# Solving a functional equation (from a problem solving textbook)

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

The problem: A real-valued continuous function satisfies for all real x and y the functional equation: $\ f(\sqrt{x^2+y^2})=f(x)f(y)$.

Prove that $\ f(x)={[f(1)]}^{x^2}$.

Attempt: I know that if we define $\ f(n)=2^{n/2}$, $\ f(1)=2^{1/2}$ resulting in $\ f(n)=2^{n/2} ={2}^{n^2/2} ?$ 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 $\ 2^{n/2},$ where n is an integer and then prove the theorem for all numbers of the form $\ m/2^{n}$, m an integer and n a nonnegative integer, how would that help me??

• Hint: the functional equation $f(x+y)= f(x) +f(y)$ is well known. You can transform yours into that. – spaceisdarkgreen Jun 20 '17 at 22:33
• @JaeKim If you found any of these answers helpful, you should probably accept one of them. :) – Frpzzd Jun 27 '17 at 17:33

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!

• Thank you! I will get back to you after my efforts to solve it :) – Jae Kim Jun 20 '17 at 22:37
• Glad I could help! If you found this helpful, don't forget to $\uparrow$! :) – Frpzzd Jun 20 '17 at 22:38
• @JaeKim Would you like another hint? – Frpzzd Jun 20 '17 at 22:44

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 $$f \bigg( \sqrt { x ^ 2 + y ^ 2 } \bigg) = f ( x ) f ( y ) \tag { 0 }$$ 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 $( 0 )$ we get $f ( | x | ) = 0$. Now letting $y = x$ if $( 0 )$, 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 $( 0 )$ 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 $( 0 )$, we have $$g ( x + y ) = g ( x ) g ( y ) \tag { 1 }$$ which yields $$g ( x ) ^ m = g \Big( \frac m n x \Big) ^ n \tag { 2 }$$ 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 $( 2 )$ 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 $f ( x ) = f ( | x | ) = f \Big( \sqrt { x ^ 2 } \Big) = g \big( x ^ 2 \big) = g ( 1 ) ^ { x ^ 2 } = f ( 1 ) ^ { x ^ 2 }$.

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 $\color{red}{f(x)=(f(1))^{x^2}}$.