Linear/Proof

Main article: Linear

Proof that Homogeneity of degree 1 (f(αx) = αf(x) for all α) follows from the additivity property (f(x + y) = f(x) + f(y)) in all cases where α is rational.

First show f(αx) = αf(x) is true for α = positive integers (p)

If

$f(x)+f(y)=f(x+y)$

then

$f(x)+f(x)=f(2x)$

$2f(x)=f(2x)$

and

$f(2x)+f(x)=f(3x)$

$2f(x)+f(x)=f(3x)$

$3f(x)=f(3x)$

and so on, so

$pf(x)=f(px)$

For α = 0

$f(0)+f(0)=f(0)$

$2f(0)=f(0)$

$f(0)=0$

For α = negative integers (n)

$f(-x)+f(x)=f(0)=0$

$f(-x)=-f(x)$

and substituting px for x

$f(-px)=-f(px)$

$f(-px)=-pf(x)$

$f(nx)=nf(x)$

For α = all integers (i)

Together, the three previous sections have shown that

$f(ix)=if(x)$

For α = 1/i

$f\left((i-1){x \over i}\right)+f\left({x \over i}\right)=f(x)$

$(i-1)f\left({x \over i}\right)+f\left({x \over i}\right)=f(x)$

$if\left({x \over i}\right)=f(x)$

$f\left({x \over i}\right)={1 \over i}f(x)$

For α = any rational number i₁/i₂

$f\left({x \over i_{2}}\right)={1 \over i_{2}}f(x)$

$i_{1}f\left({x \over i_{2}}\right)={i_{1} \over i_{2}}f(x)$

$f\left({i_{1} \over i_{2}}x\right)={i_{1} \over i_{2}}f(x)$