Integrating a Periodic Function

Question

Problem: Suppose $f:\mathbb R→\mathbb R$ is continuous and has period $p$, so that $f(x+p)=f(x)$ for all $x\in \mathbb R$. Show that $\int_{x}^{x+p}f(t)dt$ is independent of $x$ in that, for all $x,y$

$$\int_x^{x+p}f(t)dt=\int_y^{y+p}f(t)dt.$$

Show then, that $\int_0^p[f(x+a)-f(x)dx]=0$ for any number $a$. Conclude that for any number $a$, there is $x$ such that $f(x+a)=f(x)$.

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I do not know how to start this problem. I thought about defining a function $g(x)=f(x+p)-f(x)$. This is not equal to $\int_x^{x+p}f(t)dt$ right? Even though there seems to be a close relationship by the FTC. This is a clear thing when discussing $\cos(x)$ or $\sin(x)$, trig functions since they do have this periodicity. How am I supposed to show this is $x$ independent? Then it seems like we're supposed to integrate the original integral.

Answer 1

The derivative of $\int_x^{x+p} f(t) dt $ w.r.t. $x$ is $f(x+p)-f(x)$ which is $0$ by hypothesis. Hence this function is a constant and $\int_x^{x+p} f(t) dt =\int_y^{y+p} f(t) dt $ for any $x,y$. In particular $\int_x^{x+p} f(t) dt =\int_0^{p} f(t) dt $. Make the substitution $u=t-x$ to get $\int_0^{p} f(u+x) du =\int_0^{p} f(t) dt $. This can be written as $\int_0^{p} [f(t+x)-f(t)] dt=0$. Now the integrand here (which is continuous) cannot take only positive values or only negative values since the integral won't be $0$ in such cases. Hence it must attain the value $0$ at some point.

( I proved that for every $x$ there exists $t$ such that $f(x+t)=f(t)$ which is same as what we are asked to prove with slight change of notations).