I am trying to prove that this function is bounded for all $x$ $$\frac{e^{-ix}-1}{x}$$

I found that :

$|\frac{e^{-ix}-1}{x}| \le \frac{|cos(x)|}{|x|}+\frac{|sin(x)|}{|x|}+\frac{1}{|x|}$

So when $x \rightarrow (-)\infty $, the function goes to $0$

and in $x=0$, we have that : $\lim_{x \to 0} |\frac{e^{-ix}-1}{x}| = \lim_{x \to 0} |\frac{e^{-ix}-e^(i0)}{x-0}| = (e^{-ix})'(0) = 0$

and by continuity of the function in $]0,\infty[$ it cannot diverge between $0$ and $(-) \infty$, so it must be bounded

Is it correct ?

  • 4
    $\begingroup$ Try expanding $e^{ix}$, but you should show some of your own work and then ask about specifically what you are confused about. $\endgroup$ – Nalt Jun 3 at 14:46
  • $\begingroup$ The derivative of $e^{-ix}$ evaluated at zero is $-i$, not $0$. You should add a bit to your argument explaining why continuity implies it cannot "diverge" between $0$ and $\pm \infty$ given the limits you've computed. $\endgroup$ – jawheele Jun 3 at 16:04

I'll write $f(x)= \frac{e^{-ix}-1}{x}$, $x \in \mathbb{R} \backslash \{0\}$. As you've noted, a bound like $$|\frac{e^{-ix}-1}{x}| \leq \frac{|e^{ix}|+1}{|x|} = \frac{2}{|x|}$$ implies that $f$ is bounded at infinity. In particular, say, $|f(x)| \leq 2$ for $|x| \geq 1$. In addition, you've effectively noted that $$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{-ix}-e^{-i0}}{x-0} = -ie^{-ix}|_{x=0} = -i,$$ and this implies that $\exists \epsilon > 0$ such that $|x| < \epsilon \implies |f(x)+i| < 1$, which by the triangle inequality implies $|f(x)| \leq 2$.

So, $|f(x)| \leq 2$ for $|x|<\epsilon$ and $|x|>1$. Meanwhile, the set $A=\{x \in \mathbb{R} \backslash \{0\} : \epsilon \leq |x| \leq 1\}$ is compact, so continuity of $x \mapsto |f(x)|$ implies that it achieves its maximum value $M \geq 0$ on $A$. Hence, $\forall x \in \mathbb{R} \backslash \{0\}$ we have the bound $|f(x)| \leq M+2$.

A similar argument yields that any continuous function on $\mathbb{R}^n \backslash \{p\}$ for which the limits as $x \to \infty$ and $x \to p$ exist is bounded. More generally, the existence of the $x \to \infty$ limit may be relaxed to $f$ being bounded outside of a sufficiently large sphere.


Hint: $$ \left|e^{ix} - 1\right|\le \min(2, |x|) $$


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