# 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

• It should say non-decreasing function. The function f(x) = 0 is continuous on [a, b] but $f^{*}(x) = 0$ is clearly not an increasing function.
– Paul
May 20 '14 at 9:03
• I agree with you but the book uses both increasing and non-decreasing it keeps alternating. Same with decreasing and non-increasing. But yes good point May 20 '14 at 9:36
• The none decreasing bit should be easy. If w<z simply write down the definition of $f^{*}(w)$ and go from there. For the second part note that a continuous (real valued?) function on a closed interval is bounded and attains its bound i.e. there is a $y_0$ in the closed interval [a, x] for which $f^{*}(x) = f(y_0)$.
– Paul
May 20 '14 at 11:06
• Jun 28 '16 at 14:20

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 $(c - h, c]$).

• Why the downvote? Mar 20 '16 at 5:16
• This is also very nice. Considering cases like $f(c)\lt M$ handles the proof phenomenally well.
– Koro
May 19 '21 at 4:26

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

$$|f^*(x) - f^*(y)| \leq |o(f)|$$