Suppose that $\Lambda \subset C^{0}$ is equicontinuous and bounded.
(a) Prove that $\sup\{f(x)\mid f \in \Lambda\}$ is a continuous function.
(b) Show that (a) fails without equicontinuity.
(c) Show that this continuous-sup property does not imply equicontinuity.
(d) Assume that the continuous-sup property is true for each subset $\mathcal{F} \subset \Lambda$. Is $\Lambda$ equicontinuous? Give a proof or counterexample.
My attempt.
(a) We have that
$$\sup f(x) = \sup (f(x)-f(y) + f(y)) \leq \sup f(y) + \sup(f(x)-f(y)),$$
then
$$\sup f(x) – \sup f(y) \leq \sup (f(x)-f(y)) \leq \Vert f(x) – f(y) \Vert$$
and
$$\sup f(y) – \sup f(x) \leq \sup (f(y)-f(x)) \leq \Vert f(x) – f(y) \Vert,$$
that is,
$$|\sup f(x) – \sup f(y)| \leq \Vert f(x) – f(y) \Vert.$$
Since $\Lambda$ is equicontinuous, given $\epsilon > 0$ there is $\delta > 0$ such that $|x – y| < \delta$ implies $|f(x) – f(y)| < \epsilon$ for each $f \in \Lambda$. Therefore,
$$|x – y| < \delta \Longrightarrow |\sup f(x) – \sup f(y)| < \epsilon$$
(b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) – f(y)| < \epsilon$. But I cannot find a counterexample.
(c) Consider $C^{0}([0,1],\mathbb{R})$ and $\Lambda = \{f_{n}(x) = x^{n}\}_{n}$. Note that
$$\sup_{x \in [0,1]} f(x) = 1\quad \forall f \in \Lambda.$$
Thus, $g = \sup f$ is continuous, but $\Lambda$ is not equicontinuous.
(d) By Mindlack's observation, the same example of (c) works in (d).
Can someone help me?
Best Answer
$b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.
$d$: have you tried re-using $c$?