Let f(x + y) = f(x) + f(y), for x, y ∊ R Put x = 0 = y ⇒ f(0) = f(0) + f(0) ⇒ f(0) = 0 Now. f ' (0) =
lim
h→0
f(0+h)−f(0)
h
f '' (0) =
lim
h→0
f(h)
h
Now, f (x) =
lim
h→0
f(x+h)−f(x)
h
=
lim
h→0
f(x)+f(h)−f(x)
h
⇒ f ' (x) =
lim
h→0
f(h)
h
= f ' (0) ⇒ f (x) = x f ' (0) + C But f (0) = 0 ∴ C = 0 Hence, f(x) = x f ' (0), for all x ∊ R Clearly, f(x) is everywhere continuous and differentiable and f ' (x) is constant for all x ∊ R