Question

Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR such that f(x) = kx, for every XER

Answer 1

Answer with explanation:

It is given that:

f: R → R is a continuous function such that:

`f(x+y)=f(x)+f(y)------(1)` ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

• f(0)=0

( Since,

f(0)=f(0+0)

i.e.

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

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

• f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

• Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

`f(1)=f((1)/(n)+(1)/(n)+.......+(1)/(n))=f((1)/(n))+f((1)/(n))+....+f((1)/(n))\\\\\\i.e.\\\\\\f((1)/(n)+(1)/(n)+.......+(1)/(n))=nf((1)/(n))=f(1)=k\\\\\\i.e.\\\\\\f((1)/(n))=k\cdot (1)/(n)`

Also,

• when x∈ Q

i.e.
`x=(p)/(q)`

Then,

`f((p)/(q))=f((1)/(q))+f((1)/(q))+.....+f((1)/(q))=pf((1)/(q))\\\\i.e.\\\\f((p)/(q))=p(k)/(q)\\\\i.e.\\\\f((p)/(q))=k(p)/(q)\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q`

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq
`<x_n>` belonging to Q such that:

`<x_n>\to x` )

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence
`a_n` such that:

`a_n\ belongs\ to\ Q`

and

`a_n\to \alpha`

`f(a_n)=ka_n`

( since
`a_n` belongs to Q )

Let f is continuous at x=α

This means that:

`f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha`

This means that:

`f(\alpha)=k\alpha`

This means that:

f(x)=kx for every x∈ R