Functional inequation on $\mathbb{R}$: $f\left(x+y^2\right)-f(x)\geq y$

Question

I have the following equation:

$$f:\mathbb{R}\rightarrow\mathbb{R}$$ $$\forall (x,y)\in\mathbb{R}^2,\ f\left(x+y^2\right)-f(x)\geq y$$

$f$ is not necessarily differentiable/continuous/... (in fact, we can prove that it's not differentiable at all).

I need to prove if there are solutions, and if there are, I need to give one.

How would I do that?

What we have found so far:

• $f$ is not differentiable nor continuous anywhere (@nayrb and @rlartiga 's comments)
• $f$ is strictly increasing on $\mathbb{R}$ (proof : @rlartiga 's comment)

Answer 1

For any $b>a$ and $n$, let $\Delta x = (b-a)/n$ and $x_{k} = a + k\Delta x$. Then

$$ f(b) - f(a) = \sum_{k=0}^{n-1} f(x_k + \Delta x) - f(x_k) \geq \sum_{k=0}^{n-1} \sqrt{\Delta x} = \sqrt{\smash[b]{(b-a)n}},$$

which diverges as $n\to\infty$, a contradiction! So there is no such a solution.

… and I think I'm late :(

Answer 2

Hint: Using $f(x+y^2) \geq f(x) + y$,

$$f \left(\frac{\pi^2}{6} \right) = f \left(\sum_{k=0}^\infty \frac{1}{k^2} \right) = f \left(\sum_{k=2}^\infty \frac{1}{k^2} + 1 \right)\geq f \left(\sum_{k=2}^\infty \frac{1}{k^2} \right) + \sqrt{1}.$$ Similarly, $$f \left(\sum_{k=n}^\infty \frac{1}{k^2} \right) = f \left(\sum_{k=n+1}^\infty \frac{1}{k^2} + \frac{1}{n^2} \right)\geq f \left(\sum_{k=n+1}^\infty \frac{1}{k^2} \right) + \frac{1}{n}.$$

Use induction to get a lower bound on $f(\pi^2/6)$ as a function of $n$. Finally, use $f(y^2) \geq f(0) + |y| \geq f(0)$ to get a contradiction. I can fill in more details if you are still stuck.

P.S. If you want to show $f(x)$ is undefined for any $x > 0$, use $x = c \sum_{k=1}^\infty 1/k^2$ for the appropriate $c$ instead.

P.P.S. This answer can be generalized to show that no solutions exist for $f(x+y^p)-f(x) \geq y$ when $p>1$.