Let $f$ and $f_1$ two $\mathcal{C}^\infty$ complex-valued real functions such that $$|f(x)-f_1(x)|<e^{-x},\ \forall\ x\in \mathbb{R}^+ $$

Assume that the equation $f_1(x)=0$ has an infinitely many discrete solutions: $0<x_1<x_2<...<+\infty$.

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

  • $\begingroup$ If $f$ is not continuous, it can avoid any root. $\endgroup$ – Yves Daoust May 11 '19 at 12:03
  • $\begingroup$ Thank you for your remark but my functions are smooth (i ve reedited my question). $\endgroup$ – rihani May 11 '19 at 12:07
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    $\begingroup$ The example $f_1\equiv 0, f(x)=e^{-x}$ shows that zeros of $f$ cannot be guaranteed by zeros of $f_1$. $\endgroup$ – Kavi Rama Murthy May 11 '19 at 12:07
  • $\begingroup$ $f_1(x)=0$ must have a discret set of solution $x_1<x_2...$ $\endgroup$ – 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 $$|f(x)-f_1(x)|=\left|\left(\frac12-\frac{\sin x}{3}\right)e^{-x}\right|\leq \frac{5}{6}e^{-x}<e^{-x}.$$ Thus, $f$ and $f_1$ satisfy the given conditions but $f(x)=0$ does not have a solution.

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