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.

  • $\begingroup$ Where is your question from? $\endgroup$ – 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 $x$.

  • $\begingroup$ Hello,if $f(x^2)\le -\dfrac{1}{4}$ for each $x$,so I think you only $f(x)\le -\dfrac{1}{4}$ for $x\ge 0$ $\endgroup$ – inequality Apr 23 at 9:48
  • 1
    $\begingroup$ @inequality you started by saying $f$ is defined on $[0,1)$. $\endgroup$ – mathworker21 Apr 23 at 9:51

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