Yes.. the way to prove is below
Scalar addition
f(u) = <(1,2), u> = 1*u + 2*u = u + 2u = 3u
f(g) = <(1,2), g> = = 3g
f(u+g) = <(1,2), (u,g)> = 1*u + 1*g + 2*u + 2*g = 3u + 3g
f(u) + f(g) = f(u+g)
Scalar multiplication
f(kx) = <(1,2), kx> = kx*1 + 2*kx = 3kx
kf(x)...