Combating the perceived ubiquity of linearity

How many times have you seen a student claim that `sin(2x) = 2sin(x),`  or that `sin(x+y) = sin(x) +sin(y)` ?

A common example is `sqrt(x^2+9)`  that they want to be the same as `x+3.` 

If this were true then `sqrt(25) = 7` 

since `sqrt(25) = sqrt(16+9) = 4+3 = 7` 

On several occasions I have asked a class if anyone could give me an example of a function that actually satisfied the additional property, f(x+y) =f(x)+f(y). After a bunch of wrong guesses and a little bit of time someone will suggest f(x) =kx, where k is any constant. Then I challenge them to find another, or prove that there are no others. 

In this talk we will accept that challenge and see where it leads us. We will do the work in general real vector spaces, and then see how it results apply to real functions of one real variable.

Let `x,`  `y,`  and `z`  be elements of a real vector space `V.` 

Let `f`  be a transformation from `V`  to another vector space `W` .

Let `a,` `b,` `c,`  and `k`  be real numbers.

Homogeneity Property

`f(k*x) = k*f(x)`  for all `x in V`  and `k`  in the real numbers.

Additivity Property

`f(x+y) = f(x) + f(y)` for all `x` and `y`  in V.

What does each of these properties imply?

Homogeneity: Assume `f(x)`  has the homogeneity property.

Theorem 1: `f(0) = 0` 

Proof

`f(0) = f(2*0) =2f(0)` 

Now subtract `f(0)`  from both sides.

`f(0) - f(0) =2f(0) - f(0)` 

`0 = f(0)` 

Corollary 1: The graph of `f(x)`  passes through the origin.

Note: the graph of `f(x)`  is the set of points in `V times W`  of the form `(x,f(x)).` 

Theorem 2: `f(x+y) = f(x) +f(y)`  in any 1-dimensional subspace of `V.` 

Proof

Let `z`  be a basis element of the 1-dimensional subspace.

if `x`  and `y`  are any two elements of the subspace then `x=a*z`  and `y=b*z`  for some real numbers `a`  and `b.` 

 `f(x+y) = f(a*z+b*z)= f((a+b)*z) = (a+b)f(z) = a*f(z) +b*f(z) = f(a*z) + f(b*z) = f(x) +f(y)` 

Corollary 2: Any function on the reals that satisfies homogeneity also satisfies additivity.

Theorem 3: The graph of `f(x)`  restricted to a 1-dimensional subspace of V is a one dimensional subspace of `V times W.` 

Proof

Let `z`  be a basis element of the subspace of `V.` 

Let `x`  be any element in the subspace of V.

`x=a*z`  for some real number `a.` 

`f(x) = f(a*z) = a*f(z)` 

So the point `(a*z,a*f(z))`  is on the graph of `f.` 

`(a*z,a*f(z))= a*(z,f(z))` 

This is a point in the Span of `(z,f(z))` 

Likewise the proof works in reverse to show that anything in the span of `(z,f(z))`  is on the graph of `f(x).` 

Corollary 3: If `f`  is a function from `R^1`  to `R^1`  that satisfies homogeneity then the graph of `f(x)`  is a straight line through the origin.

Corollary 4: `sin(2x) ne 2sin(x)` 

Theorem 4: In general, homogeneity does not imply additivity.

Proof

Consider `f:R^2 to R^1`  defined as

 `f(langle x,y rangle) = (3x^2y-y^3)/(x^2+y^2)` 

 `f(k langle x,y rangle) = f(langle kx,ky rangle )` 

`= (3(kx)^2(ky)-(ky)^3)/((kx)^2+(ky)^2)` 

`= (k^3(3x^2y-y^3))/(k^2(x^2+y^2))` 

`=k(3x^2y-y^3)/(x^2+y^2)` 

 `=k*f( langle x,y rangle )` 

So, `f`  has the homogeneity property.

But `f(langle 1,0 rangle ) + f(langle 0,1 rangle) = 0/1+(-1)/1 = -1` 

and `f(langle 1,0rangle +langle 0,1 rangle ) = f(langle 1,1 rangle ) = (3-1)/(1+1) = 1` 

So, `f`  does not have the additivity property.

Take a look at the graph.

  

Additivity: Now we assume that f(x) has the property of additivity.

Theorem 5: `f(0) =0` 

Proof

`f(0)=f(0+0)=f(0)+f(0) = 2f(0)` 

Now subtract f(0) from both sides.

`0 = f(0).` 

Corollary 5: The graph of `f(x)`  passes through the origin.

Theorem 6: f(-1*x) = -1*f(x)

Proof

 `0 = f(0) = f(x-x) = f(x+(-1)*x) = f(x) +f(-1*x)` 

So, `f(-1*x) =-f(x) = -1*f(x)` 

Theorem 7: If f(x) satisfies the additivity property then for any integer `n,`  `f(n*x) = n*f(x)`  for all `x in V.` 

Proof by induction forward and backward

Let `x`  be any element in `V.` 

Case 1: `n=0.` 

`f(0*x) = f(0)=0=0*f(x)` 

Case 2: Assume `ngt0`  and that `f(n*x) = n*f(x)` 

`f((n+1)*x)= f(n*x+1*x) ` 

              `= f(n*x)+f(1*x) ` 

              `=n*f(x) + f(x) ` 

              `=(n+1)*f(x) ` 

Case 3: Assume `nlt0`  and that `f(n*x) = n*f(x)` 

`f((n-1)*x)= f(n*x-1*x)` 

              `= f(n*x+(-1)*x)` 

              `= f(n*x)+f(-1*x)` 

              `=n*f(x) +(-1) f(x)` 

              `=(n-1)*f(x)` 

Therefor `f(n*x) = n*f(x)`  for all integers `n.` 

Theorem 8: If `f(x)`  has the additivity property then for all non-zero integers `n`  we have `f(1/n*x) = 1/n*f(x).` 

Proof

`f(x) = f(1*x) = f(n/n*x) = f(n*(1/n*x))=n*f(1/n*x)` 

Multiply both sides by `1/n`  to get

`1/n*f(x) = f(1/n*x)` 

Corollary 6: If `f(x)`  has the additivity property then for any rational number `p/q`  we have `f(p/q*x) = p/q*f(x)`  for all `x in V.` 

Corollary 7: If `f(x)`  is a function on real numbers that has the additivity property then for any rational number `p/q`  we have `f(p/q) = p/q*f(1).` 

Theorem 9: If `f(x)`  is continuous and satisfies the additivity property then `f(x)`  satisfies the homogeneity property.

Proof

From corollary 6 we have that for all rational values of `k`  we have `f(k*x) = k*f(x)`  for all `x`  in the domain space.

We now need to show homogeneity for irrational values of `k.`  

So, let `k`  be any irrational number.

There exists a sequence of rational numbers, `k_1` , `k_2,`  ... such that `k = lim_(n to oo) k_n` 

let `x`  be any element of the domain of `f` 

`f(k*x) = f(x*lim_(n to oo) k_n)` 

        `= f(lim_(n to oo)k_n*x)` 

       `= lim_(n to oo) f(k_n*x)`     This is because we are assuming f is continuous.

       `= lim_(n to oo) k_n*f(x)`     This is because of corollary 6

       `= (lim_(n to oo) k_n)*f(x)` 

       `= k*f(x)` 

Now I know how not to lie to my students ...

Me: Is your function continuous?  Them: yes

Me: Is the graph a line through the origin?  Them: no.

Me: Then your function does not satisfy either homogeneity of additivity.

This leaves us with the question: Are there any functions that satisfy additivity but not homogeneity?

If we switch to complex vector spaces the answer is yes.

Example: `f(a+bi) = a-bi`    

This satisfies additivity

 `f((a+bi)+(c+di)) = f((a+c)+(b+d)i) `

                                           `= (a+c)-(b+d)i` 

                                           `= (a-bi)+(c-di)` 

                                           `= f(a+bi)+f(c+di)` 

but not homogeneity

`f(i*(a+bi)) = f(-b+ai) = -b-ai` 

 `i*f(a+bi) = i*(a-bi) = b+ai` 

Let's stick with reals, and "construct" an example.