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what are the continuous and differentiable functions that follow $f(x+y) > f(x-y)f(y-x)$ where the domain of the function is the real numbers

I can't think of any functions that could follow this.

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    $\begingroup$ Well, any constant $c$ with $0<c<1$ works. $\endgroup$ – lulu Jan 23 at 18:59
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    $\begingroup$ Is there any reason to imagine that this has a pleasant answer? Without thinking very hard about it, I'd expect all sorts of things to work. Something like $\frac 12 + \frac {\sin(x)}{1000}$ for example. $\endgroup$ – lulu Jan 23 at 19:09
  • $\begingroup$ The given inequality is equivalent to $f(z+2y)>f(z)f(-z)$ for any chosen $z,y\in\mathbb{R}$. Choosing $y=0$ we get the condition $f(z)>f(z)f(-z)$, so $f\neq 0$ for any $z\in \mathbb{R}$ and so we find further the condition $f(z)\in(-1,1)$ for any chosen $z$. $\endgroup$ – Masacroso Jan 23 at 19:10
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    $\begingroup$ Putting $x=y$ in the given inequality, we can further note that $f(2x)\gt f(0)^2$, so $f(z)\gt 0~\forall~z\in\Bbb R$, so we can refine to $f(z)\in (0,1)$ for any chosen real $z$ $\endgroup$ – Prasun Biswas Jan 23 at 19:12
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    $\begingroup$ Can you clarify your question? As it stands there are lots and lots of solutions, but there really doesn't seem to be anything very interesting about them. Is there any context here? $\endgroup$ – lulu Jan 23 at 19:23
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You can use substitution $a=x+y$, $b=x-y$. Then we can reverse it by $x=\frac{a+b}{2}$, $y=\frac{a-b}{2}$, so this is a bijection. Putting this into our inequality, we get $f(a)>f(b)f(-b)$ for any $a,\ b \in \mathbb{R}$.

$b=0$ gives $f(a)>f(0)^2 \geqslant 0$, so $f$ attains only positive values. Moreover, if $f(b) \geqslant 1$ for any $b$, we will have $f(b)f(-b) \geqslant f(-b)$, which is impossible. Therefore $f(\mathbb{R}) \subset (0,1)$.

Then $f(x)$ and $g(x):=f(x)f(-x)$ are some continuous and differentiable functions, and it seems that nothing more can be said about $f$ than that infimum of $f$ is larger or equal to supremum of $g$. One large family of examples are functions whose set of values is contained in some interval $[p,q]$ with $q^2<p$. But we can choose any interval $[p,q]$ for image of $f$ (provided that $0<p<q<1$): just take glue together constant function equal to $p$ on nonpositive numbers, with any function $h: [0,\infty) \to [p,q]$ such that $h(0)=p$ and $h'(0)=0$.

So, I guess, no further results can be proven and we will have various examples.

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