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Determine all $a\in[0,1]$ such that for ${\it every}$ continuous function $f:[0,1]\to \mathbb{R}$ with $f(0)=f(1)$ there exists at least one $x$ where $f(x) = f(x+a)$.

First of all, $a=0,1/2,1$ are obviously good values. Secondly, no $1/2<a<1$ works (look at $f = \sin(2\pi x)$).

I remember trying to solve this several years ago. I think I started by dividing $[0,1]$ into regions where $f\geq 0$ and $f<0$. Then studied each interval and all pairs of intervals with the same sign. However, this seems like a problem that should have a simple solution and I would love to see it.

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2 Answers 2

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The set you're looking for is $$ \{0\} \cup \left\{\frac{1}{n}\right\}_{n\in \mathbb N}. $$

My proof (not a really "simple" one!) is going to be somewhat "visual"; I'm also going to only consider (WLOG) functions such that $ f(0) = f(1) = 0 $.

If $ a = 1/n $ for a natural number $ n $, you can partition the interval $ [0,1] $ into $ n $ equally broad "bands". Let $ f $ be any continuous function; one can show that if the function is not "overall strictly increasing" ("o.s.i."; I made up this terminology) over each band, in the sense that $ f(ka) < f((k+1)a) \; \forall k $, then $$ (*)\qquad f(x) = f(x+a) $$ for some $ x $. To see this, draw the graph of $ f(x+a) $, by moving the graph of $ f(x) $ forward by $ a $, and notice that the condition $ (*) $ means that the graph of $ f(x) $ should cross that of $ f(x+a) $, which will necessarily happen if the "o.s.i." condition isn't fulfilled (you can convince yourself this is true by either visually picturing the described situation or applying the intermediate value theorem to $ f(x+a) - f(x) $). But clearly, a function satisfying this o.s.i. condition cannot simultaneously satisfy $ f(0) = f(na) = f(1) $. Therefore $ a $ belongs to your set.

On the other hand, if $ a \ne 1/n $ for a natural $ n $, you should be able to explicitly conjure up a function $ f $ for which $ f(x) \ne f(x + a) $ for any $ x \in [0,1] $. I came up with the following: let $ 1 = na + \delta $, with $ 0 < \delta < n $ by the hypothesis on $ a $. I'm going to define the value of $ f $ at certain points, and then $ f $ may be build by simply joining them with line segments. I'm choosing an arbitrary $ \varepsilon > 0 $ and then setting $$ \begin{cases}f(ka + \delta) = (n-k)\varepsilon \quad&\mathrm{for}\; k = 0,1,\ldots,n \\ f(ka) = -k\varepsilon \quad&\mathrm{for}\; k = 0,1,\ldots,n.\end{cases} $$ This function should not cross with its own shifted version. The gist is that in this case the above "band partition" isn't able to account for the whole $ [0,1] $ interval, so you can avoid crossing the shifted-function's graph by following a suitably oscillating path.

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  • $\begingroup$ Thank you for your answer! I will study it later tonight, but right now I don't see how your last construction can work. Lets try to rule out $a=0.49$. We take $n=2$ and $\delta = 0.02$. The constructed function is stricktly negative on $(0,0.5)$ and therefore all values $a\in (0,0.5)$ are achieved right? $\endgroup$
    – Winther
    May 28, 2014 at 17:59
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    $\begingroup$ @Winther: my bad, index $k$ should run in 0 thru $ n $ for both formulas; if you draw and join the points it should be clear that $ f $, if $a=.49$, is positive on $(0,2(a-\varepsilon)/3) $, then switches sign. Also, it doesn't meet $ f(x+a) $, since the latter's graph is just the former one's shifted upward by $\varepsilon$ ($f$ is in fact tailored so as to have this property). $\endgroup$
    – derpy
    May 28, 2014 at 18:47
  • $\begingroup$ I have read it now and agree with your solution. In fact, your proof cannot be simpler:) Dropping the unnecessary packaging of it we're left with: i) a simple application of the IVT and ii) a construction of a counter example for the remaining values. It can be compressed to a stringent, but still easy to read, 10 line proof:) Nitpick/cosmetics: $\epsilon$ is not needed and the $f(1)=f(0)=0$ is not needed (never used). Again, thanks for the effort:) $\endgroup$
    – Winther
    May 28, 2014 at 20:08
  • $\begingroup$ @Winther: you're right about $ f(0) = 0 $ being unnecessary (I chose it WLOG in order to have an easier-to-write counterexample); how do you get rid of the epsilon, however? You need to pick a positive value for the graph of $ f(x+a) $ to be upshifted by, don't you? $\endgroup$
    – derpy
    May 29, 2014 at 8:46
  • $\begingroup$ I meant that $\epsilon$ can be put to $1$ without taking away anything. $\endgroup$
    – Winther
    May 29, 2014 at 14:52
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Decomposition into Fourier series \begin{eqnarray*} f(x) &=&\sum_{k}f_{k}\exp [2\pi ikx],\;x\in \lbrack 0,1] \\ f(x+a) &=&\sum_{k}f_{k}\exp [2\pi ik(x+a)] \end{eqnarray*} We must have \begin{equation*} \exp [2\pi ika]=1 \end{equation*} for some $k.$ Fix $k$. Then this is the case for $a=1/n$ for some suitable integer $n$. Then \begin{equation*} f_{k}(x)=f_{k}\exp [2\pi ikx] \end{equation*} fills the bill. For real $f(x)$ take the real part of $f(x)$ \begin{equation*} f(x)=\sum_{k}\{f_{k}\exp [2\pi ikx]+\bar{f}_{k}\exp [-2\pi ikx]\} \end{equation*}

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  • $\begingroup$ I don't think you have read the question properly. I'm looking for the range of $a$ values for which the property holds for all $f$. $\endgroup$
    – Winther
    May 28, 2014 at 12:14
  • $\begingroup$ Yes, I misunderstood the question. $\endgroup$
    – Urgje
    May 31, 2014 at 18:31

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