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# Functional Equations(over reals)

A couple of doubts:
Q1)$$f(x+2y)=f(x)+2f(y)$$ given that $$f(1)=2,f(0)=0$$
Q2)$$f(x-f(y))=f(f(y))+xf(y)+f(x)-1$$

Note by Siddharth G
3 years, 1 month ago

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Q1. Putting $$y' = 2y$$, we get $$f(x+y') = f(x) + f(y')$$, which is the Cauchy Functional Equation.

Q2 is IMO 1999/Q6

- 3 years, 1 month ago

Thanks for the second! And I made a typo in the first one, I just edited it. Sorry

- 3 years, 1 month ago

It's almost the same anyways.

Putting $$x = 0$$, we get $$f(2y) = 2f(y)$$.

Now we have $$f(x+y) = f(x+2\frac{y}{2}) = f(x) + 2f(\frac{y}{2}) = f(x) + f(y)$$ , from the above statement. Cauchy's Functional Equation all over again.

- 3 years, 1 month ago