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I have three related functions on the closed interval $[0, 1]$:

  1. $f(x) = \max(f_S(x), f_C(x))$
  2. $f_S(x) = x+\frac 1 2$
  3. $f_C(x) = (1-x)f(\frac {x+1} 2)$

We are also given that for $0 \leq x \leq 1$, we have $\sup f(x) \leq 2$, that is, that $2$ is an upper bound of $f$ on that interval, but not necessarily the least upper bound.

I have been trying to find when $f_S(x) \geq f_C(x)$.

For $x = \frac 1 2$

we have $f_S(\frac 1 2) = 1$ and $f_C(\frac 1 2) = \frac 1 2 f(\frac 3 4) \leq \frac 1 2 \sup f(x) \leq \frac 1 2 \cdot 2 = 1$, so we have

$$f_S(\frac 1 2) = 1 \geq f_C(\frac 1 2)$$

For $x = \frac 1 2 + \epsilon$

Now suppose we have $x = \frac 1 2 + \epsilon$ with $\epsilon > 0$ which is just above $\frac 1 2$. Then

  • $f_S\big(\frac 1 2 + \epsilon\big) = 1 + \epsilon$
  • $f_C\big(\frac 1 2+ \epsilon\big) = (\frac 1 2 - \epsilon) \cdot f\big(\frac 3 4 + \frac \epsilon 2\big) \leq (\frac 1 2 - \epsilon) \cdot \sup f(x) \leq (\frac 1 2 - \epsilon) \cdot 2 = 1 - 2\epsilon$

So we have

$$f_S\bigg(\frac 1 2 + \epsilon\bigg) = 1 + \epsilon \geq 1 - 2\epsilon \geq f_C\bigg(\frac 1 2 + \epsilon\bigg)$$

For $x = \frac 1 2 - \epsilon$

Now suppose we have $x = \frac 1 2 - \epsilon$ with $\epsilon > 0$, which is just below $\frac 1 2$. Then

  • $f_S\big(\frac 1 2 - \epsilon\big) = 1 - \epsilon$
  • $f_C\big(\frac 1 2 - \epsilon\big) = (\frac 1 2 + \epsilon) \cdot f\big(\frac 3 4 - \frac \epsilon 2 \big) \leq (\frac 1 2 + \epsilon) \cdot \sup f(x) \leq (\frac 1 2 + \epsilon) \cdot 2 = 1 + 2\epsilon$

So we have

$$f_S\bigg(\frac 1 2 - \epsilon\bigg) = 1 - \epsilon \leq 1 + 2\epsilon \geq f_C\bigg(\frac 1 2 - \epsilon\bigg)$$

Since there was no constraint on the magnitude of $\epsilon$, I think this lets us reason that $f_S(x) \geq f_C(x)$ for $x \geq \frac 1 2$; therefore,

$$ f(x) = \begin{cases} \max(f_S(x), f_C(x)) &\text{if } 0 \leq x < \frac 1 2\\ f_S(x) &\text{if } \frac 1 2 \leq x \leq 1 \end{cases} $$

joseville
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    $\lim\limits_{n\to\infty} \dfrac {2^{n-1} + 1} {2^n} = \dfrac12$ exactly. I don't think that is the definition of $\dfrac12^+$ that you want. – David K Mar 28 '23 at 16:41
  • Maybe $\frac 1 2 + \epsilon$ with $\epsilon > 0$ – joseville Mar 28 '23 at 17:49
  • Edited to use more standard $\pm \epsilon$ notation. – joseville Mar 28 '23 at 18:08
  • Used to be related to: https://math.stackexchange.com/q/4668170/833760 – joseville Mar 28 '23 at 18:10
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    From the given relations, $$f(0)\ge f_S(0) = \frac12\ f(-1) \ge f_C(-1) = (1+1)f\left(\frac{-1+1}2\right)\ge1\ f(-3) \ge f_C(-3) = (1+3)f\left(\frac{-3+1}2\right)\ge4$$ So the given $\sup f(\mathbb R) \leq 2$ doesn't hold? – peterwhy Mar 28 '23 at 18:25
  • @peterwhy You're right. I should have restrained the functions to $[0, 1]$. Edited the question to reflect this. – joseville Mar 28 '23 at 18:43
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    Now you have a closed form of $f(x)$ for $\frac12 \le x \le 1$, then you can get closed forms of both $f_C(x)$ and $f(x)$ for $0\le x<\frac 12$ too. – peterwhy Mar 28 '23 at 19:20
  • @peterwhy Thanks! I see now! – joseville Mar 28 '23 at 22:20

1 Answers1

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$$ f(x) = \begin{cases} \max(f_S(x), f_C(x)) &\text{if } 0 \leq x < \frac 1 2\\ f_S(x) &\text{if } \frac 1 2 \leq x \leq 1 \end{cases} $$

Since $\frac 1 2 \leq x \leq 1$ is solved, let's focus on the unsolved interval $0 \leq x < \frac 1 2$ as @peterwhy suggested.

Recall $f_C(x) = (1-x)f\big(\frac {x+1} 2\big)$, but note that on the unsolved interval, we have $\frac 1 2 \leq \frac {x+1} 2 < \frac 3 4$ so $f\big(\frac {x+1} 2\big) = f_S\big(\frac {x+1} 2 \big) = \frac {x+1} 2 + \frac 1 2 = \frac x 2 + 1$; therefore, on the unsolved interval $f_C(x) = (1-x)\big(\frac x 2 + 1\big) = -\frac {x^2} 2 - \frac x 2 + 1$ and

$$f(x) = \max\bigg(x + \frac 1 2, -\frac {x^2} 2 - \frac x 2 + 1\bigg) \quad 0 \leq x < \frac 1 2$$

Setting $x + \frac 1 2 = -\frac {x^2} 2 - \frac x 2 + 1$ and gathering all the terms on one side, we have $0 = \frac {x^2} 2 + \frac 3 2 x - 1$, which has zeros at $x_{\pm} = \frac {-3 \pm \sqrt{13}} 2$. $x_-$ is outside the interval, so we ignore it.

In conclusion

$$ f(x) = \begin{cases} -\frac {x^2} 2 - \frac x 2 + 1 & 0 \leq x < x_+\\ x + \frac 1 2 & x_+ \leq x \leq 1 \end{cases} $$

where $x_+ = \frac {-3 + \sqrt{13}} 2 \approx 0.303.$

https://www.desmos.com/calculator/phngo4khea

Strategy:

  1. Solve $f$ on interval $[\frac 1 2, 1]$
  2. $f$ on interval $[\frac 1 2, \frac 3 4]$, allows solution of $f_C$ on interval $[0, \frac 1 2]$
  3. which allows solution of $f$ on the remaining interval $[0, \frac 1 2]$
joseville
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