Thursday, 13 November 2014

Solving the equation f(x+t)=f(x)+f(t)+2sqrtf(x)sqrtf(t)



I am trying to solve the equation f(x+t)=f(x)+f(t)+2f(x)f(t) - as in find a function that satisfies this equation. I notice that the RHS is (f(x)+f(t))2 but I am stuck after this.


Answer



The presence of f(x) in your functional equation implies that the range of f is nonnegative.




If you are looking for a continuous function, Simon's comment shows that g must be linear. (With real functions, continuity and additivity imply full linearity. This is mentioned here although I'd prefer to show a link to an actual proof.)



Therefore f(x)=ax



for some a.




  1. If a=0, then f is the zero function.

  2. If a>0, then f is only defined for x0, and f(x)=a2x2.

  3. If a<0, then f is only defined for x0, and f(x)=a2x2.




That is, there are three families of solutions. They all have the same form f(x)=a2x2, but either you must restrict the domain to non-negatives, non-positives, or a is necessarily 0.


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