# If $f(x)f(y)+f(xy)\le -\frac{1}{4},\forall x,y\in[0,1)$, show that $f(x)=-\frac{1}{2}$

Let $$f:[0,1) \to \mathbb{R}$$ be a function such that $$f(x)f(y)+f(xy)\le -\dfrac{1}{4} \quad \forall\, x,y\in[0,1).$$

Show that $$f(x)=-\dfrac{1}{2} \quad \forall\, x \in[0,1).$$

I have proved that $$f(0)=-\dfrac{1}{2}$$: if $$x=y=0$$, we have $$f^2(0)+f(0)\le-\dfrac{1}{4}\Longrightarrow \left( f(0)+\frac{1}{2} \right)^2\le 0\Longrightarrow f(0)=-\dfrac{1}{2}.$$

But I can't prove $$f(x)$$ be constant. Thanks.

• Where is your question from? – user574848 Apr 25 at 11:33

Plugging in $$y=0$$ gives $$f(x) \ge -\frac{1}{2}$$ for each $$x$$.
Let $$y=x$$ to get $$f(x)^2+f(x^2) \le -\frac{1}{4}$$. This implies $$f(x^2) \le -\frac{1}{4}$$ for each $$x$$, and so $$f(x) \le -\frac{1}{4}$$ for each $$x$$. But then $$f(x)^2+f(x^2) \le -\frac{1}{4}$$ implies $$f(x^2) \le -\frac{1}{4}-(\frac{1}{4})^2 = -\frac{5}{16}$$, and so $$f(x) \le -\frac{5}{16}$$ for each $$x$$. Doing this again gives $$f(x) \le -\frac{1}{4}-(\frac{5}{16})^2 = -\frac{89}{256}$$ for each $$x$$. If we keep doing this, we see that, for any $$\epsilon > 0$$, $$f(x) \le -(\frac{1}{2}-\epsilon)$$ for each $$x$$. Therefore, $$f(x) \le -\frac{1}{2}$$ for each $$x$$.
• Hello,if $f(x^2)\le -\dfrac{1}{4}$ for each $x$,so I think you only $f(x)\le -\dfrac{1}{4}$ for $x\ge 0$ – inequality Apr 23 at 9:48
• @inequality you started by saying $f$ is defined on $[0,1)$. – mathworker21 Apr 23 at 9:51