Problem 5

Let

f

be a real-valued function defined for all real numbers

x

such that, for some positive constant

a

, the equation

f(x+a)=\frac12+\sqrt{f(x)-f(x)^2}

holds for all

x

. (a) Prove that

f

is periodic. (b) For

a=1

, give an example of a non-constant function with the required property.

Step 3 of 5: A second iteration returns to f(x)

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In plain words

The square root creates an absolute value, but the global lower bound from Step 1 chooses its sign. One application moves the value through a nonlinear map; two applications act as the identity, which is exactly periodicity with period $2a$ .

f(x+2a)=\frac12+\sqrt{f(x+a)-f(x+a)^2}=\frac12+\left|f(x)-\frac12\right|=f(x)

Detailed analysis

Apply the original equation with $x$ replaced by $x+a$ . The identity from Step 2 gives $f(x+a)-f(x+a)^2=(\frac12-f(x))^2$ , hence $f(x+2a)=\frac12+\sqrt{(\frac12-f(x))^2}=\frac12+\left|f(x)-\frac12\right|$ . By Step 1, $f(x)\ge\frac12$ , so the absolute value is $f(x)-\frac12$ and $f(x+2a)=f(x)$ for every $x$ .