Let $C$ and $\alpha$ be two fixed positive real numbers. Define $K$ to be the set of all real-valued functions on $[0,1]$ satisfying $|f(x)-f(y)|\leq C|x-y|^{\alpha}$.

Is $K$ equicontinuous? Closed? Bounded? Compact?

I am confused about the notion of a set of functions. And I can't seem to find much useful information about the subject online. How can I try to picture the set $K$?

Despite my confusion this is what I've started so far: With what we are given we can find a $\delta$ for each function to show that it is continuous. We can then take the minimum of these deltas to show equicontinuity. I couldn't come up with some sound proof to show that it is also closed and bounded; but I picture it such that if it is not bounded then some function must not satisfy the hypothesis. That for some function as we approach $0$ or $1$ we would have to be going off to infinity. And for closed we would show that taking a sequence $\lbrace x_0\rbrace $ and looking at $f(\lbrace x_0 \rbrace)$ its limit point would have to be in the set $K$. These are just my thoughts so far, but I am still confused.

  • $\begingroup$ When you ask about 'closed', 'bounded' and 'compact' you need to specify in which space you want to verify these properties. I do assume it is $C^0([0,1])$. $\endgroup$ – Thomas Mar 25 '14 at 18:31
  • $\begingroup$ I believe your assumption is correct, but I was just unsure of where $K$ would reside. I see $K$ is in the set of real-valued functions on $[0,1]$ but that is all. $\endgroup$ – Ryan Smith Mar 25 '14 at 18:42

This is of course an equicontinuous set: take $\delta$ such as $$ C\delta^\alpha= \epsilon$$

Let us check the Ascoli theorem hypothesis: the set is not uniformly bounded. Hence it is not compact.


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