assume f''(x) = 0, prove k * f(x / k) = f(x)

f(x) = C_1 * x + C_2

assume C_2 is 0

k * f(x / k) = k * C_1 * x / k = C_1 * x = f(x)



assume k * f(x / k) = f(x), prove f''(x) = 0

assume f''(x) != 0

f(x) = C * g(x), where g != identity

k * f(x / k) = k * C * g(x / k)

As g != identity, the scalar of k is modified to some new value, k_2, where k_2 != 1 / k

k * f(x / k) = k * C * g_2(x) * k_2, where k_2 != 1/k

g_2 is not a function of k.  Thus, k * f(x / k) != f(x), as k * k_2 does not cancel.  Thus, f''(x) = 0