Is there a function from $ \Bbb R^3 \to \Bbb R^3$ such that $$f(x + y) = f(x) + f(y)$$ but not $$f(cx) = cf(x)$$ for some scalar $c$?  Is there one such function even in one dimension? I so, what is it? If not, why?  I came across a function from $\Bbb R^3$ to $\Bbb R^3$ such that $$f(cx) = cf(x)$$ but not $$f(x + y) = f(x) + f(y)$$, and I was wondering whether there is one with converse.  Although there is another post titled Overview of the Basic Facts of Cauchy valued functions, I do not understand it. If someone can explain in simplest terms the function that satisfy my question and why, that would be great.
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?
55720
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.
4795
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f\left( f(x)^2+f(y) \right)=xf(x)+y$ for all real numbers $x$ and $y$.   Clearly $f(x)=x$ is a solution, check by substitution.  I'm at a loss as to how to show this is the only one or find others.  Any help would be very much appreciated.
7061
What are some good examples of a function  $f : \mathbb{R} \to \mathbb{R}$ where its derivative is equal to its inverse? I attempted to find a monomial that satisfied it by starting with $f(x) = ax^b$ and showing that $f^{-1}(x) = f'(x) \implies b-1=\frac{1}{b} \implies b=\phi$ and got $$f(x) = \frac{x^\phi}{\sqrt[\phi]{\phi}}$$ Which seems to work according to WolframAlpha, but I'm having trouble double-checking it. Any other ideas?
85996
To find all functions $f$ which is a real function from $\Bbb R \to \Bbb R$ satisfying the relation  $$f(x^2 + yf(x)) = xf(x+y)$$  It can be easily seen that the identity function $i.e.$ $f(x)=x$ and $f(x)=0$ (verified just now) satisfies the above relation!! And putting $y=0$ I have got $f(x^2)=xf(x)$.  Help needed to find other functions satisfying the relation.
1226
$f:\mathbb{R}^3\to\mathbb{R}, f(rx)=r^{\alpha}f(x)$ for some $\alpha&gt;0$, $\forall x\in\mathbb{R}$, for any $r\in\mathbb{R}$, could any one tell me which of the following is true?  $1.$ If $f(x)=f(y)$ whenever $||x||=||y||=\beta&gt;0$, then $f(x)=\beta ||x||^{\alpha}$  $2.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)= ||x||^{\alpha}$  $3.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)=c||x||^{\alpha}$ for some constant $c$  $4.$ If $f(x)=f(y)$ whenever $||x||=||y||$, then $f(x)$ must be a constant function. Thank you.
73757
Let $\Bbb R $ be the set of real numbers .Determine all functions $f:\Bbb R \rightarrow \Bbb R $ satisfying the equation   $$f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$$ for all real numbers $x$ and $y$.   My work  $$f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$$ Setting $y=0$ $$f(x + f(x)) + f(0) = x + f(x )$$ $$\implies f(x + f(x))  = x + f(x ) -f(0)$$ $$\implies f(t)  = t -f(0)$$ Setting $x=0$ $$f(f(y)) + f(0) = 0 + f(y) +yf(0)$$ $$f(f(y)) =  f(y) +f(0)(y-1)$$  Setting $y=1$ $$f(x + f(x + 1))  = x + f(x + 1)$$
51565
Find all continuous functions $f:\mathbb R^{+}\to\mathbb R^{+}$ such that $f(x)^2=f(x^2)$ for all positive reals $x$.  If $f$ is a constant function, then $f(x)=1$. If $f$ is non-constant, I'm suspecting the only solutions are of the form $f(x)=x^k$, where $k$ is a constant, but I have no idea how to prove this. Thanks in advance.
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
64559
My lecturer was talking today (in the context of probability, more specifically Kolmogorov's axioms) about the additive property of functions, namely that:  $$f(x+y) = f(x) + f(y)$$  I've been trying to find what functions satisfy this. Intuition says that, for functions over $\mathbb{R}$, the only functions should be of the form $f(x) = ax$ for some real a. Unfortunately I've only shown this is true when the domain of the function is the rational multiples of a given real number.   My question is if it is possible to extend this result (that $f(x) = ax$ given additivity) to the real numbers, possibly without assuming the continuity of f. It seems to me that additivity introduces so many constrains on a function that nothing but the trivial case would be able to sneak through. The following is a summary of my thoughts to date, though they're obviously long and not 'compulsory reading'. :)  When x is rational - Preliminary Investigation  It is not hard to see that:  $$f(x + x + x) = 3f(x)$$  and more generally, for $a\in \mathbb{N}$,  $$f(ax) = af(x)$$ It is not too hard to prove (well, it took half a bus trip ... ) that this also applies first for $a\in \mathbb{Z}$ and then for $a\in \mathbb{Q}$, (for the latter you just need to consider $a=m/n$ and then note that:  $$f\Big(\frac{m}{n}x\Big)=mf\Big(\frac{x}{n}\Big)=\frac{m}{n}\cdot nf\Big(\frac{x}{n}\Big)=\frac{m}{n}\cdot f\Big(n\frac{x}{n}\Big)=\frac{m}{n}\cdot f(x)$$   The reason this little equation is cool is that we can set $x = 1$ and get:  $$f(a)=a\cdot f(1)$$  which is equivalent to what was expected intuitively, namely (after changing $a$ to $y$ and $f(1)$ to $a$)  $$f(y) = a\cdot y$$  as long as y is rational   $y$ is a rational multiple of a real number  But we can do a bit better than that. If we substitute in $x = \sqrt{2}$ or any other real number in $f(ax) = af(x)$ (which we know for rational $a$), you can conduct the exact same argument above and show that, for instance   $$f(y) =  \Big(\frac{f(\sqrt{2})}{\sqrt{2}}\Big)\cdot y=a\cdot y$$  Whenever $y = \frac{m}{n}\sqrt{2}$ i.e. whenever $y$ is a rational multiple of $\sqrt{2}$. Note however, that the value of the coefficient $a$ (i.e. the slope of the line) is apparently completely unrelated to the value taken in the case where $y$ is purely rational.   What I'm actually asking  We still haven't shown that $$f(x) = ax$$ for all $x \in \mathbb{R}$, as the slope of the line may change depending on what real number we are taking rational multiples of. As far as I've shown now, we might have $f(x) = x$ when $x$ is rational, $f(x) = 3x$ when $x$ is a rational multiple of $\sqrt{2}$, etc.  I still feel that $f(x) = ax$ for all $x \in \mathbb{R}$. One reason for thinking this comes from noting that $$f(2) = f(2-\sqrt{2})+f(\sqrt{2})$$  $2$, $2-\sqrt{2}$ and $\sqrt{2}$ are not rational multiples of each other, however the equation above gives a restraint on the slopes of the lines formed by their rational multiples (which we'll call $a_1, a_2$ and $a_3$ for the slopes on the rational multiples of $2, 2-\sqrt{2}$ and $\sqrt{2}$ respectively). We have $2a_1 = (2-\sqrt{2}) a_2 + \sqrt{2} a_3$  There's so many constraints here - all the rational multipes have the same coefficient, whenever 2 (or more) numbers which aren't rational multiples of each other are added together we get another constraint on their coefficients. The trivial solution is just that$$f(x) = ax$$  over $x \in \mathbb{R}$ and I really struggle to see how any other solution could possible squeeze through all these constraints.   Is there an additive function on $\mathbb{R}$ not of the form $f(x) = ax$?
31554
Let $n&gt;1$ , consider $\mathbb R^n$ with standard volume form $\omega _0 =dx_1 \wedge ...\wedge dx_n$ ; let $r,R$ be two positive real numbers with $R&gt;r$ . Then is it true that for every $a,b \in B(0;r)$ , $\exists$ a diffeomorphism $f:\mathbb R^n \to \mathbb R^n $ such that $f(a)=b ; f(x)=x , \forall x \in \mathbb R^n $ with $||x||&gt;R$ and $f^*\omega _0=\omega _0$ ?   As $f^*(dx_1 \wedge ...\wedge dx_n)=(det f' )(dx_1 \wedge ...\wedge dx_n)$ , so basically given $R&gt;r&gt;0$ and $||a||,||b||&lt;r$ , we have to find a diffeomorphism $f : \mathbb R^n \to \mathbb R^n  $ with $f(a)=b$ ; $f|_{\mathbb R^n \setminus B[0;R]}=Id$ and $det f'(p)=1 , \forall p \in \mathbb R^n $ .  Definitely this cannot work out in $\mathbb R$ , because the only differentiable function satisfying $det f'(p)=1 , \forall p$ and $f(a)=b$ is $f(x)=x+b-a$ and this function never gives identity unless $a =b$
31366
Find all functions $f\colon \mathbb R\rightarrow \mathbb R$ such that for all reals $x$ and $y$:   $$f(f(x)f(y))+f(x+y)=f(xy).$$   It was six hours ago in IMO 2017 (problem 2).  I tried  the standard way: $x=0$, $x=y$, $x=1$,etc. but without any success.   Let $f(0)=a$. Hence, for all $x\in\mathbb R$ we have $f(a(f(x))+f(x)=a$. What is the rest?   Also we have $(x-1)(y-1)-1+x+y-1=xy-1$ and  $1-(1-x)(1-y)+1-x-y=1-xy$.
45702
Be $f:\mathbb R^ +\mapsto\mathbb R$ a function that satisfies the following conditions:  a)$ f(f(f(x)))+2x=f(3x)$ for every $x\gt 0$;  b)  $\lim_{x \to \infty} (f(x)-x)=0$.  This was proposed by Gabriel Dospinescu  at a romanian competitions about a decade ago. No idea how to approach it. Thank you!
28500
Is there a simple way to prove this fact :      For all holomorphic functions $f : \mathbb C \to \mathbb C$, there is an holomorphic function $\psi : \mathbb C \to \mathbb C$ such that $$\psi(z+1) = \psi(z) + f(z) $$   The solution I know use Galois covering spaces and summand of automorphy.  Thanks in advance.
107853
Assuming $x,y\in\mathbb C^n$,consider:  $$f(x,y)=\sup_{θ,φ}\{||e^{i θ }x+ e^{i φ }y||^2: θ,φ\in\mathbb R\}$$  Which of the following is/are correct? 1.$ f(x,y)$≤$||x||^2+||y||^2+2|(x,y)|$ 2. $f(x,y)$= $||x||^2+||y||^2+2Re(x,y)$ 3. $f(x,y)$= $||x||^2+||y||^2+2|(x,y)|$ 4. $f(x,y)$> $||x||^2+||y||^2+2|(x,y)|$    How can I solve this problem ,I am completely stuck on it . can anyone help me?thanks
44546
Consider the normed linear space $\Bbb R^2$ with this norm $\|(x,y)\|=|x|+|y|$. Define $f$ to be a bounded linear functional on $X=\{(x,y):x=y\}$ by $f(x,y)=3x$. If $g$ is a Hahn-Banach extension of $f$ on $\Bbb R^2$ given by $g(x,y)=\alpha x+\beta y$ then find $\alpha-\beta$.  My try: Since $g$ is an extension of $f$ so $g|_X=f\implies g(x,x)=(\alpha+\beta)x=3x\implies \alpha+\beta =3$ Now $\|f\|=\sup \{f(x,x):\|(x,x)\|=1,(x,x)\in \Bbb R\}=3\implies \|g\|=3$ So $\sup \{g(x,y):\|(x,y)\|=1,(x,y)\in \Bbb R\}=3\implies \alpha x+\beta y\le 3$ where $|x|+|y|=1$. Take $x=0,y=1\implies \alpha x+\beta y\le 3\implies \beta\le 3$. How to find the value of $\alpha,\beta $  from above. Please give some hints.
120179
I'm having trouble with this functional equation, to find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$: $f(x) + f(x+f(y)) = y + f(f(x) + f(f(y)))$ I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity. So 0 progress. Tried with $x=y=0, x=0, y=0, x=-f(y), y=f(x)$, and even with some assumptions on bijectivity, but nothing helped. I don't know the source but seems like a pretty difficult problem, maybe even impossible to solve.