Sunday 27 May 2018

real analysis - I want to show that $f(x)=x.f(1)$ where $f:Rto R$ is additive.











I know that if $f$ is continuous at one point then it is continuous at every point.
From this i want to show that $f(x)=xf(1).$
Can anybody help me to proving this?


Answer



HINTS:





  1. Look at $0$ first: $f(0)=f(0+0)=f(0)+f(0)$, so $f(0)=0=0\cdot f(1)$.


  2. Use induction to prove that $f(n)=nf(1)$ for every positive integer $n$, and use $f(0)=0$ to show that $f(n)=nf(1)$ for every negative integer as well.


  3. $f(1)=f\left(\frac12+\frac12\right)=f\left(\frac13+\frac13+\frac13\right)=\dots\;$.


  4. Once you’ve got it for $f\left(\frac1n\right)$, use the idea of (2) to get it for all rationals.


  5. Then use continuity at a point.



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