# Proof of statement on derivatives of periodic functions.

Let $$y(x)$$ be a continuous function that is also continuous in all derivatives, and periodic in $$x$$ with period of length $$L$$, i.e. $$y(x)$$ and all its derivatives have the same value at $$x$$ as they do at $$x+L$$. $$dy/dx$$ at some $$x$$ between $$0$$ and $$L$$ has the value $$y^\prime$$. Prove that at some point or points between $$0$$ and $$L$$, $$dy/dx$$ has the value $$-y^\prime$$.

Not entirely sure that this is a true statement, but it seems to me intuitively that it has to be.

I thought this would be equivalent to saying some continuous function $$f$$ integrates to zero over some range $$L$$. So, if it has non-zero value $$f_0$$ somewhere, it must have value $$-f_0$$ somewhere else. Which also seems unavoidably true to me, but AFAICT is no easier to prove than the first statement.

So I'm stuck. Any ideas?

• Do you mean the following claim: If $x\in [0,L]$, then there exists $x'\in [0,L]$ such that $y'(x') = -y'(x)$? – amsmath Oct 10 at 18:17
• @amsmath Yes, thanks. – bob.sacamento Oct 10 at 19:19

It's not necessarily true. Consider if $$f$$ is a sawtooth-like function, going slowly up, then quickly down, repeating periodically (this is very possible to make smooth, even though that example is not). In that case, the large negative derivative at the "quickly down" portion is never met with a correspondingly large positive derivative in the "slowly up" portion.
As a concrete smooth example, take any monotonically increasing, smooth function $$f:\Bbb R\to \Bbb R$$ with the property that $$f(x) = 0$$ for $$x<0$$ and $$f(x) = 1$$ for $$x>1$$. Let $$L = \frac32$$, and define $$y(x)$$ on $$[0, \frac32)$$ to be the following: $$y(x) = \cases{f(x)& if 0\leq x<1\\ 1-f(2x - 2) & if 1 with periodic extension to all of $$\Bbb R$$.
• Yes. I see your point. Just one tweak with your second example ... $f$ would have to be chose s.t. $f^\prime=0$ at 0 and 1 for the $y$ to be continuous in the first derivative. – bob.sacamento Oct 10 at 19:31
• Right now we have $y(0) = 0$ and $y(3/2) = 1$... – amsmath Oct 10 at 19:42
• @bob.sacamento That's already taken care of. I said that $f$ is smooth. "Smooth" means all derivatives are continuous everywhere. Maybe I should've clarified that. – Arthur Oct 10 at 19:54
• @amsmath You're right. I mirrored the graph of $f$ both vertically and horizontally. That is now fixed. – Arthur Oct 10 at 19:55