$$f(xy) \le yf(x) + f(f(x))$$for all real $x, y$.

Question

parthkohli · Answer

Could anyone prove/disprove that $f(x) = 0$ (a trivial choice) is the only function satisfying this?

ganeshie8 · Answer

My first thoughts are to make use of Jenson's inequality and mess around

parthkohli · Answer

oh hmm I'm not too familiar with that :(

ganeshie8 · Answer

Okay
f(x) = 0 is the only constant function that satisfies the given inequality

imqwerty · Answer

putting y=0 in the equation we get this-
$f(0) \le f(f(x))$     => $0 \le f(x)$ 

so f(x) is never negative

and in this equation-
$f(xy) \le yf(x) + f(f(x))$  

well if y is set equal to 1 we get -
$f(x) \le f(x) +f(f(x))$ 

so rhs is always greater than lhs for all f(x) >= 0

parthkohli · Answer

$$f(0) \le f(f(x)) \implies 0 \le f(x)$$Doesn't that involve the assumption that $f(x)$ is nondecreasing?

kainui · Answer

$x=0$ implies:

$$f(0) \le y f(0) + f(f(0))$$

since I can just rearrange this arbitrary constant:

$$f(0)-f(f(0)) \le y f(0)$$

That must be true for all y, so we require that $f(0)=0$ to remove the arbitrary y from giving a contradiction.

kainui · Answer

Similarly, let $y=\frac{C}{x}$ for some arbitrary constant C. We now have:

$$f(C)\le C \frac{f(x)}{x} + f(f(x))$$

Maybe useful maybe not lol idk

kainui · Answer

Oooh using the fact that f(0)=0 we can plug in y=0 to the original equation to get:

$$0 \le f(f(x))$$

kainui · Answer

since $f(f(x))$ is positive I thought I'd run through some inequalities just for the data.

(x,y) : inequality

$$(\oplus, \oplus) : f(\oplus ) \le \oplus* f(\oplus) +\oplus$$$$(\oplus, \ominus) : f(\ominus ) \le \ominus* f(\oplus) +\oplus$$$$(\ominus, \oplus) : f(\ominus ) \le \oplus* f(\ominus) +\oplus$$$$(\ominus, \ominus) : f(\oplus ) \le \ominus * f(\ominus) +\oplus$$

I hope my notation isn't too confusing. Now if you assume f is positive or negative you can draw conclusions possibly. I dunno.

proheiper · Answer

^

bobo-i-bo · Answer

I DID IT!!! :D

Applying the inequality twice:
$$f(xyz) \leq xf(yz)+f^2(yz) \leq x(yf(z)+f^2(z)) +f^2(yz)=xyf(z)+xf^2(z)+f^2(yz)$$
Letting y=0,
$$f(0) \leq xf^2(z)+f^2(0) $$
$$ \Rightarrow f(0)-f^2(0) \leq xf^2(z) $$
This implies that $f^2(z)=0$ for all z.
Therefore the orginal inequality,
$$ f(xy) \leq yf(x) + f^2(x)$$
can be simplified to:
$$ f(xy) \leq yf(x)$$
If x=1, then for all y:
$$ f(y) \leq yf(1)$$
If, instead, assuming $x > 0$, letting $y = \frac 1 x$,
$$f(1) \leq \frac 1 x f(x)$$
$$\Rightarrow xf(1) \leq f(x) $$
From the last two inequalities, it follows that for $x > 0$,
$$f(x)=f(1)x$$
But $(f(1))^2x=f^2(x)=0$ implies that $f(1)=0$ so $ f(x)=0 $ for $x>0$.

I have managed to prove but leave as an exercise, proving that $f(x)=0$ for $x \leq 0$ unless it is requested. :P Also If there's anywhere which isn't clear, I can clarify :D

kainui · Answer

@Bobo-i-bo Awesome I'm halfway through understanding it I really like that trick of plugging it in twice like that with xyz really cool.

So the part I'm stuck at is what'd you do here when you plugged in y=1/x? It looks like you should end up with $f(\tfrac{1}{x} ) \le \tfrac{1}{x} f(1)$ so I don't understand what's going on there.

kainui · Answer

Ah ok I see I was looking at plugging into the wrong equation got it

bobo-i-bo · Answer

I'm not plugging $y=\frac 1 x$ into 
$$f(y) \leq yf(1) $$
I'm plugging it into
$$f(xy) \leq yf(x)$$
Sorry, that is a bit confusing :P

kainui · Answer

Awesome thanks, makes sense now fancy haha.

parthkohli · Answer

!

bobo-i-bo · Answer

:D :D You should try proving it for the negative bit, it's not really simple :P