Let $f$ be an entire function on the complex plane $\mathbb C$, assume that $$|f(z)|\le e^{|z|}.$$ Does the property $$|f(x)|\le e^{-|x|}, \qquad x\in\mathbb R,$$ imply $f\equiv 0$?
More generally, characterize the class of majorants $w$ on the real line, for which $|f(x)|\le w(x)$ yields $f\equiv 0$.
The estimate on the real line implies that $f\in L^2(\mathbb R)$. Thus, $f$ belongs to the Paley-Wiener space $PW_1$, which is the Fourier-image of $L^2(-1, 1)$. Therefore, the function $e^{iz}f$ belongs to the Hardy class $H^2$ in the upper half-plane $\mathbb C_+$, which is the Fourier image of $L^2(0, +\infty)$. If $f\not\equiv 0$, this yields the condition $$\int_{\mathbb R}\frac{|\log |f(x)||\,dx}{1+x^2}<\infty.$$ By the assumption, $\log |f(x)|\le -|x|$, which contradicts the preceding formula. Hence $f\equiv 0$.
This argument works if $w\in L^2$, $w\le 1$, and $\int_{\mathbb R}\frac{\log\frac{1}{w(x)}\, dx}{1+x^2}=\infty$.