Vanishing symmetric second derivative forces an affine function (${\lim }_{h\to 0}[F(t+h)-2F(t)+F(t-h)]/h^2=0$)

If a continuous real function on an interval has the displayed limit zero at every interior point, it is affine. On any compact subinterval subtract its endpoint chord. If the difference were positive somewhere, adding $\epsilon (t-a)(t-b)$ for sufficiently small positive $\epsilon$ would retain a positive interior maximum. At a maximum every symmetric second difference is nonpositive, but its normalized limit would be $2\epsilon >0$. Thus the function lies below its chord; applying the same argument to its negative gives equality. Real and imaginary parts give the complex-valued version. Existence of ordinary second derivatives is unnecessary.