Show that $K := \{x \in C[0,1] : x(0) \in [-3,4], |x(t)-x(s)| \leq d |t^2-s^2|, \forall t,s \in C[0,1]\}$ is compact.
I already know that $C[0,1]$ is a compact set. So it is only necessairy to show that $K$ is closed.
Is anyone is able to prove it or give me a good hint?
On
Suppose $\;\{x_n\}_{n=1}^\infty\subset K\;$ is such that $\;x_n\xrightarrow[n\to\infty]{}x\;$ . We must show that $\;x\in K\;$.
By continuity:
$$x(0)=\lim_{n\to\infty}x_n(0)\in[-3,4]\;,\;\;\text{as this last is a closed interval}$$
Also, again by continuity, for all $\;t,s\in[0,1]\;$ , we have
$$|x(t)-x(s)|=|\lim_{n\to\infty}\left(x_n(t)-x_n(s)\right)|=\lim_{n\to\infty}|x_n(t)-x_n(s)|\le\lim_{n\to\infty}d|t^2-s^2|=d|t^2-s^2|$$
Hint:
Say that you have a sequence $x_n\to x$ such that $x_n(0) \in [-3,4]$ and $|x_n(t)-x_n(s)| \leq d |t^2-s^2|$ for all $t$ and $s$ (you mean in $[0,1]$), for each $n$.
What happens in these two properties if $n\to\infty$?