# “Rising sun” function

Let $f:[0,1] \to \mathbb R$ be bounded and $$f_\odot :[0,1] \to \mathbb R: x \mapsto \sup \{f(y) : y \in [x,1] \}$$ This is well defined since $f$ is bouned.

Claim: If $f$ is continuous then $f_\odot$, too.

I begun as follows: Let $x_0 \in [0,1]$ and $\epsilon > 0$. First: Find $y_0 \in [x_0,1]$ s.t. $f_\odot(x_0) - \epsilon < f(y_0) \leq f_\odot(x_0)$. Then using continuity of $f$ I get a $\delta > 0$ s.t. $\forall a \in (y_0-\delta,y_0+\delta) : f(y_0) - \epsilon < f(a) < f(y_0) + \epsilon$ and thus also $|f(a)-f_\odot(x_0)|<2\epsilon$ for $a \in (y_0-\delta,y_0+\delta)$. Now I get suck. I think that this $\delta$ also works for $f_\odot$ somehow but can't figure it out.

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Does this function have a house? – Asaf Karagila Nov 8 '13 at 18:51
What do you mean ? – Andre Nov 8 '13 at 19:05
@Andre I think he's referring to the House of the Rising Sun. – IBWiglin Nov 8 '13 at 19:21

Note that by continuity of $f$ we have $\sup=\max$ on compact intervals. Thus we may assume that in fact $f_\odot(x_0)=f(y_0)$.

If $f(x_0)<f_\odot(x_0)$, then $y_0>x_0$ and we have $f(x)<f_\odot(x_0)$ in some $\delta$-neighbourhood of $x_0$. Without loss of generality, $\delta<y_0-x_0$ and hence $f_\odot(x)=f(y_0)$ for $x\in(x_0-\delta,x_0+\delta)$.

If on the other hand $f(x_0)=f_\odot(x_0)$, then $f(x)\le f_\odot(x)\le f(x_0)$ for $x\ge x_0$ and if we pick $\delta>0$ such that $|f(x)-f(x_0)|<\epsilon$ for $|x_0-x|<\delta$, then $f(x_0)-\epsilon<f(x)\le f_\odot(x)\le f_\odot(x_0)$ for $x_0\le x<x_0+\delta$ and also $f(x_0)\le f_\odot(x)<f(x_0)+\epsilon = f_\odot(x_0)+\epsilon$ for $x_0-\delta<x\le x_0$.

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Another method to prove this is the following:

1) Show that for any function $f$ (continuous or not), that for $x_1, x_2 \in [0,1]$ such that $x_1 < x_2$, the following equality holds $$0 \leq f_{\odot}(x_1)-f_{\odot}(x_2) = \sup\{f(y) : y \in [x_1,1]\}-\sup\{f(y) : y \in [x_2,1]\} \\\leq \sup\{f(y) : y\in [x_1,x_2]\}-\inf\{f(y):y \in [x_1,x_2]\}$$

2) Assuming that $f$ is continuous then it is uniformly continuous on $[0,1]$. So take $\epsilon >0$ and find a $\delta$ such that for any $x_1, x_2 \in [0,1]$ with $|x_1 - x_2|<\delta$ then $|f(x_1)-f(x_2)| < \epsilon$.

3) Apply part 2 to part 1 and use that when $f$ is continuous, there is some $a \in [x_1,x_2]$ and $b\in [x_1,x_2]$ such that $f(a) = \sup\{f(y) : y\in [x_1,x_2]\}$ and $f(b) =\inf\{f(y):y \in [x_1,x_2]\}$. Therefore, you will have if $|x_1 - x_2|<\delta$ $$|f_{\odot}(x_1) - f_{\odot}(x_2)| \leq |f(a) - f(b)| < \epsilon$$ since $|a-b|< |x_1-x_2|<\delta$.

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This is also a nice method. I just have to ckeck point 1). – Andre Nov 8 '13 at 19:08
Yeah, there's a little bit of work to do here. The method I wrote down is doubtfully the fastest, but more just to understand the "nature" of the function $f_{\odot}$ (ie. it was a fun function to play around with!). It was a good question. – Tom Nov 8 '13 at 19:39
Indeed, it is not hard to show that 1) is true. For interested people: Call the quantities in the second line $\alpha$ resp. $\beta$. Then we want that $f_\odot(x_1) \leq \alpha-\beta + f_\odot(x_2)$. Let $y \in [x_1,1]$. First if $x_1 \leq y \leq x_2$ then $f(y) \leq \alpha$ and $f_\odot(x_2) \geq \beta$ since $\beta \leq f(x_2) \leq f_\odot(x_2)$. If $x_2 < y \leq 1$ then $f(y) \leq f_\odot(x_2)$. Then $\alpha - \beta \geq 0$ which is obvoius. Thus $f(y) \leq \alpha - \beta + f_\odot(x_2)$ for all $y \in [x_1,1]$ which proves 1). – Andre Nov 8 '13 at 19:50