Question: Let $\sigma: [0,1] \to \mathbb{R}^{\ge 0}$ be a nonnegative, continuous function such that $\sigma(0) =0$. For each real $\lambda \ge 0$ define the following set: $$ F_ \lambda = \left\{ f \in C[0,1] : |f(0)| \le \lambda |f(1)| \text{ and } |f(x)-f(y)| \le \sigma\left(\left|x-y\right|\right) \text{ for all } x,y \in [0,1]\right\} $$ Characterize the $\lambda$ for which $F_\lambda$ is a compact subset of $C[0,1]$.
Solution: (Let $\epsilon>0)$. I'm trying to use Arzela Ascoli. First, equicontinuity. Since $\sigma$ is continuous and $\sigma(0)=0$, there is some $\delta$ so that $\sigma(x) < \epsilon$ whenever $0<x<\delta$. Then, for $|x-y|<\delta$ $$ |f(x)-f(y)| \le \sigma(|x-y|) <\epsilon $$ so $F_\lambda$ is equicontinuous for all $\lambda$.
Closedness is straightforward . . . Let $\|f_n - f\|_\infty \to 0$ where $(f_n) \subset F_\lambda$. The limit $f$ is continuous, and continuity of $f$ ensures that the other two criterion of $F_\lambda$ are satisfied. (Take limits.)
The difficulty is with (equi)boundedness. Since $\sigma$ is continuous on a closed interval $[0,1]$, we have $\| \sigma\|_\infty \le M < \infty$ by the Extreme Value Theorem.
If $\lambda=0$, then $|f(x)| \le |f(x)-f(0)| + |f(0)| \le \sigma(x) + 0 \le M$, so $F_0$ is uniformly bounded.
If $1 \le \lambda<\infty$, then the constant functions $f_n(x) = n$ are in $F_\lambda$, but they have no uniform bound.
My main troubles are with $0 < \lambda <1$ . . . I have been trying to construct specific sequences that are unbounded and satisfy $|f_n(0)| \le \lambda |f_n(1)|$, but I am never sure if $|f_n(x)-f_n(y)| \le \sigma(|x-y|)$ without knowing more about $\sigma$. (Maybe that's the point)
Thank you.
Suppose $\lambda \in [0,1)$, and fix $f \in F_{\lambda}$.
Then $|f(1)| \leq |f(1)-f(0)|+|f(0)| \leq \sigma(1)+\lambda |f(1)|$.
Consequently, $(1-\lambda)|f(1)| \leq \sigma(1)$, so that $|f(1)| \leq \frac{\sigma(1)}{1-\lambda}$.
Thus for all $x \in [0,1]$ we see that $$|f(x)| \leq |f(x)-f(0)| +|f(0)| \leq \sigma(x)+\lambda|f(1)| \leq \|\sigma\|_{\infty} + \frac{\lambda \sigma(1)}{1-\lambda}$$
This proves equi-boundedness when $\lambda \in [0,1)$.
If $\lambda \geq 1$, your counterexample works.