Prove that there is no function $f:\Bbb{R}\to\Bbb{R}$ with $f(0)>0$ such that $\forall x,y\in\Bbb{R}, f(x+y)\geq f(x)+y f(f(x))$

Solution 1:

DISCLAIMER: This answer is incorrect as it stands, namely in the $k>0$ section is claims that: $$ f(0)=f(x-x)\geq f(x)-x f(f(x)) $$ implies $f(x)<f(0)$, which is false.

This disclaimer will stay here until the problem is fixed.

From the condition $f(0)>0$ it follows that with $x=0$ we have $$ f(0+y)-y f(f(0))\geq f(0)>0 $$ so it follows that $f(y)>ky$ where $k=f(f(0))$ is some non-zero constant. It has to be non-zero since otherwise $0=k=f(f(x))>k f(x)=0$. Now we have to divide into cases.

If $k>0$

If $k>0$ it follows that $x f(f(x))>kx f(x)>k^2 x^2\geq 0$. So in that case we have $$ \underbrace{f(x-x)}_{f(0)}\geq f(x)-\underbrace{x f(f(x))}_{\text{strictly greater than }0}\\ \iff\\ f(x)<f(0) $$ But the above statement is absurd by all means. For one thing $f(0)<f(0)$ and moreover $f(0)>f(x)>kx\rightarrow\infty$ which is also absurd.

If $k<0$

If we apply the beginning of what Denis showed, namely that $f(x)\rightarrow\infty$ for $x\rightarrow\infty$, then we get in addition to that from my analysis that for $k<0$ we have $f(-x)\geq -kx\rightarrow\infty$ for $x\rightarrow\infty$. So we see that $$ f(0)=f(-x+x)\geq f(-x)+x f(f(-x))\rightarrow\infty $$ for $x\rightarrow\infty$ too. A contradiction since we have just shown that the constant $f(0)$ is greater than or equal to an expression that tends to infinity. So $k<0$ is impossible too.

Solution 2:

Let $a=f(0)>0$, $b=f(a)$ and $c=f(b)$.

We know that $f(0+b)\geq a+b^2$, so we get $c>0$.

We also have for all $x$, $f(x+a)\geq f(a)+cx$, so in particular $\lim_{x\to\infty} f(x)=\infty$ and more precisely $f(x)=\Omega(x)$ when $x\to\infty$.

This means that for $x$ big enough, $f(x)>0$, and therefore, $f(x+1)\geq f(x)+f(f(x))>f(f(x))$. This gets us $f(f(x-1))<f(x)$.

Moreover, for $x$ big enough, we have $f(x-1)>x$ (since $f(x+1)=\Omega(x^2)$).

Let us choose $x_0$ big enough, and define $x_{n+1}=f(x_n-1)$.

The sequence $(x_n)$ is strictly increasing, and for all $n$, we have $f(x_{n+1})<f(x_n)$. We reached a a contradiction with $\lim_{x\to\infty} f(x)=\infty$.