Solving the functional equation $ f \left( \sqrt { x ^ 2 + y ^ 2 } \right) = f ( x ) f ( y ) $

Question

I happened to run across this problem in Larson's Problem Solving through Problems and I wanted to ask how you would approach it.

A real-valued continuous function satisfies for all real $ x $ and $ y $ the functional equation $$ f \left( \sqrt { x ^ 2+ y ^ 2 } \right) = f ( x ) f ( y ) \text . $$ Prove that $ f ( x ) = { [ f ( 1 ) ] } ^ { x ^ 2 } $.

Attempt:

I know that if we define $ f ( n ) = 2 ^ { \frac n 2 } $, $ f ( 1 ) = 2 ^ { \frac 1 2 } $ resulting in $ f ( n ) = 2 ^ { \frac n 2 } = 2 ^ { \frac { n ^ 2 } 2 } $. But this doesn't seem to make sense for me. Any clues?

EDIT:

If I were to use the hint given in the book where I am asked to first prove the theorem for all numbers of the form $ 2 ^ { \frac n 2 } $, where $ n $ is an integer and then prove the theorem for all numbers of the form $ \frac m { 2 ^ n } $, with $ m $ an integer and $ n $ a nonnegative integer, how would that help me?

Answer 1

HINT: Consider the cases in which $y=x$. Then you have that $$f(y\sqrt{2})=f(y)^2$$ Then observe the following pattern: $$f(\sqrt2)=f(1)^2$$ $$f(2)=f(1)^4$$ $$f(2\sqrt2)=f(1)^8$$ And so on. Using induction, we can say that $$f(\sqrt2^n)=f(1)^{2^n}$$ Can you use this to prove what you are after?

If you need another hint, just ask!

Answer 2

let $x=y$ \begin{eqnarray*} f(\sqrt{2}y)=(f(y))^2 \end{eqnarray*} now let $x=\sqrt{2}y$ and so on \begin{eqnarray*} f(\sqrt{3}y)&=&(f(y))^3 \\ f(\sqrt{4}y)&=&(f(y))^4 \\ f(\sqrt{5}y)&=&(f(y))^5 \\ &\vdots& \\ f(\sqrt{n}y)&=&(f(y))^n \end{eqnarray*} Now let $n=x^2$ and $y=1$ & we have $\color{red}{f(x)=(f(1))^{x^2}}$.

Answer 3

Note that the constant zero function satisfies the equation. So here you must accept that $ 0 ^ x = 0 $, for all real numbers $ x $ ( which is problematic when $ x $ is close to $ 0 $). Otherwise your claim that $ f ( x ) = f ( 1 ) ^ { x ^ 2 } $ is not correct.

Letting $ x = y = 0 $ in $$ f \bigg( \sqrt { x ^ 2 + y ^ 2 } \bigg) = f ( x ) f ( y ) \tag 0 \label 0 $$ we have $ f ( 0 ) \big( f ( 0 ) - 1 \big) = 0 $ and thus $ f ( 0 ) = 0 $ or $ f ( 0 ) = 1 $. If $ f ( 0 ) = 0 $, then letting $ y = 0 $ in \eqref{0} we get $ f ( | x | ) = 0 $. Now letting $ y = x $ in \eqref{0}, we have $ f ( x ) ^ 2 = f \big( \big| \sqrt 2 x \big| \big) $ and hence in this case $ f $ is identically zero. So, from now on, we assume that $ f ( 0 ) = 1 $. In this case, letting $ y = 0 $ in \eqref{0} we get $ f ( | x | ) = f ( x ) $ which means $ f $ is an even function. Also since $ f ( | x | ) = f \Big( \frac x { \sqrt 2 } \Big) ^ 2 $, we have $ f ( x ) \ge 0 $. For convenience, we define $ g : [ 0 , + \infty ) \to [ 0 , + \infty ) $ with the equation $ g ( x ) = f \big( \sqrt x \big) $. Substituting $ \sqrt x $ for $ x $ and $ \sqrt y $ for $ y $ in \eqref{0}, we have $$ g ( x + y ) = g ( x ) g ( y ) $$ which yields $$ g ( x ) ^ m = g \Big( \frac m n x \Big) ^ n \tag 1 \label 1 $$ for every positive integers $ m $ and $ n $, using a simple induction. Using continuity of $ f $ at $ 0 $, one can show that for large enough $ n $, $ g \big( \frac x n \big) > 0 $, which by \eqref{1} gives us $ g ( x ) > 0 $. By continuity, this lets us generalize $ g \big( \frac m n \big) = g ( 1 ) ^ { \frac m n } $ to $ g ( x ) = g ( 1 ) ^ x $ for every non-negative real number $ x $. Finally we have $ f ( x ) = f ( | x | ) = f \Big( \sqrt { x ^ 2 } \Big) = g \big( x ^ 2 \big) = g ( 1 ) ^ { x ^ 2 } = f ( 1 ) ^ { x ^ 2 } $.