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Let $ g:\mathbb{R} \to \mathbb{R}$ be a continuous function: $g(x) = g(x+1)$. Show that $g$ satisfies the equation:

$g(x)=\frac{1}{4} \left(g(\frac{x}{2})+g(\frac{x+1}{2})\right)$ for all $x$.

Is it enough to first assume that $g$ satisfies the equation and to check that $g(x)=g(x+1)$ if it does?

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  • $\begingroup$ If $g$ is the constant $1$ function and you plug it in your equation for $f$ you get $1=\frac14(1+1)=\frac12$. From here you could prove everything. $\endgroup$ – Mirko Nov 29 '15 at 1:28
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It is not enough to assume that $g$ satisfies $$g(x)=\frac{1}{4}\left(g(\tfrac{x}{2})+g(\tfrac{x+1}{2})\right),$$ and then check that $g(x)=g(x+1)$. This is the converse of what you are being asked to prove, and it is not true in general.

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