# How to use asymptotic exponential approximation of a complex-valued real function?

Let $$f$$ and $$f_1$$ two $$\mathcal{C}^\infty$$ complex-valued real functions such that $$|f(x)-f_1(x)|

Assume that the equation $$f_1(x)=0$$ has an infinitely many discrete solutions: $$0.

Is there any condition to impose on $$f$$ in order to prove that $$f(x)=0$$ has a solution?

• If $f$ is not continuous, it can avoid any root. – Yves Daoust May 11 '19 at 12:03
• Thank you for your remark but my functions are smooth (i ve reedited my question). – rihani May 11 '19 at 12:07
• The example $f_1\equiv 0, f(x)=e^{-x}$ shows that zeros of $f$ cannot be guaranteed by zeros of $f_1$. – Kavi Rama Murthy May 11 '19 at 12:07
• $f_1(x)=0$ must have a discret set of solution $x_1<x_2...$ – rihani May 11 '19 at 12:13

Take $$f(x)=\frac{1}{2}e^{-x}\qquad\text{and}\qquad f_1(x)=\frac{1}{3}e^{-x}\sin(x).$$ Then we have $$f_1(x)=0$$ for all $$x=n\pi$$ where $$n\in\mathbb Z$$ and $$|f(x)-f_1(x)|=\left|\left(\frac12-\frac{\sin x}{3}\right)e^{-x}\right|\leq \frac{5}{6}e^{-x} Thus, $$f$$ and $$f_1$$ satisfy the given conditions but $$f(x)=0$$ does not have a solution.