Given $f$ continuous, $T>0$ and supposed $\int_x^{x+T} f(t)dt=\int_0^T f(t)dt$, show that $f$ is periodic with period $T$

Question

Given $T>0$ and supposed $f$ continuous for all $x\in\mathbb{R}$, show that if $\int_{x}^{x+T} f(t)dt=\int_0^T f(t)dt$ then $f$ is periodic for all $x\in\mathbb{R}$ with period $T>0$.

My approach is the following: since by hypothesis $f$ is continuous, then by the fundamental theorem of calculus its integral function is differentiable; so, since by hypothesis $\int_x^{x+T}f(t)dt=\int_0^T f(t)dt$, taking derivatives both sides it is $$\frac{d}{dx}\int_x^{x+T}f(t)dt=\frac{d}{dx}\int_0^T f(t)dt \implies f(x+T)-f(x)=0 \implies f(x+T)=f(x)$$ That is, $f$ is periodic with period $T>0$.

1. Is my approach correct? I am unsure because I know that a period is defined as the minimum $T>0$ such that $f(x+T)=f(x)$, so how do I know that the equality $f(x+T)=f(x)$ I've found is satisfied for the minimum $T$? Or in this case the problem is asking just a period and not the minimum one?

2. I believe that this result holds for non continuous function as well (just integrable ones), so for those functions the proof has to change because I can't use the fundamental theorem of calculus anymore, so maybe I'm using too much regularity to prove this.

3. I know that if $f$ is continuous for all $x\in\mathbb{R}$ and periodic with period $T>0$, then $\int_x^{x+T}f(x)dx=\int_0^T f(x)dx$, combining this with the result in this problem is correct to affirm that "Given $f$ continuous and periodic for all $x\in\mathbb{R}$, $f$ is periodic if and only if $\int_x^{x+T} f(t) dt=\int_0^T f(t) dt$"? If yes, can this be taken as an equivalent definition of periodic functions?

4. My textbook writes $\int_x^{x+T} f(t)dt=\int_0^x f(t)dt+\int_x^T f(t)dt+\int_T^{x+T} f(t)dt$ and proceeds in another way. Why he assumes that $x$ is "after" $0$ in the integration interval and $x$ is lesser than $T$? Can he do this WLOG because they are generic values and if it is not like this the proof works the same just by inverting them or is there another reason?

Answer 1

1. Yes, your approach is fine. But you cannot prove that $T$ is the least positve number with $f(x+T)=f(x)$. Very often Mathematicians say $f$ is periodic with period $T$ just to mean that $f(x+T)=f(x)$.

2. There is a theorem of Lebsgue which extends FTC to locally integrable functions. So, as long as the integrals in the question exist we can prove that $f(x+T)=f(x)$ holds for almost all $x$.

3. The given property is indeed equivalent to the condition $f(x+T)=f(x)$ for continuous functions.

4. $\int_a^{b}f(x)dx$ is defined as $-\int_b^{a}f(x)d$ when $ a>b$, so your book is not assuming that $0 <x<T$.

Answer 2

Some comments;

1. The question intends you to show $T$ is one possible period, but does not imply $T$ is the least such. Indeed, the property is true when applied to constant $f(x)$ for all $T$ or $f(x)=\sin x$ for $T=2\pi, 4\pi, 6\pi, $ etc.
2. It is also true when $f$ is merely integrable over finite intervals of length $T$. Taken in the sense of the Lebesgue integral, the conclusion $f(x)=f(x+T)$ is true for almost all $x$ (i.e. the set on which it is not true has zero measure). That follows from the fact that if $f(x)$ is integrable over $[a,b]$, $\int_a^x f(t) dt $ is differentiable and its derivative is equal to $f(x)$ for almost all $x, a \leqslant x\leqslant b$ (see for example, Real Analysis, Stein and Shakarchi, Princeton University Press to see definitions and proof). Therefore the same argument as in (1) can then be applied.
3. Your statement is correct but could be better phrased as, "Given $f$ is continuous [integrable .. ] on every finite interval of length $T$ then $f$ is periodic with a period $T$ if and only if $\int_x^{x+T} f(t)dt = \int_0^T f(t) dt$ for every $x$.
4. The equality in you text book does not appear to be correct regardless of the sign of $x$ unless $x$ happens to be zero.