Find all functions $f:\mathbb{R}\to\mathbb{R}$ s.t. for all $x,y\in\mathbb{R}$, we have $$yf(x)+f(y)\ge f(xy)$$

Problem from the my math olympiad training problem set few weeks before.

Functional equations (and inequalities) are mostly on topic.

  • $\begingroup$ While I agree this is on-topic, you would receive better answers if you had posted it in MSE rather than PSE. $\endgroup$
    – Anonymous
    Feb 5, 2021 at 5:05

3 Answers 3


I will be building upon Bubbler's progress.

Specifically, Bubbler's answer shows

$f(1)=0$, and $f(x)+f(-x)=C$ for all non-zero $x$.

Let us use $a$ for $f(0)$. Note that

Plugging $x=0$ in the actual inequality gives $$f(y)\ge a-ay.$$ If we further assume $y$ is non-zero, then replacing $y$ with $-y$ gives $$f(-y)\ge a+ay.$$

But we know that

$f(-y)=C-f(y)$ ($y$ was non-zero), so this rewrites as $$f(y)\le C-a-ay.$$

Thus we have the bounds

$a-ax\le f(x)\le C-a-ax$ for all nonzero $x$.

Next, if we

replace $x\to -x$ in the original inequality and add it to the original equality, we get $$y((f(x)+f(-x))+2f(y)\ge f(xy)+f(-xy).$$With the added assumption $x,y$ are non-zero, this gives $$yC+2f(y)\ge C\implies f(y)\ge \frac C2-\frac C2y.$$But teher's already an upper bound on $f(y)$, namely $C-a-ay$, so these bounds must be compatible. In other words, $$C-a-ay\ge \frac C2-\frac C2y.$$This rearranges into $\left(\tfrac C2-a\right)(y+1)\ge 0$. But this has to hold for all non-zero $y$, even though $y+1$ can have any sign! Thus $C/2-a$ has to be zero.; i.e., $C=2a$.

Now our upper and lower bounds for $f(x)$

reduce to the same thing! In other words, we have $f(x)=a-ax$ for $x\ne 0$. This is true for $x=0$ too, because $a=f(0)$ by definition. Thus this is the only possible form; the fact that this indeed works is simple to check.

  • 1
    $\begingroup$ Nice solution! (Perhaps mines is more complex than this) Checkmark incoming! $\endgroup$ Feb 5, 2021 at 13:16
  • $\begingroup$ The original proof had a small hole (while bounding $yf(x)$, I implicitly assumed $y$ is positive). I think it's fixed now. $\endgroup$
    – Ankoganit
    Feb 8, 2021 at 6:09
  • $\begingroup$ Do you think my method with Vector space can be saved? $\endgroup$
    – nonuser
    Feb 8, 2021 at 17:45
  • $\begingroup$ @Greedoid I think the vector space argument can be skipped altogether. If f is a solution, then so is g(x)=f(x)-f(0)+f(0)x, and g satisfies g(0)=0. Then the rest of your very neat reasoning applies. $\endgroup$
    – Ankoganit
    Feb 9, 2021 at 3:00
  • 1
    $\begingroup$ Ah, true, I missed that completely. Shame, I was really hoping your approach is salvageable. $\endgroup$
    – Ankoganit
    Feb 9, 2021 at 16:53

Here are the properties of $f(x)$ I identified so far. I doubt it's enough to get any credit on a math olympiad, but anyway...

  1. The value of $f(1)$:

$$\begin{align}&x=y=1 & \Rightarrow \quad& f(1)+f(1) \ge f(1), & f(1) \ge 0 \\ &x=y=-1 & \Rightarrow \quad& -f(-1) + f(-1) \ge f(1), & f(1) \le 0 \end{align}$$

$$\therefore f(1) = 0$$

  1. A relationship between $f(x)$ and $f(-x)$, and some knowledge about $f(0)$

$$yf(x) + f(y) \ge f(xy) \\ -yf(x) + f(-y) \ge f(-xy) \\ \therefore f(y)+f(-y) \ge f(xy) + f(-xy)$$ $$xyf\left(\frac{1}{x}\right) + f(xy) \ge f(y) \\ -xyf\left(\frac{1}{x}\right) + f(-xy) \ge f(-y) \\ \therefore f(xy)+f(-xy) \ge f(y) + f(-y), \text{ given } x\ne 0$$

Therefore the two sides are equal. Since any two sums in the form of $f(x)+f(-x)$ ($x$ nonzero) are equal by above, we can say that it equals some constant $C$. $$\exists C \in \mathbb{R}, \; \forall x \ne 0, \; f(x)+f(-x)=C$$

Since $f(1) = 0$, $f(-1) = C$ holds, and we can simplify the above to $$ \forall x \ne 0, \; f(x)+f(-x)=f(-1)$$

The first inequality still holds for $x=0$, so $$ f(0) \le \frac{C}{2}$$ (we can't say they are equal because $f$ is not necessarily continuous.)

  • 1
    $\begingroup$ I think the quantifiers in the statement about $C$ are the wrong way round. $\endgroup$
    – Ankoganit
    Feb 5, 2021 at 12:49
  • $\begingroup$ @Ankoganit Thanks, fixed. $\endgroup$
    – Bubbler
    Feb 7, 2021 at 23:44
  • $\begingroup$ This might get you 1 or 2 points, since it is important. $\endgroup$
    – justhalf
    Feb 9, 2021 at 8:25

Specific values:

  • Putting $x=1$ gives $yf(1)\geq0$ for all $y\in\mathbb{R}$, so we must have $f(1)=0$.

  • Putting $x=0$ gives $f(y)\geq (1-y)f(0)$ for all $y\in\mathbb{R}$, so $f$ is bounded below on any finite subset of $\mathbb{R}$.

Properties of the function $f$:

$yf(x)+f(y)\geq f(xy)$ implies that $f(xy)-f(y)\leq yf(x)$ for all $x,y$.

For any $y>0$, this means $\frac{f(xy)-f(y)}{y}\leq f(x)$, so $$\lim_{x\to1^+}\frac{f(xy)-f(y)}{y}\leq\lim_{x\to1^+}\frac{f(x)}{1-x},\quad\quad\lim_{x\to1^-}\frac{f(xy)-f(y)}{y}\geq\lim_{x\to1^+}\frac{f(x)}{1-x}.$$
For any $y<0$, we have $\frac{f(xy)-f(y)}{y}\geq f(x)$, so $$\lim_{x\to1^-}\frac{f(xy)-f(y)}{y}\leq\lim_{x\to1^+}\frac{f(x)}{1-x},\quad\quad\lim_{x\to1^+}\frac{f(xy)-f(y)}{y}\geq\lim_{x\to1^+}\frac{f(x)}{1-x}.$$
So, if $f$ is differentiable, then $f'(y)=\lim_{x\to1^+}\frac{f(x)}{1-x}$ is constant for all $y$, so $f$ is linear. Say $f(x)=ax+b$, so we know $a+b=0$ and the original inequality becomes $y(ax-a)+(ay-a)\geq (axy-a)$, which is always true as the LHS and RHS are identical.

So the general differentiable solution is

$f(x)=ax-a$, $a\in\mathbb{R}$.

But must $f$ be differentiable according to the given condition? This part I haven't quite figured out yet.

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    $\begingroup$ I frankly doubt it's legal to talk about the limits. $f$ is simply given as a real function, with no guarantees about being continuous or having certain limits. $\endgroup$
    – Bubbler
    Feb 5, 2021 at 6:41
  • $\begingroup$ @Bubbler My use of the notation $\lim_{x\to1}$ implicitly assumes that the limit exists. It's like "we don't know if this limit exists, but if it does, it satisfies this inequality ..." Then the assumption that $f$ is differentiable justifies all that. $\endgroup$ Feb 5, 2021 at 6:52

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