It is easy to prove that if f(x) is periodic with period T,then $f^'(x+T)=f^{'}(x)$. However I don't know how to prove T is the shortest period for $f^{'}(x)$.
Can anyone help me? Or if it is not true,can anyone give me a counterexample?
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It seems that the hint I have given before was too obscure. So I now give a full solution. By assumption $f'$ is the derivative of a $T$-periodic function $f$, where $T>0$ is a primitive period of $f$. Therefore $f'$ has period $T$ as well. If $f'$ were a constant then this constant would be $0$ (or $f$ would not be periodic). As a consequence $f$ would be constant and would not have a primitive period. Therefore $f'$ has a primitive period $p>0$, which then necessarily is of the form $p={T\over n}$ for some $n\geq1$. Put $\int_0^p f'(t)\ dt=:c$. Then $$\int_x^{x+p}f'(t)\ dt=c$$ for any $x$; furthermore one has $$0=f(T)-f(0)=\int_0^T f'(t)\ dt=n\int_0^p f'(t)\ dt=n\, c\ .$$ Therefore $c=0$, and this in turn implies that $$f(x+p)-f(x)=\int_x^{x+p} f'(t)\ dt=0$$ for all $x$; whence $f$ has period $p$. This implies $p= m T$ for some $m\geq1$, so that we in fact have $p=T$. |
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