# Functional inequality?

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.

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

I will be building upon Bubbler's progress.

$$f(1)=0$$, 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.

• Nice solution! (Perhaps mines is more complex than this) Checkmark incoming! Feb 5, 2021 at 13:16
• The original proof had a small hole (while bounding $yf(x)$, I implicitly assumed $y$ is positive). I think it's fixed now. Feb 8, 2021 at 6:09
• Do you think my method with Vector space can be saved? Feb 8, 2021 at 17:45
• @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. Feb 9, 2021 at 3:00
• Ah, true, I missed that completely. Shame, I was really hoping your approach is salvageable. 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.)

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

• 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. Feb 5, 2021 at 6:41
• @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. Feb 5, 2021 at 6:52