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:


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 $g(x,y,z)≥0$.


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 $g>0$.

  • $\begingroup$ @ Paul Hurst: $f$ has infinitely many real zeros. $\endgroup$ – DER Jun 4 '14 at 9:33
  • $\begingroup$ $f$ 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. $\endgroup$ – Paul Hurst Jun 4 '14 at 20:52
  • $\begingroup$ @ Paul Hurst: That fact is a part of the hypothesis. yes, we can neglect it and using only $f$ being non-constant and entire. $\endgroup$ – DER Jun 5 '14 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.