Show that $f^*(x) = \sup \{ f(y) : a \leq y \leq x \}$ is a non-decreasing continuous function I am currently working on a problem and stuck on it. Here is the problem (it comes form Elementary analysis, the theory of Calculus by K. Ross P.153):
Q: Let $f$ be a continuous function on [a,b]. Show the Function $f^*(x) = \sup \{ f(y) : a \leq y \leq x \}$  for $x$ in [a,b], is an non-decreasing continuous function on [a,b].
-I know that since the interval is closed then the function must be uniformly continous over the interval but that won't help (at least I don't think).
-Also I don't think the epsilon-delta definition of continuity would help either.
-So I am leaning towards the sequence definition of continuity and this is what I have:
$f$ is continuous on [a,b], so there exists a sequence $(x_n)$ in [a,b] where $x_n$ converges to $x_0$ such that $f(x_n)$ converges to $f(x_0)$. WTS: $f^*(x)$ is an non-decreasing continuous function.
but I don't know how to proceed from here. 
Any help will be appreciated
Thank you for your help!
 A: First we can see that as $x$ increases the interval $[a, x]$ also increases in size and the only possibility is that the $\sup f(x)$ on $[a, x]$ increases or remains constant. So we can see that $f^{*}(x)$ is non-decreasing.
For continuity part we need to take a point $c \in (a, b)$ and analyze the behavior of $f^{*}(x)$ near $x = c$. Let $f^{*}(c) = M$ so that $M = \sup_{a \leq x \leq c} f(x)$. If $f(c) < M$ then for all sufficiently small $h$ we have $f(x) < M$ for all $x \in (c - h, c + h)$ (this happens because of continuity of $f$) and hence $f^{*}(x) = M$ for all $x \in (c - h, c + h)$. If on the other hand $f(c) = M$ then for any given $\epsilon > 0$ we can find an $h > 0$ such that $M - \epsilon < f(x) < M + \epsilon$ for all $x \in (c - h, c + h)$ and this means that $M - \epsilon \leq f^{*}(x) \leq M + \epsilon$ for all $x \in (c - h, c + h)$. Noting that $f^{*}(c) = M$ it follows that $$|f^{*}(x) - f^{*}(c)| \leq \epsilon$$ for all $x \in (c - h, c + h)$ in all cases. We thus conclude that $f^{*}$ is continuous at $c$. The argument can be modified easily when $c = a$ (here we deal with intervals like $[c, c + h)$) or $c = b$ (here we deal with intervals like $(c - h, c]$).
A: This also follows from the triangle inequality and the oscillation characterization of continuity, giving a cleaner answer. 
Namely, it is quite easy to prove that a function is continuous iff the oscillation of $f(x)$, tends to $0$. 
$$o(f) = \lim_{\delta \to 0}\sup_{x \in (x+\delta,x-\delta) } f(x) - \inf_{x \in (x+\delta,x-\delta) } f(x) = 0$$
Note that $$|o(f)| = \sup_{t,s \in (x-\delta,x+\delta)} |f(t)-f(s)|,$$
and the proof follows from manipulating the triangle inequality, since it can be shown that
$$|f^*(x) - f^*(y)| \leq |o(f)|$$
