Prove that any bounded sequence in $C^{1/2}([0,1])$ admits a convergent subsequence in $C^{1/3}([0,1]),$ where we say that $f \in C^{\alpha}([0,1])$ if $f$ is Holder continuous of order $\alpha.$ The norm defined on $C^{\alpha}([0,1])$ is $$ ||f||_{C^\alpha} = \sup{\{|f(x)| : x \in [0,1]\}} + \sup{\{\frac{|f(x) - f(y)|}{|x-y|^\alpha} : x,y \in [0,1] \text{ and } x \neq y \}}. $$
I'm pretty new to these types of problems, I've tried to work things out and write down what I think everything means:
Letting $\{f_n\}_{n=1}^\infty$ be a sequence in $C^{1/2}([0,1]),$ the sequence is bounded if for all $n$, $||f_n||_{C^{1/2}} < M$ for some constant $M < \infty.$
Letting $g_n = f_{\sigma(n)}$ (the function $\sigma(n)$ just picks out the subsequence) be the elements of the subsequence of $\{f_n\}_{n=1}^\infty,$ this subsequence $\{g_n\}_{n=1}^\infty$ is convergent in $C^{1/3}$ if for $\epsilon > 0, \exists N \text{ such that for } n,m>N, ||g_n - g_m||_{C^{1/3}} < \epsilon.$ In other words, its required that $$ \sup{\{|g_n(x) - g_m(x)| : x \in [0,1]\}} + \sup{\{\frac{|g_n(x) - g_m(x) - g_n(y) + g_m(y)|}{|x-y|^\alpha}}\} < \epsilon $$ for all $n,m > N.$
Im having a hard time seeing how boundedness in one space permits a convergent subsequence in another space. Any hints/solutions or explanations of why this makes sense to begin with would be greatly appreciated.
