# Spivak, Ch. 14, Prob 12d : Prove if $f'$ is periodic with period $a$ and $f$ is periodic (with some period not necessarily $=a$), then $f(a)=f(0)$.

In Spivak's Calculus, there is the following problem in Ch. 14 "Fundamental Theorem of Calculus"

1. A function $$f$$ is periodic, with period $$a$$, if $$f(x+a)=f(x)$$ for all $$x$$.

*(d) Prove that if $$f'$$ is periodic with period $$a$$ and $$f$$ is periodic (with some period not necessarily $$=a$$), then $$f(a)=f(0)$$.

The solution manual solution (with my filling in some intermediate steps) is as follows

Let $$g(x)=f(x+a)-f(x)$$.

Then $$g'(x)=f'(x+a)-f'(x)=0$$, because $$f'$$ is periodic.

Therefore, $$g$$ is constant and $$g(x)=g(0)$$, so

$$f(x+a)-f(x)=f(a)-f(0)$$

$$f(x+a)=f(x)+f(a)-f(0)$$

$$f(na)=f[(n-1)a]+f(a)-f(0)$$

$$=f[(n-2)a+a]+f(a)-f(0)$$

$$=f[(n-2)a]+2f(a)-2f(0)$$

$$=f(0)+nf(a)-nf(0)$$

$$=n(f(a)-f(0))+f(0)$$

Since $$f$$ must be bounded (since it is periodic), then $$f(a)=f(0)$$, otherwise $$f$$ would be unbounded.

Is there an interesting interpretation for this result?

f apparently can have some period $$b\neq a$$ which we didn't need to make use of in the proof above. Doesn't the fact that $$g$$ is constant mean that $$f$$ has a period of $$a$$?

$$g(x)=f(x+a)-f(x)=0, \text{ for all } x$$

$$\implies f(x+a)=f(x), \text{ for all }x$$

When I tried to solve this problem I started on the assumption that $$f$$ had a period $$b\neq a$$

$$f'(x+a)=f'(x)$$

$$f(x)=\int_0^x f' + f(0)$$

$$f(x+b)=\int_0^{x+b}f’ +f(0)=f(x)$$

But didn't get much further than this.

• I think that the question actually tells you that under these circumstances $f$ is periodic with period $a$, doesn't it? Commented Jul 7, 2022 at 4:18
• "Therefore $g$ is constant..." So for all $x$ we have $g(x)=g(0)$. So for all $x$ we have $f(x+a)-f(x)=g(x)=g(0)=f(a)-f(0)=0.$ Commented Jul 7, 2022 at 6:24
• Here is a different problem: Suppose $f'$ is periodic and has arbitrarily small positive periods. Prove that $f$ is constant, i.e. $f'=0$. Commented Jul 7, 2022 at 6:39

At no point do we know that $$g(x)=0$$ for all $$x$$. We only know that $$g'(x)=0$$ for all $$x$$, which means that $$f(x+a)-f(x)=c$$ for some $$c \in \mathbb R$$. Notice that if $$f$$ was just a line, then this property would also hold (and, obviously, unless the line has a slope of $$0$$, said line is not periodic).

I would note that Spivak's claim relies on the assumption that periodicity implies $$f$$ is bounded. In general, this is not true (https://math.stackexchange.com/questions/1004426/does-a-periodic-function-have-to-be-bounded#:~:text=This%20function%20is%20defined%20for,R%20since%20it%20is%20periodic).

In order for periodicity to imply boundedness, we also need to know that $$f$$ is continuous. However, because this problem makes use of $$f'$$, implicitly, we must have that $$f$$ is continuous. Therefore, the periodicity assumption of $$f$$ implies boundedness.

An equivalent result could have been generated by instead stating that $$f$$ is a bounded function (in addition to the other assumptions about $$f'$$). In fact, it is not clear to me why Spivak supplied the 'more specific' assumption that $$f$$ is periodic because, as you have pointed out, no where in the proof do we make use of that fact outside of the boundedness property that comes with periodicity of a continuous function.

Now, to your point about 'does this imply that $$f$$ has a period of $$a$$'...yes, it does. Here is why:

Consider the derivative $$f'$$ on the interval $$[0,a]$$. Because this function repeats itself for every $$n$$ interval of the form $$[(n-1)a,na]$$, it should be clear that in order to correspond to a function $$f$$ that is bounded, we must have that there is no net growth (or loss) of the function $$f$$ across each of these intervals.

Suppose that $$f'$$ carries more positive weight across the $$[0,a]$$ interval. Then, the function $$f$$ will have net growth as we jump from $$[0,a]$$ to $$[a,2a], \cdots$$. i.e. $$f$$ would have no upper bound (similarly for if $$f'$$ carries more negative weight across the $$[0,a]$$ interval...which would lead to net loss as we jump from $$[0,a] \to [a,2a] \to \cdots$$ and prevent $$f$$ from having a lower bound). This is equivalent to stating that our assumptions force us to have $$\displaystyle \int_0^a f'=0$$.

Now, by the second fundamental theorem of calculus, this implies that $$f(a)-f(0)=0 \implies f(a)=f(0)$$. Of course, the above logic applies to any interval $$[x,x+a]$$, which means that $$\displaystyle \int_x^{x+a} f'=0 \implies f(x+a)-f(x)=0 \implies f(x+a)=f(x)$$