Respuesta :
Answer with explanation:
It is given that:
f: R → R is a continuous function such that:
[tex]f(x+y)=f(x)+f(y)------(1)[/tex] ∀ 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,
[tex]f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}[/tex]
Also,
- when x∈ Q
i.e. [tex]x=\dfrac{p}{q}[/tex]
Then,
[tex]f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q[/tex]
(
Now, as we know that:
Q is dense in R.
so Э x∈ Q' such that Э a seq [tex]<x_n>[/tex] belonging to Q such that:
[tex]<x_n>\to x[/tex] )
Now, we know that: Q'=R
This means that:
Э α ∈ R
such that Э sequence [tex]a_n[/tex] such that:
[tex]a_n\ belongs\ to\ Q[/tex]
and
[tex]a_n\to \alpha[/tex]
[tex]f(a_n)=ka_n[/tex]
( since [tex]a_n[/tex] belongs to Q )
Let f is continuous at x=α
This means that:
[tex]f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha[/tex]
This means that:
[tex]f(\alpha)=k\alpha[/tex]
This means that:
f(x)=kx for every x∈ R
