The following question is causing me some problems.
Consider the metric space $(X, d)$, where $X$ = $C([0, 1], \mathbb R)$ is the set of continuous functions
$f : [0, 1] \to \mathbb R$, and $d$ = $d_{\infty}$ is the distance which is given by
$$d(f,g)=\Vert f−g\Vert_{\infty}=\sup\{|f(t)−g(t)|:t\in[0,1]\}.$$
(a) $L=\{f\in X \mid \forall x,y\in [0,1] ,f(0) = 0, \text{ and } ∀x, y ∈ [0, 1] : |f(y) − f(x)| \leq |y − x|^{1/4}\}$
i. Is $L$ closed?
ii. Is $L$ compact? Prove your answer
Any help would be greatly appreciated, also apologies for poor notation. I'm still new to the method of composing a question using this site. Thanks in advance
$L$ is closed. First, notice that $L = L_0 \cap \bigcap\limits_{x, y \in [0, 1]} L_{x, y}$, where $L_0 = \{ f \in X : f(0) = 0 \}$ and $L_{x, y} = \{ f \in X : |f(y) - f(x)| \leq |y - x|^{\frac{1}{4}} \}$. So it is enough to show that each of these sets is closed.
For $L_0$, pick $f \notin L_0$, so $f(0) = x \neq 0$. If $d(f, g) < |x|$, then $|g(0)| \geq |f(0)| - d(f, g) > 0$, so $g \notin L_0$.
For $L_{x, y}$, pick $f \notin L_{x, y}$, so $|f(y) - f(x)| - |y - x|^{\frac{1}{4}} =: \epsilon > 0$. If $d(f, g) < \epsilon / 2$, then $|g(y) - g(x)| - |y - x|^{\frac{1}{4}} \geq |f(y) - f(x)| - 2 d(f, g) - |y - x|^{\frac{1}{4}} > 0$, so $g \notin L_{x, y}$.
We conclude that $L$ is closed.
$L$ is compact. Recall the Arzela-Ascoli theorem: a subspace $Y \subset X$ is compact if and only if it is closed, uniformly bounded and equicontinuous. We already know that $L$ is closed. It is uniformly bounded, since for any $x \in [0, 1]$ we have $|f(x)| = |f(x) - f(0)| \leq |x - 0|^{\frac{1}{4}} \leq 1$. We are left to show equicontinuity.
Let $\epsilon > 0$, and $\delta := \epsilon^4$. Now let $x, y \in [0, 1]$ be such that $|y - x| < \delta$. Then $|f(y) - f(x)| \leq |y - x|^{\frac{1}{4}} < \delta^{\frac{1}{4}} = \epsilon$ for any $f \in X$.
We conclude that $L$ is compact.