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

It is given that:

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

 ∀  x,y ∈ R

Now, let us assume f(1)=k

Also,

(  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,

i.e.

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

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

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,

Also,

i.e.  

Then,

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

)

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence such that:

and

( since belongs to Q )

Let f is continuous at x=α

This means that:

This means that:

                       This means that:

                    f(x)=kx for every x∈ R

-60.75

step-by-step explanation:

just add the numbers then put the negative sign on it