# Supremum of a Continuous Function is Continuous

I'm working on this problem from Elementary Analysis by Ross which is intuitive when sketched but keeps stymieing me when I try to write it out.

Let $f$ be a continuous function on $[a,b] \subset \mathbb{R}$. Define $f^\star (x)$ as:

$$f^\star(x) = \sup \{f(y)\mid a \leq y \leq x \}$$ Prove that $f^\star$ is a continuous increasing function on $[a,b]$.

Things I've figured out: (these are mostly trivial)

• $f$ is uniformly continuous on $[a,b]$.
• Since $[a,b]$ is closed and bounded there exists some $c \in [a,b]$ such that $f(c) \geq f(x)\ \forall x \in [a,b]$. (In words: the supremum is actually attained.)
• If $c \in [a,b]$ is as above, then $f^\star$ is constant (and hence continuous) on $[c,b]$ (which is possibly a singleton).

This seems completely obvious when you actually draw a continuous function but translating that to a formal proof eludes me...

Continuity is easy at points where $$f(x_0), because we will still have $$f(x) for $$|x-x_0|<\delta$$ for some $$\delta$$, meaning $$f^\star$$ will be constant on a neighborhood of $$x_0$$. Then trivially, $$f$$ is continuous at these points.

So suppose $$f^\star(x_0)=f(x_0)$$. Given $$\epsilon$$, choose $$\delta$$ so $$|x-x_0|<\delta\implies|f(x)-f(x_0)|<\epsilon$$. Then for $$x_0-\delta, $$f^\star(x_0)\ge f^\star(x)\ge f(x)>f(x_0)-\epsilon=f^\star(x_0)-\epsilon$$ and for $$x_0, $$f^\star(x_0)\le f^\star(x)=\sup_{[a,x]}f(y)\stackrel{1}{=}\sup_{[x_0,x]} f(y)\stackrel{2}\le f(x_0)+\epsilon=f^\star(x_0)+\epsilon$$ 1 follows since $$f(x_0)$$ is the maximum of $$f(y)$$ for $$y\in[a,x_0]$$ (recall we assumed $$f(x_0)=f^\star(x_0))$$, so the region $$[a,x_0]$$ is redundant. 2 follows since $$|y-x_0|<\delta$$ when $$y\in[x_0,x]$$.

The two displayed inequalities imply $$|f^\star(x)-f^\star(x_0)|<\epsilon$$ for $$|x-x_0|<\delta$$, so $$f^\star$$ is continuous.

f* is non-decreasing because f*(x) <= f*(y) if x < y due to the sup being taken over a larger set.

f* is continuous because consider a point x in [a,b] and assume f obtains its maximum value of y in [a,x]

Assume y < x, then |f*(x) - f(y)| = 0 < eps if x is within neighborhood |x-y| of x. If y = x then as f is continuous by definition we can find a neighborhood d of x such that |f*(x) - f(y)| < eps for any eps > 0.

For all $\epsilon>0, x\in[a,b]$, since $f$ is continuous at $x$, $\exists \delta>0$, s.t. $|x-y|<\delta$, $|f(x)-f(y)|<\frac{1}{2}\epsilon$

$\forall y, |x-y|<\delta$, WLOG assume $y>x$.

Let $x_0\in[a,x],y_0\in[a,y]$ obtains the supremum due to uniform continuity. If $y_0\in [a,x]$, we are done since in this case $f^\star(y)=f^\star(x)$. Hence WLOG assume $y_0\in(x,y]$ then $f(y_0)>f(x_0)\geq f(x)$

$f^\star(y)-f^\star(x)=f(y_0)-f(x_0)<f(y_0)-f(x)\leq\frac{1}{2}\epsilon<\epsilon$. So $f^\star$ is continuous.

Increasing is easy just because we take supremum over a larger set.

• It doesn't affect the conclusion, but you cannot conclude the strict inequality $f(y_0)>f(x_0)$, even when $y_0\in(x,y]$. Jun 3, 2018 at 0:25

Today I faced exactly the same problem and that is my plan of proof :

1) $$f(x) \in C(x_0) \Leftrightarrow \omega_f(x_0) = 0$$, where $$\omega_f(x_0)$$ is oscillation of $$f$$ at $$x_0$$

2) $$\omega_f(x_0, x_0+\delta) \geq \omega_{f^\star}(x_0, x_0+\delta)$$

First statement is actually an equivalent definiton of continuous function, second can be proved by contradiction and implies continuity of $$f^\star$$.