# Find some sets in the real line where the function $g$ is positive

Let $f\colon\mathbb C\to\mathbb C$ be an entire non constant function. We consider its values on the real line. Let us consider the function $g\colon ℝ^3→ℝ$ defined by $$g(x,y,z)=f(x)-\left(\frac{\ln f(y)+\ln f(z)}{\ln2}\right)+2$$ defined in the set where $f$ is positive, i.e., $f(x)≥0$, $f(y)>0$ and $f(z)>0$.

The function $f$ has infinitely many real zeros and there is infinitely many real solutions to the equation $f(s)=w$ for any real number $w$.

My question is: Can we find some sets in the real line where the function $g$ is positive?

My answer is as follow:

Proof:

For each discrete set containing $(y_{i},z_{i})∈ℝ²$, let $w_{i}=((ln f(y_{i})+ln f(z_{i})))/ln2)-2$, then the equation $f(x)=w_{i}$ has infinitely many real solutions $x_{i}$ (the set of $x_{i}$ is discrete since $f$ is entire). So, $g(x,y,z)$ has infinitely many real zeros of the form $(x_{i},y_{i},z_{i})$, i.e., $g$ changes its sign. Since $g$ is continuous, then there exist open set $V$ containing a point $(x_{i},y_{i},z_{i})$ such that $g(x,y,z)≥0$.

## 1 Answer

You need the conditions $f(y_i) > 0$ and $f(z_i) > 0$. Perhaps you can explain how you get from $f$ being entire to there existing such points. From there, why don't you just look for a value of $x_i$ that gives you $f(x_i) > w_i$. Then use continuity of $f$ on $\mathbb{R}$ to get a neighborhood of $(x_i, y_i, z_i)$ where $g>0$.

• @ Paul Hurst: $f$ has infinitely many real zeros. – DER Jun 4 '14 at 9:33
• $f$ doesn't necessarily have infinitely many real zeros. Take $f(z) = z$, for example. The restriction of $f$ to the reals is just $f(x) = x$, which only has one real zero. I think what you mean is that $g(x,y,z)$ has infinitely many real zeros. Or is that part of the hypothesis? Because I think you can get the result based only on $f$ being non-constant and entire. – Paul Hurst Jun 4 '14 at 20:52
• @ Paul Hurst: That fact is a part of the hypothesis. yes, we can neglect it and using only $f$ being non-constant and entire. – DER Jun 5 '14 at 7:47