"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.
 A: 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 $x_0-\delta<x\le x_0$.
A: 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 $|a-b|< |x_1-x_2|<\delta$.
