Linear Functions A function f is linear if f(ax + by) = af(x) + bf(y) Or equivalently f is linear if 1.) f(ax) = af(x) and 2.) f(x + y) = f(x) + f(y) Theorem: If f is linear, then f(0) = 0 Proof: f(0) = f(0 · 0) = 0 · f(0) = 0 Example 1.) f : R → R, f(x) = 2x Proof: f(ax + by) = 2(ax + by) = 2ax + 2by = af(x) + bf(y) Example 2.) f : R^{2} → R^{2},

f((x_{1}, x_{2})) = (2x_{1}, x_{1} + x_{2})

Proof: Let x = (x_{1}, x_{2}), y = (y_{1}, y_{2}) ax+by = a(x_{1}, x_{2})+b(y_{1}, y_{2}) = (ax_{1}, ax_{2})+(by_{1}, by_{2}) = (ax_{1} + by_{1}, ax_{2} + by_{2}) f(ax_{1} + by_{1}, ax_{2} + by_{2})

= (2(ax_{1} + by_{1}), ax_{1} + by_{1} + ax_{2} + by_{2})
= (2ax_{1} + 2by_{1}, ax_{1} + ax_{2} + by_{1} + by_{2})
= (2ax_{1}, ax_{1} + ax_{2}) + (2by_{1}, by_{1} + by_{2})
= a(2x_{1}, x_{1} + x_{2}) + b(2y_{1}, y_{1} + y_{2})
= af((x_{1}, x_{2})) + bf((y_{1}, y_{2}))

Example 3.) D : set of all differential functions → set of all functions, D(f) = f′

Proof: D(af + bg) = (af + bg)′ = af′ + bg′ = aD(f) + bD(g)

Example 4.) Given a, b real numbers, I : set of all integrable functions on [a, b] → R , I(f) = ∫_{a}^{b} f

Proof: I(sf + tg) = ∫_{a}^{b} sf + tg = s ∫_{a}^{b} f + t ∫_{a}^{b} g = sI(f) + tI(g)

Example 5.) The inverse of a linear function is linear (when the inverse exists). Suppose f^{−1}(x) = c, f^{−1}(y) = d. Then f(c) = x and f(d) = y and f(ac + bd) = af(c) + bf(d) = ax + by. Hence f^{−1}(ax + by) = ac + bd = af^{−1}(x) + bf^{−1}(y). Example 6.) D : set of all twice differential functions → set of all functions, L(f) = af′′ + bf′ + cf Proof: L(sf + tg) = a(sf + tg)′′ + b(sf + tg)′ + c(sf + tg)

= saf′′ + tag′′ + sbf′ + tbg′ + scf + tcg = s(af′′ + bf′ + cf) + t(ag′′ + bg′ + cg)

= sL(f) + tL(g)

Consequence 1: If ψ_{1}, ψ_{2} are solutions to af′′ + bf′ + cf = 0, then 3ψ_{1} + 5ψ_{2} is also a solution to af′′ + bf′ + cf = 0, Proof: Since ψ_{1}, ψ_{2} are solutions to af′′+bf′+cf = 0, L(ψ_{1}) = 0 and L(ψ_{2}) = 0. Hence L(3ψ_{1} + 5ψ_{2}) = 3L(ψ_{1}) + 5L(ψ_{2})

= 3(0) + 5(0) = 0.

Thus 3ψ_{1}+5ψ_{2} is also a solution to af′′+bf′+cf = 0 Consequence 2: If ψ_{1} is a solution to af′′ + bf′ + cf = h and ψ_{2} is a solution to af′′ + bf′ + cf = k, then 3ψ_{1}+5ψ_{2} is a solution to af′′+bf′+cf = 3h+5k, Since ψ_{1} is a solution to af′′+bf′+cf = h, L(ψ_{1}) = h. Since ψ_{2} is a solution to af′′+bf′+cf = k, L(ψ_{2}) = k. Hence L(3ψ_{1} + 5ψ_{2}) = 3L(ψ_{1}) + 5L(ψ_{2})

= 3h + 5k.

Thus 3ψ_{1} + 5ψ_{2} is also a solution to af′′ + bf′ + cf = 3h + 5k