will do. Regards.
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Suppose $f(x)$ is continuous on the closed interval $[a,b]$. Define $m(x)=\max_{a\leq s\leq x}\, f(s)$, $a\leq x\leq b$. Is $m(x)$ continuous necessarily? Thank you.