Is there a function satisfying the following conditions:

Question

I was wandering if it is possible to find a continuous function, besides $f(x)=1$, satisfying the following conditions: \begin{align*} &f(x)\geq 0, \text{for all } x\in[0,1),\\ &f(0)=1, \\ &f(x)\geq f(2x), \text{for all } x\in[0,1)\\ &f(x)=f\left(x+\frac{1}{2}\right). \end{align*} Given that such a function has to be $1/2$-periodic, I have been trying to do this by looking for trigonometric polynomials of form \begin{equation} f(x)=\frac{a_{0}}{2}+\sum_{k=1}^{N}(a_{k}\cos(4\pi kx)+b_{k}\sin(4\pi kx)), \end{equation} but so far, the only function I can come up with is $f(x)=1$. I think the reason for this is that satisfying conditions 3 and 4 simultaneously is very difficult, if not impossible?

Answer 1

We can find lots of points where $f$ is constant by arguments such as the following: $$f(0.1) \ge f(0.2) \ge f(0.4) \ge f(0.8) = f(0.3) \ge f(0.6) = f(0.1),$$ so we must have $f(0.1)=f(0.2)=f(0.3)=f(0.4)=f(0.6)=f(0.8)$.

So let's extend this to a proof that if $f$ is also continuous, then $f(x)=1$ for all $x$, which is essentially making sure that we can get these chains of equal values to come arbitrarily close to every element of $[0,1)$. This is a bog-standard analysis proof, except with some number theory thrown in.

Choose an arbitrary $x^* \in [0,1)$, and let $\epsilon>0$ also be arbitrary. By continuity of $f$, we can find $\delta_1, \delta_2$ such that $$|x - 0| < \delta_1 \Rightarrow |f(x) - f(0)| < \frac{\epsilon}{2}$$ and $$|x-x^*| < \delta_2 \Rightarrow |f(x) - f(x^*)| < \frac{\epsilon}{2}.$$

Next, pick $k$ large enough that $\frac1{2^k-1} < \min\{\delta_1,\delta_2\}$. Let $n = \left\lfloor \frac{2^k}{x^*}\right\rfloor$ or $n = \left\lceil \frac{2^k}{x^*}\right\rceil$, whichever is odd. Either way, $|n - \frac{2^k}{x^*}| < 1$, which means that $$\left|\frac{2^k}{n} - x^*\right| < \left|\frac{2^k}{2^k/x^* - 1} - x^*\right| = \left|\frac{(x^*)^2}{2^k-x^*}\right| < \frac1{2^k-1} < \delta_2.$$ We also have that $|\frac1n| < \frac1{2^k-1} < \delta_1$.

Now begin the chain $$f(\tfrac1n) \ge f(\tfrac2n) \ge f(\tfrac4n) \ge \dots \ge f(\tfrac{2^k}{n}) \ge \dots$$ which, because $n$ is odd, eventually gets back to $f(\tfrac1n)$, making all of these values equal. In particular, $f(\tfrac1n) = f(\tfrac{2^k}{n})$, and we have $$|f(x^*) - f(0)| \le |f(x^*) - f(\tfrac{2^k}{n})| + |f(\tfrac{2^k}{n}) - f(\tfrac{1}{n})| + |f(\tfrac{1}{n}) - f(0)| < \frac{\epsilon}{2} + 0 + \frac{\epsilon}{2} = \epsilon.$$ But this is true for arbitrary $\epsilon>0$, so we must have $f(x^*) = f(0)$, as desired.