VJIMC 2017, Category I, Problem 1 : Functional Equation
by adityaguharoy, Apr 6, 2017, 3:59 AM
Problem : (The problem that appeared after translation )
Let
be a continuous function satisfying
for every
. Prove that
is constant.
Solution
Let

![\[f(x+2y)=2f(x)f(y)\]](http://latex.artofproblemsolving.com/4/b/b/4bbc0d1c4233494baa8d98cab6ec3c49c2118672.png)


Solution
(Thanks to Michael Tang and Tintarn (Tintaarn actually showed the official solution)
Switching the roles of
and
we find that
for all
. Thus, taking
, we find
for all
, which is enough to show that
is constant.
Furthermore, if
be the assertion that
then using
gives us
and thus
.
So, the required solutions are :
or
.
Done !!
(See that we have not used the continuity).
But here is the official solution :
Official solution
Switching the roles of








Furthermore, if





So, the required solutions are :


Done !!
(See that we have not used the continuity).
But here is the official solution :
Official solution
By taking
we obtain
for any
. If
, it follows that
for all
and
is a constant function.
Therefore, in the rest of the proof assume that
. After taking
in
we get
for any
.
Let
. One can easily check that
for all
. Therefore, since
is continuous, we obtain that
and the proof is complete.







Therefore, in the rest of the proof assume that





Let




![\[f(a)=\lim_{n \to \infty} f(a)=\lim_{n \to \infty} f\left(\frac{a}{2^n}\right)=f\left(\lim_{n \to \infty} \frac{a}{2^n}\right)=f(0)=\frac{1}{2}\]](http://latex.artofproblemsolving.com/e/5/1/e519431d0d56a80396e44a498c0b367c0e8092e8.png)