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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??

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

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

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

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