Superstability of the exponential equation

Remarks

Write $K = \frac{1 + \sqrt{1 + 4\delta}}{2}$, so that $K^2 - \delta = K$ and $K \ge 1$. If $\abs{f(a)} = K + \eta$ with $\eta > 0$, then $\abs{f(2a)} \ge (K + \eta)^2 - \delta \ge K + 2\eta$, and iterating shows that $f$ is unbounded. If $f$ is unbounded, compare $f(x + y + z)$ with $f(x + y) f(z)$ and with $f(x) f(y + z)$ to get $$\abs{f(x + y) - f(x) f(y)} \abs{f(z)} \le 2\delta + \abs{f(x)} \delta,$$ and let $\abs{f(z)} \to \infty$. So an approximate solution of Cauchy's exponential equation is either bounded or an exact solution. This is called superstability: it is stronger than Hyers–Ulam stability, since an unbounded approximate solution must be exact. Example: on $S = (\R, +)$, the constant $f = 1/2$ satisfies the hypothesis with $\delta = 1/4$, and it is bounded by $K = (1 + \sqrt{2})/2$ but not exponential. The result was first proved by J. Baker, J. Lawrence and F. Zorzitto in 1979 for real functions on a vector space.

Going deeper

For a student

For the additive Cauchy equation, approximate solutions are near exact ones but need not be exact. The exponential equation behaves differently because it is multiplicative: an error is amplified by large values of $f$. Similar superstability results hold for d'Alembert's equation and other multiplicative equations.