Aoki–Rassias theorem

Remarks

The proof follows Hyers' theorem. Setting $x = y$ gives $\norm{f(2x) - 2f(x)} \le 2\theta \norm{x}^p$, so consecutive terms of $2^{-n} f(2^n x)$ differ by at most $\theta \, 2^{n(p-1)} \norm{x}^p$. Because $p < 1$, these bounds form a convergent geometric series with sum $\frac{2\theta}{2 - 2^p}\norm{x}^p$. For $p = 0$ the hypothesis is a constant bound $2\theta$, and the theorem gives back Hyers' theorem with $\delta = 2\theta$. Example: $f(x) = x + \sqrt{\abs{x}}$ on $\R$ satisfies the hypothesis with $\theta = 1$ and $p = 1/2$, because $0 \le \sqrt{\abs{x}} + \sqrt{\abs{y}} - \sqrt{\abs{x + y}} \le \sqrt{\abs{x}} + \sqrt{\abs{y}}$; here $A(x) = x$. The hypothesis $p < 1$ cannot be dropped: Z. Gajda showed that the statement fails for $p = 1$, already for maps $\R \to \R$. T. Aoki proved the additive case in 1950, and Th. M. Rassias proved the linear case in 1978.

Going deeper

For a student

The theorem started the study of stability with a control function, often called Hyers–Ulam–Rassias stability. The same rescaling argument works for $p > 1$ with $A(x) = \lim_n 2^n f(2^{-n} x)$, and the excluded case $p = 1$ marks the boundary where the geometric series diverges. Similar results hold for the Jensen, quadratic and many other functional equations.